1) THE FORMATION OF ESCAPE AND ORBITAL VELOCITIES
2) GRAVITY IN TERMS OF SATELLITES
3) ESCAPE VELOCITY AND ORBITAL VELOCITY
4) THE ENTRANCE VELOCITY
5) ORBITS AND SUPPORT FOR QUARK THEORY
6) ORBITAL ENERGIES AND PLANETARY DYNAMICS
7) EXPLAINING THE ORBITS OF COMETS
8) THE SOLSTICE GAP
9) MEASURING THE DISTANCE TO THE MOON10) THE NATURE OF ELLIPTICAL ORBITS
11) THE RINGS OF SATURN
1) THE FORMATION OF ESCAPE AND ORBITAL VELOCITIES
We know that the kinetic energy of the gravitational mass which comes together to create a star is a redirection of the energy of the Big Bang, which threw matter outward to begin the universe as we know it. Just as the kinetic energy in a ball that comes back down is a redirection of the energy used to throw the ball upward.
If matter can be brought together by gravity, the energy of the Big Bang which originally thrust it apart can be transformed into kinetic energy which crunches smaller atoms together into larger ones so that the leftover energy from the crunching process is released, and this energy is what causes the star to shine. Energy from the Big Bang, with regard to things on earth, is indirect since it has been through at least one star formation and then supernova, as the star exploded and scattered it's atoms across space which later coalesced by gravity to form the earth and Solar System.
What I want to show here is that matter which was thrown outward by the Big Bang cannot coalesce without the reversal of energy from the Big Bang, and this reversed energy is manifested as gravity. This means that the earth's gravity can be expressed entirely as the energy that went into forming the earth, including it's atoms having been crunched together from the primordial hydrogen and helium, the lightest elements, following the Big Bang.
When a ball is thrown into the air, if more energy is put into throwing the ball relative to the energy which went into creating the atoms of the ball when they were within a previous star that exploded in a supernova, which is the mass of the ball relative to the energy that went into the earth, the mass of the earth, the ball will reach what is known as escape velocity and will leave the earth entirely. This is why a planet has a certain escape velocity.
Mass and density, which are the roots of gravity, are a function of the kinetic energy of the mass being pulled together and the atoms being crunched together, which is simply a reversal in the direction of the energy of the Big Bang. Our throw of the ball brings into play the proportions of the energy it took to create the earth and create it's atoms, which is equal to the energy which was overcome relative to the energy put into the ball, and the mass of the earth relative to the mass of the ball.
The energy which crunched the atoms together, from previous smaller atoms, must be counted because higher density means higher surface gravity because a more dense planet will have a lower starting point. In another way of looking at it, density brings higher gravity because the planet's total gravity is spread over less of a surface area.
Put simply, we can actually measure the total energy that was redirected from that of the Big Bang to create the earth, both the kinetic energy of the earth's mass being brought together by gravity and the energy that went into crunching the atoms of the earth together from the smaller primordial atoms that formed after the Big Bang. If we have a rocket with a mass 1/x that of the earth, and it takes a given minimum amount of energy to get the rocket beyond the earth's gravity, the total energy that was redirected from the Big Bang to form the earth is that given amount of energy, multiplied by x. Of course, in practical terms, we would have to subtract the energy from the rocket that it took to move the fuel itself.
In a supernova, a star exploding from it's core, the entire mass of the star reaches escape velocity. This is only possible because the heavy atoms, which were crunched together within the star, remain intact. Such a star explosion in which heavier atoms within the star came apart into the smaller atoms that were pulled together by gravity to form the star in the first place is impossible because that would require the supernova blast to have the same amount of energy to put into the explosion that was redirected from the Big Bang to form the star. But since energy had been radiating from the star, as heat and light, during it's life cycle, that is not possible and the star can only explode in a supernova because it's heavy atoms remain intact so that the explosion uses less energy than went into forming the star to begin with.
Notice that a supernova typically does not blast matter so far apart that much of it cannot come back together again by gravity. This is why there are second-generation stars with planets, such as our Solar System. There was not enough energy in the Big Bang to throw matter so far apart that it could not, to some extent, re-coalesce by gravity. Since the energy that formed the star which exploded in a supernova is a redirection of that of the Big Bang, we should expect the same pattern in that matter thrown outward by a supernova would not go so far that none of it could re-coalesce by gravity. In this way a supernova can be seen as a kind of miniature of the Big Bang, following the same energy pattern because it was a redirecting of the energy of the Big Bang which formed the star.
This scenario is fundamental physics, and not cosmology. But just to touch on my cosmological theory, in which the matter of which we are composed was thrown over four dimensions of space by the Big Bang, but one of those dimensions is what we perceive as time. The cosmological theory can be reviewed in the posting "In Cosmology Everything Just Fell Right Into Place", on this blog.
The outward momentum from the Big Bang of the matter of our universe involves four dimensions, while we can only access three of those dimensions. This means that we cannot exert energy in the same direction as that from the Big Bang. This is why, when we move something such as throwing a ball into the air, we are actually opposing the energy of the Big Bang.
We cannot work in harmony with this energy due to our dimensional limitations. If there were to be a physics force vector in three dimensions, we could never exert a force in harmony with it if we could exert our force in only two dimensions. Thus, when we throw the ball into the air, we are working in opposition to the energy that went into putting the earth together and crunched the heavy atoms of the earth together from lighter atoms by kinetic energy, and which can ultimately be traced back to the Big Bang. This is why the ball proceeds some distance into the air, before falling back down.
Doesn't gravity seem more clear when thinking of it as redirected energy? I have never before seen gravity explained in this way.
2) GRAVITY IN TERMS OF SATELLITES
I have always thought that there is quite a bit of room for improvement in explaining gravity, and that is what I am doing here. I wrote a posting last week about the possibility of satellites in the early evening being mistaken for UFOs, and I decided during the week that it would be good to explain gravity in terms of satellites.
We know that a higher altitude represents a higher energy level, simply because it requires more energy to get an object to a greater altitude and this energy is returned to the earth on impact if the object should fall from that altitude. An object falling from a greater height will have more energy and will make a greater impact.
It is the energy of a given altitude that causes an object placed there to fall. The universe always seeks the lowest energy state, and the space immediately below the object at a given altitude is of a lower energy because the higher energy altitude direction is opposite the direction from which gravity is pulling.
A ball thrown into the air represents a simple exchange between two types of energy. Remember that energy is never created or destroyed, but only changed in form. When the ball is thrown upward, the up and then down movements of the ball are reverse mirror images of one another with equal velocity at equal altitudes. The only difference is the direction. When the ball impacts the ground, we are getting the energy back that was put into the ball in the first place.
This return of energy upon the ball's impact with the ground shows that there is energy in altitude. The thrown ball was given the energy which matched that of a certain altitude. When the ball reached that altitude, it became motionless for a moment until it began the mirror image downward journey. When the thrown ball is below the altitude that matches it's energy, some of the energy in the ball manifests as the kinetic energy of motion.
The lower the ball is in it's journey, whether traveling upward or downward, the lower the energy of that altitude, and the more the difference between the energy of the lower altitude and the energy of the altitude that the ball was given. This difference in energy is manifested as the kinetic energy of motion, which is why the ball travels faster just after it has been thrown and then just before it reaches the ground. The ball ceases motion for a moment at the maximum altitude because there is no difference between the energy of that altitude and the energy that the ball was given, because if there was a difference it would be manifested as the kinetic energy of motion.
An object falls because the space immediately below it is at a lower energy state. The difference in energy between the two levels is manifested as velocity of the falling object. But if an object is high enough so that the proportion of distance to the center of the earth means that the space immediately below does not represent a significantly lower energy state, the object will not fall downward. But is must accommodate the energy of the altitude that it is at because of the pull of the earth's gravity on it. The object will move laterally in orbit, because the space immediately above and below does not have a significant energy difference.
If an object in orbit is given twice as much energy as before, it will not orbit any faster. All objects in orbit at the same altitude will move at the same speed. This speed will be faster in a lower orbit, in accord with the earth's stronger gravity at a lower altitude, even though the object will have less energy because it is at a lower level.
The energy of a given orbital altitude is proportional to the area of space that a line from the satellite to the center of mass of the planet traverses in a given time period. This is related to Kepler's Law that such orbits tend to be ellipses, rather than circles, but the space that this line traverses per unit of time always remains constant. Remember that the line must be from the mass center of the earth and not from the surface.
This law of Kepler's means that the orbital energy of an object in orbit must always remain constant throughout the orbit. Although it has a perihelion (low Point) and an aphelion (high point), there cannot be a high energy and a low energy side because because it was a given amount of energy that put the satellite in orbit and the alternation between perihelion and aphelion is simply an exchange between altitude and velocity with the product of the two always remaining constant. The energy in an orbit is equal to velocity x altitude, which is also expressed as Kepler's line always traversing equal areas of space in equal periods of time. This is similar to the alternation between kinetic and potential energy when the ball is thrown into the air.
Kepler's Law about the line between the orbiting object and the center of the planet always sweeping across equal areas of space in equal periods of time only applies to a single object in a particular orbit. It does not apply to two objects in different orbits at different altitudes. This is because the object in higher orbit will have a higher energy, and the line from it to the center of the planet will sweep over more space in a certain period of time, in proportion to the energy of that orbit. This orbital energy in proportion to the line sweeping across areas of space illustrates how space between matter is equivalent to the energy dimension of my sheet model of the Big Bang, as described in the posting "Energy And Space" on this blog.
Close to the earth's surface, where weakening gravity due to increasing altitude is not a factor, twice the altitude means twice the energy because it would have taken twice the energy to get it there. An object falling from twice the altitude will impact the level ground at the square root of two (1.414) times the velocity, and with twice the force.
(Note- This means that those lectures about speeding are correct, a car moving at twice the speed will have actually four times the force).
If the earth's gravity was constant, as we move outward into space, similar rules would apply as do with falling objects near the earth's surface. A satellite which was initially given twice the energy of another would orbit at the same speed, but at the square root of two (1.414) times the distance. The reason that this is not what happens is that the rule represented by Kepler's Law of a line from the center of the satellite to the center of the earth sweeping over equal areas of space during equal periods of time has to be combined with the fact that earth's gravity gets weaker as we move outward, according to Newton's Inverse Square Law.
The final result is that if a satellite is given twice the energy, it will orbit at four times the distance, but with only one-half the speed, so that the line from the center of the satellite to the center of the earth will sweep over twice the area of space in equal periods of time, representing twice the energy. A satellite in higher orbit must necessarily move slower, even though it is in a higher energy orbit, in accordance with the earth's gravity getting weaker with increased altitude by the Inverse Square Law.
The energy of a given altitude is the energy that would be released, upon impact, if that object fell to the ground. Energy of altitude must match gravity because it is gravity that must be opposed to get the object to that altitude. If gravity decreases as per the Inverse Square Law as we move further from earth, the energy of an orbit in terms of altitude must follow the same pattern.
If a satellite is given twice the energy, it must orbit at four times the distance from the center of the earth, because as the distance from the earth increases the speed at which an object will orbit must also decrease. A satellite with twice the energy will orbit at four times the distance, but with only half the velocity, so that the line from the center of the satellite to the center of the earth will traverse twice the area of space in a unit of time.
(Note- Have you ever wondered what the difference between speed and velocity is? If the motion can be in any direction, it it called speed. If the motion is in a certain definite direction, it is called velocity).
At twice the distance, there will be only one-fourth the gravity because the earth's gravity will be spread over four times the area of space, in accordance with the Inverse Square Law. Remember that this does not apply to objects falling near the earth's surface because there is no significant difference in the proportion of distance to the earth's center.
Put simply, an object in orbit with three times the energy must orbit at nine times the distance, due to the Inverse Square Law, because at nine times the distance it will orbit at only one-third of the speed and this is the relationship that will bring about a line from the satellite to the center of the earth traversing three times the area of space in the same period of time. The area of space that this line sweeps across in a given period of time is the very definition of the energy of the orbit, regardless of the combination of velocity and distance which brings it about.
The energy of orbit obviously increases as we move outward from the earth, but it cannot be linear because the earth's gravity is getting weaker as per the Inverse Square Law. An object in orbit must move faster when the orbit is closer to the earth because the earth's gravity is stronger there, even though that is a lower energy orbit because it requires less energy to get an object into that orbit.
As an example I would like to use two of the most important orbits around the earth, that of the moon and the geostationary orbit. This geostationary orbit is so-called because it is so useful for communications satellites. A satellite in this orbit, which is at an altitude from the earth's surface of about 358,800 km (22,300 miles), will orbit at the same rate as the earth rotates so that it stays at the same point overhead in the sky. It's revolution period will be exactly one day.
The moon, in orbit at about 384,500 km (239,000 miles) from the earth, is more than nine times the distance from the earth's center as the geostationary orbit. If we compare these two orbits using the reasoning of the line traversing equal areas of space in equal periods of time, without regard to the higher energy of higher orbits, we find that the moon should orbit the earth in about 86 days because the line from the moon to the center of the earth will sweep over that much times the area of space as the line from the satellite in geostationary orbit to the center of the earth.
But the fact is that the moon orbits the earth in only about 29 days, which is about one third of 86. The moon is over nine times as distant from the center of the earth as the geostationary orbit. This means that it will move at only one-ninth the speed of a satellite in the geostationary orbit. Then the distance and the speed together means that the line from the moon to the center of the earth sweeps over three times the area of a line from a geostationary satellite.
This is because the orbit of the moon is at three times the energy level of the geostationary orbit. It cannot be at nine times the energy, even though it is at nine times the distance, because the earth's gravity decreases as we move outward in accord with the Inverse Square Law. The energy of orbits must operate according to the square, nine times the distance equals three times the energy, because earth's gravity decreases by the Inverse Square Law.
By the way, the tidal bulge in the earth's oceans which is brought about by the moon's gravity as the earth rotates with the moon overhead, actually takes energy from the earth's rotation and transfers it to the moon. This whips the moon into a higher energy orbit so that it moves away from the earth at a rate of about one centimeter a year. The moon was once much closer to the earth. At the same time, this gradually slows the earth's rotation.
3) ESCAPE VELOCITY AND ORBITAL VELOCITY
Here is the conclusion that I have come to with regard to orbital and escape velocity that I have never seen before.
As we saw in 1) THE FORMATION OF ESCAPE AND ORBITAL VELOCITIES, the escape velocity of the earth is determined by how much energy has gone into the atoms of the earth since the matter of the universe formed of mostly hydrogen following the Big Bang, which began the universe as we know it. It takes energy, in the centers of stars to crunch smaller atoms together into larger ones. This energy comes from the kinetic energy of matter thrown outward by the Big Bang, which is then reversed by gravity and the matter is brought together to form a star. Some of this energy goes into overcoming the electron repulsion between atoms so that smaller atoms are crunched together into heavier elements.
If we can take an object of a certain mass, such as an artillery shell, and apply energy into propelling it directly upward, it will leave the earth's gravity altogether if we can propel it to a velocity of seven miles per second (11.26 km/sec). This is known as the earth's escape velocity. If we can propel the shell to a velocity of five miles per second (about 8 km/sec), it will not escape earth's gravity but will go into orbit. This is known as orbital velocity.
(Note- when neighboring Canada converted to the Metric System when I was age 15, I learned the system also. But all of the facts that I learned before this, including the escape and orbital velocities of the earth, I remember in the units of the old system).
Escape and orbital velocities only apply to ballistic objects, and not to rockets. A rocket is moving under it's own power so all that is necessary is to keep the engine running to escape the earth's gravity, regardless of velocity. It is true that escape and orbital velocity are somewhat theoretical because propelling a ballistic projectile to such velocities has never actually been done. There was a Canadian project in the 1960s, the High Altitude Research Project, that fired research projectiles into the upper atmosphere but with not quite the velocity necessary to reach orbit.
My concept is that the energy applied to form the mass that is now the earth from crunching small atoms, mostly hydrogen from after the Big Bang, together into the heavier elements that form the earth today did not change the actual mass of the earth. The same volume of matter has the same mass regardless of whether it consists of more smaller atoms or the fewer larger atoms that they are crunched into. The energy that it took to crunch those smaller atoms together will show up as an increased escape velocity because the greater density of the resulting larger atoms will give a projectile a lower starting point and thus it will require more energy to overcome the concentrated gravity of the earth. This also includes the energy within molecular bonds in the molecules of the earth.
The escape velocity represents the the kinetic energy of matter thrown outward by the Big Bang that was reversed by gravity to bring the mass that is now the earth together within a star, that has since exploded as a supernova to scatter the matter that would be the earth across space where it would coalesce by gravity into the earth. What is now the earth's escape velocity represents the kinetic energy that has gone into the atoms of earth, within the previous star that exploded as a supernova.
If we fire a projectile upward, and if we can give it kinetic energy per mass that is greater, per the mass of the projectile in comparison with the mass of the earth, than the total energy that went into forming the earth from the primordial hydrogen after the Big Bang, the projectile will reach escape velocity and will leave the earth's gravity altogether. This is because it would be necessary to overcome the energy that has gone into forming the earth.
Suppose that we could undo all of the heavier atoms and molecules of the earth so that there was only the primordial mostly hydrogen that formed after the Big Bang. The earth would still be a sphere, but would be less dense and thus much larger than it is today. Remember the posting "Electron Repulsion And Density", on the physics and astronomy blog that explains why an equivalent mass in smaller atoms is less dense than one of larger atoms.
If the earth's heavy atoms were broken back down into the primordial hydrogen atoms from which they were formed by stellar nucleosynthesis, and those hydrogen atoms were together as a sphere in space, the escape velocity from the sphere would be close to zero. A cloud in space of the lightest element cannot really have an escape velocity.
But as energy goes into crunching those hydrogen atoms together into heavier elements, we eventually arrive at the earth that we have today. Our earth does have a definite escape velocity. It can be seen, then, that the earth's escape velocity is actually an escape from the kinetic energy that has gone into crunching lighter atoms into the earth's heavier atoms. If we can exceed this energy, in comparison of the mass of the projectile relative to the mass of the earth, the projectile escapes the earth's gravity. If we give it less than this energy, the projectile cannot escape the earth's gravity but can go into orbit, which is essentially what was once the outer limit of the proto-earth cloud of primordial hydrogen.
(Of course, the heavier atoms of the earth were actually crunched together from lighter atoms within a star, and not by simple gravitational contraction as I am portraying. But that does not matter for our scenario here).
When something reaches orbital velocity it is, in effect, moving along the surface of that imaginary sphere that the earth would be without the energy that had been put into it to overcome electron repulsion between atoms, and crunch smaller atoms together into larger ones. If we apply energy to bring an object to orbital velocity, we are overcoming the energy that has gone into crunching together the heavier atoms of the earth, in the proportion of the mass of the object to the mass of the earth, so that the object can go to what would be the surface of that sphere and can move laterally around it in what we perceive as an orbit.
NEW FORMULA FOR ESCAPE AND ORBITAL VELOCITIES
There is the standard formula for escape velocity that is given in the article about it on Wikipedia. But I have noticed another formula for escape and orbital velocities that I think reveals more about the real nature of these velocities.
The standard acceleration due to gravity near the earth's surface is 32 feet (9.8 meters) per second squared. This simply means that at the end of the first second of free-fall, the velocity of a compact object will be 32 feet per second. At the end of the second second the velocity will be 64 feet per second, and so on. If the object begins falling with a velocity of zero, and ends the first second with a velocity of 32 feet per second, that means that it fell a total of half of 32 feet, or 16 feet, during the first second.
When an object is launched upward, and falls back down, the velocity that it impacts the ground with will be the same as that it was launched with. The velocities on the way up will be identical to those on the way down, except in the opposite direction, at the same altitudes. There is an energy level at each altitude, equivalent to the energy it takes to get it there and the energy of impact on the ground of an object falling from that height.
The escape velocity of 7 miles per second must actually correspond to the energy of a certain altitude. I was surprised to find that it actually corresponds to an altitude that is exactly equal to the radius of the earth. If there was no such thing as escape velocity, and we launched an object upward with what actually is the escape velocity, it would reach an altitude equal to the radius of the earth, and then fall back to the surface.
But that cannot actually happen. If we launch a projectile with a velocity equal to the energy of the altitude that corresponds to a distance equal to that of the earth's radius, it will not fall back down but will continue into space. The same goes for any planet, it does not matter how dense the planet is. The more dense and compact the planet is, the more energy it will take to propel a projectile to the altitude equal to the radius, even though it will be a shorter distance. This energy, once again, is equivalent to the kinetic energy that went into forming the planet by gravity, nucleosynthesis and, molecular bonds, considering the mass of the projectile relative to the mass of the planet.
(I will display the mathematics that I used to arrive at this at the end of the section).
If we had a cloud of hydrogen, the lightest atom, of uneven density, the furthest outer distance at which matter will fall into the cloud, the energy of the escape velocity and, the free fall velocity into the cloud all increase as the cloud is compressed by it's mutual gravity because the gravity of the mass would become more concentrated because it is spread over less area. Both the free-fall velocity and the escape velocity of such a mass reflects both the total mass and the energy that has gone into compression.
This means that, if you throw a ball up into the air with a given amount of energy, and then measure the time which the ball remained in the air before falling back to the same level it was launched from, you are actually measuring the total energy that went into crunching the atoms of the earth together, within a star that has long since exploded in a supernova, from the primordial mostly hydrogen atoms that formed after the Big Bang which began the universe. (The exact calculation procedure is described at the bottom of this section).
Upon compression of the cloud mass by it's own gravity, the altitude equivalent to the radius decreases but the energy required to reach this altitude increases even faster, at the square of the decrease of the radius. This is because geometrically, the size of a sphere follows the Law Of Inverse Squares. Put another way, when a planet is more dense an object below orbital altitude will fall faster to the surface, but the orbital altitude will be lower. What this all comes down to is an interchange between energy and distance.
If the earth should be compressed so that it's radius decreases by half it's volume will be only one quarter, according to the Inverse Square Law. The altitude that must be reached by a projectile launched from the surface to achieve escape velocity will also be reduced by half, along with the radius. The rate of free-fall, which is now the 32 feet per second squared, will be multiplied by two, in inverse proportion to the radius. The energy required to launch the projectile to escape velocity will be multiplied by four, in inverse proportion to the volume. The time that a projectile, such as a ball or artillery shell, will remain in the air before falling back to the same level from which it was launched will be one-quarter of what it was before, in direct proportion to the volume of the compressed earth.
Put simply if the earth were compressed to one-quarter it's present volume with one-half the present radius, a projectile would have to be launched with four times the energy as before to reach an altitude that was an equivalent proportion of the escape velocity equivalent altitude (an altitude equal to the planet's radius) as before. The projectile would climb to only half the altitude as before, but it would travel upward and then fall downward twice as fast as before so that it would remain in the air for only a quarter of the time as before, in direct proportion to the volume of the earth.
Have you ever noticed something about the ring systems around planets in our Solar System? The less dense a planet is, the more likely a significant ring system will be found around it. Saturn is the least dense of the planets, and has by far the most spectacular ring system. A ring around a planet is small particles, such as pieces of ice, that interact to exchange orbital energy around the planet, first to settle on one orbital plane and then to keep all of the particles at or near one orbital energy level.
In the large outer planets of the Solar System, Jupiter, Saturn, Uranus and, Neptune, the orbital zone is much broader, and the orbital speed much slower. These outer planets have more total mass than the smaller inner planets, but their gravity is spread over a much wider area.
I found something else that surprised me about the orbital and escape velocities. The energy of impact upon falling is actually based on the square of the velocity. Twice the velocity will have four times the force. If something falls from twice the altitude, it will have twice the energy upon impact but will be traveling at a velocity of the square root of two, 1.414, times the first object.
With this in mind, what do you notice about the orbital velocity of the earth of 5 miles per second and the escape velocity of 7 miles per second?
The square of 7 (49) is almost exactly double the square of 5 (25). Keep in mind that the figures usually given for these two velocities are rounded. It is difficult to define an exact value for these two velocities because the earth's gravity is not exactly the same from one place to another. Also, if we are close to the equator then the spin of the earth assists in putting objects in orbit. Finally, the practical consideration of air resistance makes it difficult to but a precise value on either velocity.
However, I think it is safe to say that the escape velocity requires exactly twice the energy of the orbital velocity. This means that if we could launch a projectile with the energy to reach an altitude of half the earth's radius, based on the standard rate of free-fall which is the 32 feet per second squared, it would achieve orbit.
I find that the standard figures of 5 miles per second for orbital velocity and the 7 miles per second of escape velocity are rounded figures. The orbital velocity is actually .707 of the energy of the escape velocity, and the escape velocity is 1.414 the energy of the orbital velocity. These figures appear often in calculations involving cycles. These two numbers are reciprocals as well as 1.414 being double .707.
An orbit can be compared to an electrical circuit with alternating current. The mean strength of the current is .707 the maximum amplitude of the wave. Ascending against gravity can be considered as the positive side of the electrical sine wave, and falling back down as it's negative side. The energy required to launch a projectile to escape velocity is twice that required for orbital velocity, but the actual resulting velocities are 1.414 times the orbital velocity for the escape velocity or .707 times the escape velocity for the orbital velocity.
These figures are also prominent in artillery, the firing of a ballistic projectile to less than orbital velocity, so that it falls back to earth. If we fire a shell directly upward, it will reach it's maximum altitude. If we fire it at a 45 degree angle, it will achieve it's greatest range. At this 45 degrees, the shell will reach .707 of it's maximum altitude and a range of 1.414 times the maximum altitude.
So, wouldn't it be logical to expect that these two figures of .707 and 1.414 should also be prominent in the orbital and escape velocities of a planet?
Remember what we saw in the above section "Gravity In Terms Of Satellites", on this blog. An orbit has two components that define it's orbital energy. This energy is equivalent to the altitude multiplied by the orbital velocity. The escape velocity, in contrast, only has one component which is straight up. This explains why the orbital velocity, which propels it to a given altitude, has only half the energy of the escape velocity. It is because the altitude which is attained by this energy is only one of the two components of an orbit, the other is the lateral movement which does not vary in energy and so requires no additional energy to attain once orbital altitude is reached.
Remember Newton's Law that an object either at rest or in motion will continue in motion until acted upon by an outside force. We can think of an object with orbital velocity as moving in a vector between this continued motion and the earth's gravitational pull. If an object from outside has been pulled in by the gravity of a compressed mass, such as a planet, the planet's free-fall rate of an object falling to the planet's surface has increased since before it was compressed. But this must mean that the distance at which the mass can pull an object in from space must also have decreased so that an object that is pulled by the earth's gravity yet lies beyond this distance goes into orbit, rather than falling to the surface, as a compromise.
Doesn't this make it simpler? Instead of the standard algebraic formula for escape velocity, why don't we just say that escape velocity is the energy necessary to reach an altitude on any planet equal to the radius of the planet. Orbital velocity is represented by half the radius of the planet. This applies to any planet, no matter how compact or dense.
Obviously, this formula only applies to projectiles launched from the surface of the planet. If we were to launch a projectile from higher than the surface, the escape velocity would be lower because it would be at an altitude that was once the theoretical surface of the earth before as much energy had been put into it, and this is the energy that must be overcome.
We could say that the radius of a planet is equivalent to it's gravity. If we can launch a projectile to an altitude equivalent to the radius, it will escape the planet's gravity. If we can launch a projectile to an altitude equivalent to half the radius it will have attained one of the two components of an orbit, with the other following naturally, and will go into orbit.
If you want to see the mathematics of how I arrived at this, here it is:
Escape velocity: 7 miles per second = 36,960 feet per second.
36,960 / 32 feet per second = fall of 1155 seconds
36,960 x 1155 = 42,688,800
42,688,800 / 2 (because the total distance of a fall is half the final velocity because the initial velocity is zero) = 21,344,400
21,344,400 / 5,280 (feet in a mile) = 4,042.5 miles, equivalent to the radius of the earth.
Orbital velocity: 5 miles per second = 26, 400 feet per second
26,400 / 32 feet per second = fall of 825 seconds
26,400 x 825 = 21,780,000
21,780,000 / 2 (because the total distance of a fall is half the final velocity because the initial velocity is zero) = 10,890,000
10,890,000 / 5,280 (feet in a mile) = 2,062.5 miles, equivalent to half the radius of the earth.
Notice that the would-be falling time of the escape velocity projectile, 1155 seconds, is approximately the square root of two (1.414) times the would-be falling time of the orbital velocity projectile, 825 seconds. This is because the velocity which is equivalent to energy of altitude is based on a square. A height that has a falling time to the surface of the square root of two of a lower altitude will have twice the energy of altitude of that lower altitude.
To calculate the total approximate energy that has gone into compressing the atoms of the earth into the heavier atoms that they are today, from the primordial mostly hydrogen formed after the Big Bang:
Multiply the 1155 seconds fall from radius altitude by two to give the total theoretical time in the air of a projectile at escape velocity = 2310 seconds.
Divide this by the time the projectile stays in the air, the heavier the projectile and the less affected by air resistance, the more accurate the result.
Multiply that figure by the mass of the earth divided by the mass of the projectile.
Multiply this by the energy that went into launching the projectile taking, of course, any energy inefficiency into account.
This is the approximate energy that has gone into compressing the atoms of the earth, by crunching together the atoms from primordial mostly hydrogen in a former star, into the heavy atoms that make up the earth today.
The vast majority of this energy will be stored as nuclear binding energy. All other factors, such as molecular bonds and material structural bonds, are insignificant in comparison with the energy in nuclear binding energy.
4) THE ENTRANCE VELOCITY
Here is one of those questions of the ages that we rarely seem to wonder about: Why is the moon so heavily cratered by meteorite impacts, in comparison with the earth? This is particularly true of the far side of the moon, the side that always faces away from earth, where there are many cases of multiple craters within one another.
Here is a photo of the far side of the moon, in which the extensive cratering can clearly be seen: http://en.wikipedia.org/wiki/Far_side_of_the_Moon#mediaviewer/File:Back_side_of_the_Moon_AS16-3021.jpg
The usual answer that is given is that the weathering and erosion on earth would eventually erase all but the largest meteorite impact craters, and also that all but the largest meteors would burn up by friction with the atmosphere.
(Note-By the way, it is a meteor when it is in motion, it becomes a meteorite when it impacts the ground).
But my conclusion is that, if the earth had been hit by as great a concentration of impact craters as can be seen on the moon, particularly the far side which did not have the volcanic activity of the near side which might erase craters, we would not only see geological evidence of it below the surface but the earth would also have a greater concentration of minerals than it does, which come from meteorites. A meteor of any size will make it through friction with the atmosphere. The typical "falling star" that is seen shooting across the sky on a clear night might be only as big as a grain of sand, it makes a streak of light in the atmosphere because it was moving at such a high speed in space.
Plainly and simply, there was a much greater concentration of meteorite impacts on the moon then there was on earth. The question is: why?
Remember the basic scientific principle that energy can never be created or destroyed, but only changed in form. it is this seemingly unrelated principle that I find leads us to our answer. It also leads to an explanation of how orbits operate that I have never seen before.
Remember also Newton's Law that an object either at rest or in motion will remain either at rest or in motion, until it is acted on by an outside force.
We should also understand that there is no energy at all in the basic forces. Gravity is such a basic force. When you throw a ball up into the air, it comes back down with the energy of force. But none of it's energy came from gravity. You would only be getting back the same energy that you put into the ball in the first place. Gravity would have simply redirected that energy from upward to downward.
With these three facts in mind, consider the following question:
Suppose that there is a small planet, and a long distance away there is a much larger and more massive planet. A ballistic projectile, such as an artillery shell, is launched from the small planet with just enough velocity to escape the planet's gravity and continue into space. Eventually, the projectile approaches the large planet. What happens then?
Your answer might be that the projectile would simply fall into the influence of the larger planet's gravity, and impact the surface of the large planet.
But wait a minute, there is a complication that comes with that answer.
If the projectile were to impact the surface of the larger planet, after being pulled in by it's gravity, it would have to impact the planet at a faster velocity than that at which it was launched from the smaller planet. This is simply because the larger and more massive planet would have stronger gravity and thus a faster gravitational free-fall rate then the smaller planet. Remember that the projectile was launched from the smaller planet with only enough energy and velocity to overcome it's lesser gravity.
Basically, this scenario would result in the projectile impacting the surface of the larger planet with more energy than it was launched with from the smaller planet. But that would be a violation of the scientific principle that energy cannot be created out of nothing. There is no energy at all in gravity so the extra energy cannot be said to have come from the gravity of the larger planet.
(Note-There is a spaceflight technique, known as Gravitational Assist or the "Slingshot Effect", in which a spacecraft can use the gravity of a planet to change direction without using up fuel, and can go into temporary orbit around the planet to "borrow" some of the planet's orbital energy, but that would not apply in our scenario of the projectile directly approaching a planet without any such technique).
But what about an orbit around the planet? We saw in the above section, "Gravity In Terms Of Satellites" that an object in a higher orbit around a planet or star actually has more orbital energy than an object in a lower orbit, even though the object in higher orbit will move more slowly. An object with three times the orbital energy will orbit at nine times the distance from the mass center of the planet, but will move with only one-third the speed. The energy of an orbit is proportionally equivalent to the total space covered, during the orbit, by a line between the mass center of the planet and the mass center of the object in orbit.
Applying this to a meteor that might strike the earth, if an asteroid in orbit in the vicinity of the massive planet Jupiter is moving faster than Jupiter, it may have some of it's orbital energy "borrowed" by Jupiter even though Jupiter is already in a higher-energy orbit than the asteroid. This would mean that the orbit of the asteroid would have to contract, so that the line from the mass center of the sun to the mass center of the asteroid covered less distance. Because remember that the energy of an orbit is proportional the the total area of space that is covered by this line during an orbit. It is this contraction that raises the possibility that the orbit of the asteroid might cross the orbit of the earth, resulting in an impact.
Next, let's review the escape velocity of a planet. This is the velocity that a projectile would have to be launched with in order to escape the gravity of the planet altogether, and continue into space. For the earth, the escape velocity is approximately 7 miles per second (11.263 kps). Keep in mind that the escape velocity only applies to ballistic flight, and not to rockets. For a rocket to escape the earth's gravity, it does not have to move at this velocity, all that is necessary is to keep the engine running regardless of the velocity.
In the above section, "Escape Velocity And Orbital Velocity", I explained my observation that, if the escape velocity is divided by the normal rate of free-fall, which on earth is 32 feet per second squared, the escape velocity is equivalent to an altitude which is equal to the radius of the earth, about 4,000 miles (6436 km). This means simply that, if we could launch a ballistic object to an altitude equal to the planet's radius, it would not fall back down nor go into orbit but would continue indefinitely into space. As of this writing, the escape and orbital velocities are somewhat theoretical because launching a ballistic object, without it's own power source, either into orbit or beyond, has never actually been done.
The equivalence of the escape velocity of a planet with the free-fall velocity from an altitude equivalent to the radius of the planet, provides some insight as to what would happen if a projectile or meteor were to enter the earth's gravity from space. If the projectile approached directly at earth, as in the example of the projectile launched from the smaller planet, so that it directly impacted the earth, it would have to be falling at least at the velocity of the escape velocity at the point of impact. If the projectile or asteroid were not already moving at this velocity, it would have to be accelerated to that velocity, as per the rules of free-fall.
But suppose that the energy of the asteroid or projectile, as it entered the earth's gravitational field, was less than the energy that it would have if it was accelerated to this velocity by the earth's gravity. This brings us back to our complication in the example of the projectile launched from the smaller planet toward the larger planet. We cannot create energy out of nothing, and there is no energy at all in gravity.
If it is an asteroid which loses orbital energy so that it intersects earth's orbit, the same principle applies. If the asteroid impacted the earth, it would then effectively become a part of the earth's orbit around the sun. But if the orbit of the asteroid around the sun had more energy than that of the earth's orbit to begin with, it could not impact the earth, even though it would be pulled in by the earth's gravity, with more energy than the difference between it's orbital energy and that of the earth. If it did, it would be breaking the Law of the Conservation of Energy. No matter what we do, energy cannot be created out of nothing.
This is why I conclude that a planet's escape velocity could be more properly called it's entrance velocity.
Have you ever wondered why orbits exist? If we launch an object into space with less than the escape velocity, it will not leave the earth's gravity but neither will it fall back to the surface, it will go into orbit around the earth. The conclusion that I have come to is that orbits exist to uphold the Law of the Conservation of Energy.
If a projectile or asteroid approaches the earth from outside, but with less energy than the energy of the impact with the surface would be if it were pulled in by the earth's gravity, the object cannot break the Law of the Conservation of Energy and impact the earth's surface, but neither can it escape the earth's gravity. What will happen is that the object will go into orbit around the earth. The energy of this orbit would be, as we might expect, equal to the energy of the object's momentum. This must be the case because, once again, energy cannot be created out of nothing.
This means that an orbit serves the same purpose from either direction. Just as an orbit prevents an object launched from the surface from continuing into space because it lacks the energy of escape velocity, it also prevents an object from space from continuing and impacting the earth's surface because it does not have enough energy to match a free-fall that would produce a velocity equivalent to orbital velocity. The object goes into earth orbit in the same way regardless of whether the approach is from up or down.
Suppose that the object in orbit was a spacecraft with the ability to thrust a rocket in the opposite direction, a retro thrust. My logic is that the effects of the forward and retro thrusts would be reversed, according to whether the spacecraft entered orbit from above or below.
An orbit is a vector of two components, the gravitational pull toward earth and the forward momentum around the earth. For a spacecraft launched from the surface into orbit, a retro thrust with the right energy would cancel the forward momentum component of the orbit so that the spacecraft would fall back toward the surface by the remaining component, the gravitational pull toward the earth. While a forward thrust would add to the spacecraft's orbital energy and it would go into a higher orbit, with the increase in the total area of space covered by a line from the spacecraft to the mass center of the earth being proportional to the added energy.
If the spacecraft went into orbit from the outside, because it did not have enough energy to go directly to the earth's surface, the retro thrust would operate in reverse even though the result would be the same. The apparent retro thrust would actually be adding energy to the spacecraft so that now it could reach all the way to the earth's surface. A "forward" rocket thrust on the spacecraft that had entered orbit from outer space would actually act as a retro thrust would and sent the spacecraft into a higher orbit and back in the direction from where it came, even further from the surface.
Remember that it is not the size or mass of the meteor that determines whether it will reach the surface of the earth or go into orbit. It is it's velocity which is the determining factor.
Finally, let's go back to our original question of why the earth's surface has been much less impacted by meteors than that of the moon. It is because, as we see above, it requires less energy to get to the surface of the moon than it does to the surface of the earth. The great paradox is that this is actually because the earth has much stronger gravity than the moon so that the moon's escape velocity and free-fall rate are low and a meteor need only to match this energy to impact the moon.
I find that this scenario explains well how gravity, energy and orbits operate in space.
The reason that there are not more meteors in orbit around the earth is simply that the orbits that encounter the earth's gravity are still in orbit around the sun, which has much stronger gravity than the earth. The result is that the earth's gravity merely deflects the meteors, rather than holding them permanently in orbit.
5) ORBITS AND SUPPORT FOR QUARK THEORY
This scenario provides support for quark theory.
I would first like to write some more about how the nature of a large-scale structure must somehow be a reflection of it's building blocks, since this is a part of the information that has gone into it. One obvious example is how the easiest shape in which to build a house that is made of bricks is in the shape of the bricks themselves. But the most important example would certainly be the similarity between the orbitals of electrons around nuclei within atoms and the orbits, on a very large scale, of moons around planets and of planets around stars that are composed of these atoms.
We have seen above how objects can be drawn by gravity into orbit around a planet or stars only if compression of the atoms in the planet or star has taken place by nucleo-synthesis. This is the process that takes place in the centers of stars where smaller atoms are crunched into heavier ones by the tremendous heat and pressure. To accomplish this, the force of gravity within the star must be enough to overcome the electron repulsion between adjacent atoms.
The star may later explode and scatter it's component matter across space, some of which may re-condense to form planets. This means that compression of the planet by this nucleo-synthesis does not take place by it's own gravity, which would be nowhere near strong enough to overcome electron repulsion, but took place within a star which later exploded and scattered it's component matter of heavier atoms across space.
All matter was originally a cloud of mostly hydrogen, which is the lightest of all elements with only one electron in orbit around one proton in it's atom. The important thing here is that there can be no orbit around such a cloud of hydrogen. Orbits can only take place when the cloud of hydrogen condenses by it's own gravity into a star, and compression of lighter elements into heavier ones takes place by nucleo-synthesis.
A primordial cloud of hydrogen, without any gravitational compression, cannot be involved in an orbital mechanism. We could say that the space around the star in which an orbit can take place is the "ghost" of the much larger space that had been occupied by the cloud of primordial hydrogen before it was compressed by it's own gravity into a star.
Thus, whenever a body is formed by compressing it's component matter, and the body has some attractive force such as electric charge or gravity, another body that is attracted by this force will be able to go into orbit around the first compressed body. But this can only take place if both bodies have undergone such compression. The energy that has gone into this compression becomes potential orbital energy.
We know that protons, with the same like positive charge, have this like charge repulsion overcome by binding energy so that the protons, which would usually repel one another, are bound together to form the atomic nucleus. My conclusion is that this is also compression and is why electrons, attracted to the nucleus by opposite charge attraction, can go into orbitals around it. This is why atoms exist.
The orbit of a moon around a planet, or of a planet around a star, follows exactly the same pattern as the orbit of an electron around the nucleus. This is what we would expect if a large-scale structure must manifest some the nature of it's component building blocks. But then this should mean that the gravity which drives the operation of the universe on a large scale is somehow related to the electrical force that drives the orbits of electrons around the nucleus.
The universe, both space and matter together, consist of an equal number of negative and positive electric charges. Among these charges, like charges repel while opposite charges attract so that the overall forces of attraction and repulsion must be equal. But the binding energy which forms the nucleus of an atom by holding like-charged protons together against the mutual repulsion upsets the overall balance between attraction and repulsion, by overcoming some of the mutual repulsion of like charges, so that there is a net attractive force, and this is manifested as gravity.
So, the orbits of moons and planets and the orbitals of electrons in the nucleus are both related to the formation of the nucleus by binding energy, as smaller atoms are crunched together within stars. The formation of heavier atoms from lighter ones enables electrons to fill higher orbitals than in the lighter atoms. The resulting compression of the volume of the mass brings about the possibility of other objects, composed of heavier atoms, going into orbit around the compressed mass in exactly the same way as electrons around the nucleus.
It is not possible for a cloud of hydrogen, the lightest and simplest atom, to be involved in an orbital mechanism because no such compression has taken place, and thus there is no potential orbital energy. The only difference is that there is a mutual attraction by gravity, rather than a mutual repulsion of like-charged electrons, between the objects in orbit.
But this brings us to a question. We see that for electrons to orbit the nucleus or for moons to orbit planets or for planets to orbit stars, it is necessary that the crunching of lighter atoms into heavier ones must have taken place. This is accomplished by nucleo-synthesis, by the tremendous heat and pressure in the centers of stars.
But what about the lightest and simplest element of all, the hydrogen atom? The vast majority of atoms in the universe as a whole are still hydrogen. No compression could have taken place because the basic hydrogen atom consists of only one electron in orbit around one proton. How can the electron orbit the proton, as we have described here, if no such compression of lighter atoms into heavier has taken place? If my scenario is correct then some type of compression must have taken place within the single proton of the hydrogen atom.
This brings us to quark theory. This is a theory that was introduced in 1964. According to this theory, both protons and neutrons in the nucleus are composed of particles called quarks. There are several varieties of quark, but the most important are up and down quarks. Up and down quarks have partial electric charges. An up quark has a charge of + 2/3, and a down quark has a charge of -1/3. Protons and neutrons are each composed of three quarks. Two up quarks and one down quark make up a proton, which has an overall charge of +1. Two down quarks and one up quark make up a neutron, which has an overall charge of zero. It is said that if all quarks except these up and down quarks disappeared, only physicists would notice. Particles such as electrons are referred to as leptons, and are not composed of quarks. But yet leptons and quarks cannot be composed of completely different materials because an electron can be crunched into a proton to produce a neutron (known as K-capture).
This theory of orbitals of electrons and orbits of planets resulting from nucleo-synthesis provides support for quark theory. There must have been some kind of compression going into every proton or there would not be the potential orbital energy so that the electron could go into orbit around one proton to form the hydrogen atom. Quark theory provides the ideal explanation because it holds that three quarks went together to form the proton. This makes electron orbits in the lowest orbital shell possible, but the binding together of multiple protons are necessary for higher electron orbitals.
6) ORBITAL ENERGIES AND PLANETARY DYNAMICS
I was hoping that this hypothesis would provide some insight into the planetary dynamics of the Solar System.
There is energy in every orbit. The energy is proportional to the space enclosed within the orbital path. If there were two objects in orbit around the earth, with one having three times the orbital energy of the other, the one with the higher energy would orbit at nine times the distance from the center of gravity of the earth, but would move along it's orbital path at only one-third the speed. The reason for this is that the earth's gravitational pull is only one-ninth that on the object with higher orbital energy as it is on the closer object with lower orbital energy.
The rate at which an object in orbit around a planet must move in order to stay in orbit is related to the strength of the planet's gravitational pull. If the gravitational pull is only one-ninth for one object than it is for the other, yet the object experiencing one-ninth the gravity is moving with one-third the speed of the closer object with lower orbital energy, that means that the distant object with higher orbital energy must have three times the orbital energy of the closer object.
Two objects, in orbit around a third object, cannot collide or go into orbit around one another unless they have different orbital energies. If the two objects approach one another, and go into a mutual orbit, there will be energy in that new orbit that must have come from the original orbital energy around the third object that both had.
If one object has higher orbital energy than another, due to an eccentric orbit or possibly due to the gravitational pull of a passing star, and their orbits cross, what may happen depends on the difference in their orbital energies. If the energy difference is enough to supply the energy that would have to be released by the collision of one object with the other, according to the rules of gravitational acceleration due to free-fall, they can collide and both continue to orbit the third object as one. If the energy difference between the orbits of the two are not enough to supply the energy that would have to be released, according to the rules of gravitational free-fall, the two objects can still join together in an orbital pattern. But if any orbital energy difference between the two is not even enough to form an orbital relationship, the two can neither collide nor form an orbit.
This is why the orbits of planets around the sun are elliptical in form, rather than circular. The orbital energy is constant throughout the orbit because the planet moves faster when at it's closest to the sun (perigee), and slower when it is furthest from the sun (apogee), so that a line from the planet to the sun sweeps over equal areas of space in equal periods of time. But when rocky and metallic debris collects together by gravity, some of the pieces will be further from the sun than others. The orbits of all pieces must be conserved in the final collection and so the perigee of the elliptical orbit represents those pieces of the planet that were originally closest to the sun, and the apogee of the orbit represents those pieces which were furthest from the sun.
The final elliptical orbit of the planet is thus a kind of "averaging" of the orbital paths of all the collective pieces. This "span" of orbital paths of pieces of debris that collect together to form a planet serves another important purpose. The various distances within the span provides different orbital energies so that collisions can take place at all. Remember that two orbiting objects with exactly the same orbital energies cannot collide because there would be no energy difference to release as the energy that would have to be released by the collision of the two. Furthermore, two objects in orbit around the sun at exactly the same speed would never pass one another so that they could fall together by gravity.
But notice that while oddly-shaped pieces of rocky and metallic debris fall together to form the spheres of planets and moons, the spheres themselves never collide unless something highly unusual takes place, such as the gravitational influence of a passing star. Consider Neptune and Pluto, their orbits around the sun cross one another but this has been going on for billions of years without a collision or even a gravitational capture.
But this is because of the way the universe works, energy must always be conserved. We could say that spheres never collide because if they were going to collide, they would not have formed as they did by gravitational agglomeration. But the reason from an energy conservation point of view if two objects were in the same orbit and collided, they would still have the same total orbital energy, with regard to the sun. But that would leave no remaining energy either to be released upon collision if they collided, according to the acceleration of free-fall, or to be transformed into orbital energy if they went into an orbital relationship. The only other alternative is to keep passing one another without meeting.
The reason that moons orbit planets, when drawn in by the planet's gravity, instead of just crashing into the planet, is that, the larger a planet grows by the accumulation of debris in space, the higher is it's "entrance velocity" which is the velocity of a passing object in space that is necessary to collide with the planet. This is because there is energy that would have to be manifested by the collision, according to the rules of an object in free-fall due to gravity, and there must be at least that much difference in the energy of the two orbits.
Energy cannot be created out of nothing and if there is not enough spare energy in the higher orbital energy object, then the objects can form an orbital relationship with the difference in orbital energy around the third object between the two becoming the orbital energy of one around the other. This explains why Mars, right next to the asteroid belt, captured it's small moons Phobos and Deimos. But the two did not go on to impact the surface of Mars.
If there is enough difference between the orbital energies of the two astronomical objects to supply the energy that would have to be released upon collision, according to the rules of gravitational acceleration in free-fall, then a collision is what will take place. The threshold for this collision to happen is what I refer to as "The Entrance Velocity", as we saw in the section by that name above.
The impact cannot be made any lighter, or of lower energy, because it must abide by the rules of free-fall in accordance with the mass of the planet. The reason that moons of the outer planets remained moons is that they had enough difference in orbital energy to gain an orbit around the planet but, as the planet was continuously growing by agglomeration of mass it's "entrance velocity" kept climbing higher, and when the moon came close enough to the planet there was not enough difference in orbital energy to meet the "entrance velocity" because the increased mass of the planet meant that more energy would have to be released upon impact of the moon, in accordance with the rules of planetary mass and acceleration during free-fall.
This explains why the earth, the moon, Mercury and, presumably, Venus, have been pounded by meteorite and asteroid impacts but have not captured any as permanent small moons, as Mars has captured Phobos and Deimos. This is because, at the lower orbital energy levels of the inner planets, earth, Venus and, Mercury, the meteorites or asteroids would have enough difference in orbital energy with the planet to exceed the "entrance velocity" and collide with the planet instead of going into orbit. Of course, the dense atmospheres of earth and Venus would burn up all but the largest meteors by friction.
As a general rules, the entrance velocity of a planet is practically equal to the rotational velocity of the planet. If the object is so close to the planet, which represents lower orbital energy, that it has to orbit faster than the planet is rotating in order to remain in orbit, it would represent a lower energy state just to have the object crash into the surface of the planet. Notice that no moon in the Solar System orbits faster than it's planet rotates.
All of this shows that comets must be of a different orbital energy level then the rocky and metallic objects of the Solar System or they would not be able to land on planets. It is believed that the water on earth came from comets, and two have impacted Jupiter since 1994. The comets are composed mainly of ice and clearly have a much higher orbital energy level around the sun then the planets and moons.
My theory, as described in "New Thinking About The Origin Of Comets And Water", on the physics and astronomy blog www.markmeekphysics.blogspot.com , is that the star which preceded the sun and exploded in a supernova which provided the materials to form the present Solar System, first exploded as a nova in which only the outer layers are blasted off into space. This is what formed the comets, and icy objects of the Kuiper Belt.
It has a higher orbital energy and is usually further from the sun simply because it had a higher original starting point. This explains why the moons of the outer planets, as well as objects like Pluto, Sedna and, Xena, tend to be composed of ice rather than rock or metal. Notice that Pluto, next to the Kuiper Belt just as Mars is next to the Asteroid Belt, has several small icy moons just as Mars has the rocky Phobos and Deimos. This also explains why comets can collide with Jupiter but it's rocky and metallic moons can't.
7) EXPLAINING THE ORBITS OF COMETS
We know that the sun is a second-generation star. There was a previous star which exploded in a supernova, scattering it's matter across space. Much of this mass came back together by gravity to form the sun and planets of the Solar System. One way that we know the sun to be such a second-generation star is that it already contains heavy elements, the formation of which would require reaching a stage in the successive nuclear fusion process ahead of where the sun is now.
I occasionally wonder what this former star must have been like. Obviously, it was mush larger than the sun is now. We can see traces of this former star in the abundance of iron in the inner Solar System, because the usual fusion process only goes as far as iron. The fact that rare earth elements are usually found together on earth tells us that the r-process (for rapid) of fusion that takes place only with the input of energy that is possible when a supernova is actually in the process of exploding does not have time to separate the rare earth elements from one another after lighter elements are fused into these heavier elements by the burst of energy from the supernova. The energy released by the supernova explosion of this former star can be seen in the orbital energy of moons around planets and of planets around stars, in tidal energy and, in the energy of nuclear fission (uranium 235 or plutonium) which reverses the rapid fusion process which took place when the star was actually in the process of exploding, and releases some of this energy.
My understanding of how and why some large stars explode is explained in "The Mystery Of Exploding Stars", on the physics and astronomy blog www.markmeekphysics.blogspot.com .
In the posting on the physics and astronomy blog, "New Thinking About the Origin Of Comets And Water", I explained my concept that before the former star exploded from the core as a supernova, it blasted off it's outer layers as an ordinary nova. The outer layers of this star contained lighter elements, such as hydrogen and oxygen. The energy from the nova did not fuse these together into heavier elements, as the r-process does when the star explodes as a supernova, instead it fuses these lighter atoms together into molecules. This is how we get water. This would seem to also include salt, as we saw in "The Mystery of Salt" on the physics and astronomy blog, and why salt is mixed with the water on earth.
This primordial water is where comets originate, the primary component of which is water ice. In my hypothesis, the reason that comets tend to be based further from the sun than planets is that the matter which forms comets had a higher "starting point" in being hurtled out into space from the previous star. Comets originated from the matter thrown outward by the nova which blasted off the star's outer layers, while the rocky and metallic material that formed the planets was thrown outward by the supernova of the star exploding from the center.
The landing of comets seems certain to be where the water on earth came from. But the question about comets is why their orbits are so eccentric. All planets orbit the sun in elliptical orbits, with perigee being the point in the orbit closest to the sun and apogee the point furthest from the sun. But the planets tend to have orbital paths that are not far from being circular. The orbit of earth, for example, has a difference in distance of only about four percent between apogee and perigee.
The orbits of comets, in stark contrast, are extremely eccentric and elliptical. At their furthest from the sun in their orbits, comets typically extend to the far outer reaches of the Solar System. They then come very near to the sun when at the closest to the sun point of the orbit. The "tail" of a comet, which always points away from the sun, is caused by the sun vaporizing some of the ice of the comet which then reflects sunlight so that we can see it.
My reason that the planets inevitably have elliptical, rather than circular, orbits is that pieces of metallic and rocky debris come together by gravity, while all are in orbit around the sun, to form a planet. This material must coalesce together over a certain span of distance from the sun, simply because if all of the pieces were at the same distance from the sun they would be in orbit around the sun at the same velocity and so would never pass one another so that they could be drawn together by gravity.
Momentum must always be conserved, and this includes the original orbits of those pieces in orbit around the sun which coalesced together to form the planet. This necessary conservation of orbital momentum is reflected in the final orbit of the planet. The perigee represents the former orbits of those pieces that were closer to the sun, and the apogee represents the former orbits of those pieces which were furthest from the sun. In this way, the momentum of these former orbits is conserved.
The reason that the orbits of the planets in the Solar System are in close to the same geometric plane is that debris from the exploded star originally went into orbit around the central sun in all different directions until, through collision and mutual gravitational attraction of the pieces of debris, one plane came to be dominant. The same pattern can be seen today in the rings of Saturn, which are composed of pieces of ice.
The reason that the orbits of comets around the sun may not be anywhere near the same general plane of the planets is that comets were outside the range of debris from the exploded star, due to having a higher starting point when the outer layers of the star were blasted off by the nova, which coalesced to form the planets and so, as explained in my hypothesis in "New Thinking About The Origin Of Comets And Water", did not undergo this dominant geometric plane process.
The orbits of comets around the previous star, after lighter molecules such as water and salt were blasted into space by the nova, must have been nearly circular because comets are smaller than planets and so would not require any wide span of consolidation. So, how could the orbits of those comets possibly have become as eccentric and elongated as they are today?
There is energy in an orbit, and this energy has a direct relation to the mass around which the orbit takes place. Remember also that mass and energy are really the same thing, this is well-established in science. Suppose that an object, such as a comet, was in orbit around a star, such as the previous star which existed before the sun. Now suppose that the central star undergoes a sudden and drastic reduction in mass. This is what happened when that star exploded in a supernova, and only a portion of the mass came back together by gravity to form the sun.
If the mass at the center is suddenly drastically reduced, then the orbital energy must suddenly decrease as well. When that happens, the orbit of the comet must also be affected. The sharp reduction in orbital energy explains why the orbits of comets tend to be so elongated. The apogee of the orbit, the furthest out point, had to remain constant in the same way that the previous orbits of pieces of rocky and metallic debris must be reflected in the final orbit of the planet which they coalesce by gravity to form.
But the rest of the comet's orbit underwent shrinkage to accommodate the lower orbital energy when the mass of the central star was much reduced by the explosion of the supernova. The apogee had to remain constant to reflect the momentum of the previous orbit of the comet, before the supernova, the velocity could not just decrease because that would violate Newton's Law of an object remaining in continuous rest or motion.
So what happened is that the perigee, the closest point of the orbit, moved closer to the sun. This decreased the total volume of space within the orbit, and thus the orbital energy, to match that of the reduction of the mass, and thus energy, of the central star.
It would seem that an object in orbit around a star should move closer to the star if the star's mass increased, and move further away if the star's mass were to decrease. Isn't this the way the Law of Gravity would have it?
The reason that it does not work this way, the comet in orbit remains further from the central star when the star has more mass, and moves overall closer when the star's mass is drastically reduced by way of a supernova, shows the validity of my hypothesis in "Orbital And Escape Velocities And Impacts From Space". The orbital energy is proportional to the space enclosed within the orbit but this orbital energy is a function of the mass, and thus energy, in the central object around which the orbit proceeds. If this mass is changed by the star exploding as a supernova, then the space enclosed within the orbit must reflect that change. Energy must always be conserved.
This also explains why Pluto has an orbit that is much more eccentric then the planets, but not as eccentric as comets. It also has an orbital plane around the sun which is different from the planets, but not as different as that of the comets. Pluto started as a rocky object, it could have once been a moon of Neptune. But it was struck by a comet so that now it is both rocky and icy in nature, and it incorporated the momentum of the comet so that both it's eccentricity and it's orbital plane are between that of the planets and the comets. If this is the case, the added momentum of the comet could have broke Pluto free of orbit around Neptune, but evidence of this former momentum remains in how the orbit of Pluto around the sun crosses that of Neptune.
8) THE SOLSTICE GAP
We saw in the supporting document, "Why The Earth Is Tilted On It's Axis", of the posting "The Story of Planet Earth", on the geology blog, www.markmeekearth.blogspot.com , an explanation that the earth is tilted on it's axis at an angle of 23 1/2 degrees, which gives us the four seasons, because of the tectonic shift of land northward which made the northern hemisphere heavier than the southern hemisphere.
Today, the vast majority of the earth's land is in the northern hemisphere. But processes like these, being limited to the earth itself, could not change the gravitational line between the center of the earth and the center of the sun. So what happened to accommodate the change in mass distribution is that the earth had to tilt on it's axis, with the gravitational line between the center of the earth and the center of the sun remaining the same.
But this still does not explain why the earth tilted in exactly the direction that it did, out of a 360 degree circle.
My explanation is that, like the tilt on it's axis itself, the particular direction of the tilt was to achieve mechanical balance. The northern hemisphere of the earth is heavier than the southern hemisphere. Also, the earth is not a perfect sphere, with the north pole being just a little bit further from the equator than the south pole and thus adding more proportional mass to the northern hemisphere.
The solstices, the points during the year when the point at which the sun is directly overhead is furthest north or south from the equator, closely correspond with the apogee and perigee of the earth's orbit around the sun. The earth orbits the sun in an elliptical path, with apogee being the point where it is furthest from the sun, and perigee being where it is closest to the sun. The heavier northern hemisphere is tilted toward the sun when the earth is furthest from the sun, and the lighter southern hemisphere is tilted toward the sun when the earth is closest to the sun.
The perigee, the point in the earth's orbit where it is closest to the sun, occurs around January 4. This is why it is summer, at this time, in the lighter southern hemisphere, because it is the hemisphere that is tilted toward the sun.
The apogee, the point in the earth's orbit where it is furthest from the sun, occurs around July 4. This is why it is summer, at this time, in the heavier northern hemisphere, because it is the hemisphere that is then tilted toward the sun.
Think of the seasons, with the two hemispheres being unequal in mass, and the elliptical orbit of the earth around the sun, as two imbalances. The two imbalances operate in such a way as to help to cancel one another out. The heavier hemisphere tilts toward the sun when the earth is furthest from the sun, and the lighter hemisphere tilts toward the sun when the earth is closest to the sun.
That seeking of mechanical balance in the earth's orbit and it's seasons explains why the solstices closely line up with the apogee and perigee. But the two are not exactly the same. There remains a difference of about two weeks between the line between apogee and perigee and the line between the winter solstice and the summer solstice. The northern hemisphere summer solstice occurs on or around June 21, but the point of apogee is not until around July 4. The northern hemisphere winter solstice occurs on or around December 21, but the point of perigee is not until January 4.
It makes sense that the solstices would line up with the points of apogee and perigee, but why would there be this two week gap involved?
The earth, while tilted the 23 1/2 degrees on it's axis, still rotates at an even rate and shifts on it's axis between solstices at an even rate. But it does not move in it's orbit around the sun at an even rate. Kepler's Law about orbits is that planets will orbit the sun in an elliptical path, with the sun being at one of the two foci of the ellipse, and a line from the center of the planet to the center of the sun will sweep over equal areas of space in equal periods of time. This means that the earth is moving faster along it's orbital path when it is closer to the sun, at perigee in January, than it is when it is furthest from the sun, at apogee in July.
A day, and a day's shift of the earth on it's axis between the solstices, occupies a longer span of the earth's orbital path near the perigee, in January, than it does near the apogee, in July, simply because the earth is moving faster in it's orbit near perigee but is not rotating any faster. This means that, if we measure the progress of the year by the number of days that have gone by, there will be a gap in comparison to what there would be if we measured the year in terms of apogee and perigee. A line from the apogee to the perigee point of the earth's orbit, and a line from the summer solstice to the winter solstice, each divides the earth's orbit exactly in half, but the two divisions are not exactly the same.
The earth's orbit has an average distance from the sun, but we get a different figure if we measure the average distance day-by-day throughout the year than if we measure the average distance overall. This is simply because the earth moves more slowly when it is further from the sun, which means that there are more days with the earth further from the sun than the average distance than there is when the earth is closer to the sun than the average distance.
Put another way, the two week difference between each solstice and the corresponding apogee or perigee is proportional to, and a reflection of, the difference, including the effects of Kepler's Laws of Planetary Motion meaning that earth moves more slowly in it's orbital path when it is at a greater distance from the sun, in how fast the earth moves at apogee, compared to at perigee. The proportion between the difference in the earth's velocity during the course of the year will be equal to the proportional difference between the two week gap, and the entire year.
If we take the six month average of the earth's distance from the sun, since each solstice as well as each apogee or perigee occurs every six months, earth goes from the average distance from the sun to apogee or perigee and then back to the average distance again. So, the six month average distance is the average between the yearly average distance and the distance at apogee or perigee.
The earth's average distance from the sun, through the course of the year, is about 150 million km, or 93 million miles. The apogee distance, around July 4, is about 153 million km, or 95 million miles. The perigee distance, around January 4, is about 146 million km, or 91 million miles.
If we take the squares, due to the Inverse Square Law which governs gravitational force at a distance and thus how fast the earth moves in it's orbit at different distances, of the average yearly distance from the sun, and the apogee or perigee distance, the difference if we divide the lower by the higher value is about .042. If we multiply that by the number of days in a year, it gives us just over 15 days. This is almost exactly the difference between each of the solstices and the corresponding apogee or perigee.
The reason that we do not get exactly 14 days is that the difference between the two is not exactly 14 calendar days. Both the solstice and the apogee or perigee must occur at a specific moment, the sun being directly overhead at the point furthest from the equator and the earth being at the point closest to, or furthest from, the sun, and not over the course of an entire day. The first day of winter or summer may vary by one or two days, from year to year, and the point of apogee or perigee may not fall always exactly at the same date
Let's refer to this approximately two week difference between the solstices and the corresponding apogee or perigee as "The Solstice Gap".
9) MEASURING THE DISTANCE TO THE MOON
I would like to show how a calculation that I have never seen done before could have been done centuries ago, if we would have realized how information operates in the universe. The distance to the moon could have been calculated, long ago, by knowing just it's orbital period and the rate of free-fall of an object on earth. There must be a direct connection between the two because both are governed by the same gravity and, since gravity is a simple thing and there is no new information added, that connection must also be simple.
This posting has already been added to the compound posting on this blog, "Orbital And Escape Velocities And Impacts From Space".
The distance to the moon can actually be measured just by things that we can easily measure on earth. We have to know that the earth is spherical and that the moon revolves around the earth, but we do not need to know that the moon's orbital velocity operates by the Inverse Square Law.
We do know that the moon revolves around the earth every 29 days. We can determine this simply by watching the moon go through it's phases. We can also notice that the moon rises 50 minutes later each day, then it did the day before. There is 24 hours in a day, and if we divide that by 50 minutes it tells us that the moon revolves around the earth every 29 days.
We can measure the acceleration due to gravity on earth of a falling object. This acceleration is 32 feet, or 9.75 meters, per second squared. This mean that, starting with a velocity of zero, a dropped object is moving 32 feet per second at the end of the first second, 64 feet per second at the end of the second second, and so on. In the real world this falling velocity does not increase indefinitely, due to air resistance, but we can ignore that for our purposes here.
This means that we can measure the distance that an object falls, in a given period of time, by use of multiples of 16 feet (4.88 m), a unit of measure that I refer to as a "grav", for gravity. I discussed this in my book, "The Patterns Of New Ideas", and also in the posting on this blog, "The Way Things Work". In the first second, the falling object starts at a velocity of zero and the velocity continuously increases to a velocity of 32 feet (9.75 m) per second, at the end of the first second. This means that the average velocity of the object during the first second was 16 feet (4.88 m) per second and, since it fell for one second, that must mean that it fell 16 feet (4.88 m).
This calculation is just easier to do in units of feet and miles, because it gives us round figures that we will not get if we use metric.
At the beginning of the second second of fall, the velocity is 32 feet per second, and is 64 feet per second at the end of the second second. This means that the average velocity during the second second is 48 feet per second, and since the second second is one second, it fell 48 feet.
Notice that 48 feet per second is three times 16 feet per second. This is how falling works, if we ignore air resistance. In the first second, the object falls 16 feet. In the second second, it falls 3 x 16 feet. In the third second, it falls 5 x 16 feet. This means that, by using 16 feet as a unit, we only have to square the number of seconds of fall to see how far an object will fall, ignoring air resistance. In the reverse of this, we can take the square root of a distance, expressed in units of 16 feet, to tell us how many seconds it would take to fall that distance.
The moon is moving at a certain velocity as it orbits the earth. If we could just determine what that velocity was, we could tell how long it's orbital path was by multiplying it by 29 days. If we knew how long the moon's orbital path was, we could easily determine the distance to the moon by dividing it by 2 pi.
The orbit of the moon around the earth is governed by the same gravity that causes an object to drop on earth. When we drop an object, it's velocity continuously increases. At some point during the object's fall, and we can have a theoretical falling object, it will have a velocity that is identical to the average velocity of the moon in it's orbit. We only have to determine what that point is.
Gravity is a simple thing, there is nothing complicated about it. The movement of the earth in orbit and the fall of an object when dropped is governed by exactly the same gravity from the earth. This can only mean that the relationship that we are looking for must be a simple one.
The velocity of a falling object increases the further it falls. But the velocity of the moon in it's orbit decreases the further it gets from earth. Even if we did not know the Inverse Square Law, it wouldn't make sense that an object in orbit further from earth would move faster than one closer to the earth. We can see this by simply throwing a ball up in the air. It's velocity, whether on the way up or down, is greater the closer it is to the ground.
The velocity of the falling object increases according to the 32 feet, or 9.75 meters, per second, every second, as described above. But the orbital velocity of an object in orbit, such as the moon, decreases in accordance with the Inverse Square Law with increasing distance from earth. If an object is at nine times the distance from earth, it will orbit at 1/3 the velocity and will have three times the orbital energy.
This means that there must be a "crossing point" somewhere, at which the velocity of the falling object equals the velocity of the moon in it's orbit. We are trying to find that point.
The simplest and most logical relationship between these two examples of the same gravity is:
Time / Time = Distance / Distance
If the distance of one increases, the distance of the other must decrease, due to the fact that they have a "crossing point", and the same goes for the times. If the distance of the fall gets longer, the falling object will be moving faster, and if this is applied to the moon it would mean that the moon would have to be closer because the faster it moves in orbit, the closer to the earth it must be. The "crossing point" means that if one time gets longer, the other must get shorter, and the same for distance. Since time and distance are related, although differently for each, this is the only possible relationship.
The only one of the four variables that we know so far is that the moon's orbit is 29 days. Since we are expressing the velocity of the falling objects in seconds, let's convert the moon's orbit into seconds 29 days = 2,505,600 seconds.
Let's express the above equation as A,B,C, D. A and B are the two time measurements, the time of free-fall of the object and the time of the moon's orbital period. C and D are the two distance measurements.
A = time of fall on earth to reach velocity equal to the moon's orbital velocity.
B = the moon's orbital period of 2,505,600 seconds.
C = the distance that the object on earth must fall to reach the velocity equal to the moon's orbital velocity.
D = the distance of the moon's orbital path.
Now, let's start by picking a number of seconds for the object to fall on earth. How about trying 20 seconds of fall, and see where it gets us?
If an object theoretically falls for 20 seconds, it will fall 20 squared x 16 feet = 400 x 16 feet = 6400 feet. This would bring the velocity of the object to 640 feet per second (32 feet per second x 20).
So, if this might be correct, it would mean 20 / 2,505,600 = 6400 / D. With D being the distance around the moon's orbital path, in feet. This would give us 80,179,200 feet as the moon's orbital path.
But this would not work because it would give us the velocity of the moon in it's orbit as only 32 feet per second, which falls far short of the falling object's velocity of 640 feet per second after falling for 20 seconds. But we can see how we can calculate the moon's velocity in orbit if we can get those two numbers to match.
Let's try 100 seconds of free fall for the falling object. 100 squared x 16 feet = 160,000 feet. This would bring the velocity of the falling object to 3200 feet per second (32 feet per second x 100).
So, if this might be correct, it would mean 100 / 2,505,600 = 160,000 / D. With D being the distance around the moon's orbital path, in feet. This would give us 40,008,960,000 feet as the moon's orbital path.
But this would not work either because it would give us the velocity of the moon in it's orbit as 16,000 feet per second, which is far higher than 3200 feet per second of free-fall velocity. Thus, 100 seconds is too high.
Our first guess, 20 seconds of free fall, was too low and our second guess, 100 seconds of free fall, was too high. But this gives us something to go on as we see that our first guess was too low by a factor of 20 (640 / 32), and our second guess was too high by a factor of 5 (16,000 / 3200). This gives us a clue that the answer that we are looking for, the number of seconds of free fall on earth that it takes to equal the moon's orbital velocity, is much closer to 100 seconds than to 20 seconds.
In fact, the number must be four times (20 / 5) closer to 100 seconds than to 20 seconds. The difference between 20 and 100 divided by 5 gives us 16. 100 - 16 = 84. Let's try 84 seconds of free fall on earth.
An object that has been in theoretical free-fall for 84 seconds will have a velocity of 84 x 32 = 2688 feet per second. During those 84 seconds it would fall 112,896 feet (84 squared x 16 feet, as described above).
We know that the moon has the orbital period of 2,505,600 seconds. If the moon were indeed moving at 2688 feet per second, this would mean that the length of it's orbital path was 6,735,052,800 feet.
Calculating the moon's orbital velocity with 84 seconds of free-fall of the object on earth gives us 84 / 2,505,600 = 112,896 / D, with D being the distance around the moon's orbital path in feet. This gives us an orbital path of 3,367,526,400 feet.
This falls short of 6,735,052,800, but notice that 3,367,526,400 is exactly half of our target figure.
We have arrived at our answer but have to consider that the moon's orbit goes all around the earth, while our falling object is only affected by the gravity from one direction of earth. This is why the calculation gives us the length of half the total orbital path of the moon, rather than the complete orbital path.
Dividing this by pi (3.14), which is the relationship between the circumference of a circle and it's diameter, or the relationship of half the circumference to the radius that we are dealing with here, we get a distance to the moon of 1,072,460,637 feet. Dividing this by 5,280 feet in a mile, we get 203,117 miles as the distance to the moon.
This answer is not exactly correct, we know today that the average distance to the moon is about 239,000 miles. But the answer by this method here is within about 16% of it. The reason that our answer is short is that we, on the surface of the earth, are some distance from the center of gravity of the earth. The moon is affected by this center of gravity but our falling object would take less time to fall a distance if we were closer to the center of gravity.
This would decrease the value of A relative to C, in the equation above, and thus make the moon's orbital path longer and the moon further away. It is the same reason that you would weigh less at the top of a mountain then at sea level.
But I think you can see how this concept of how information flows through the universe could have been used, centuries ago, to have effectively measured the distance to the moon. Gravity is simple so the relationship that would enable us to measure the distance to the moon must be simple. The falling of an object on earth, and the revolution of the moon around the earth are both governed by the same gravity, and there is no added information from anywhere. If the scale of the distance to the moon could have been measured so long ago, it would have revolutionized the understanding of the scales of the universe, at a much earlier time.
10) THE NATURE OF ELLIPTICAL ORBITS
Here is a question that I have never seen before. Remember my principle that the way to discovery is often not to answer questions that no one else has answered, but to ask questions that no one else has asked.
When understanding orbits, such as that of planets around stars, what happens if the mass of the star changes?
Why are astronomical orbits elliptical? An elliptical orbit contains more information than a circular orbit. Where does the additional information come from?
The mass of the sun must be gradually decreasing over time. Solar flares regularly erupt from the sun, which eject matter into space. Charged particles are continually emitted from the sun, in what is known as the "Solar Wind". Most of all, the vast amount of energy radiated from the sun must be equivalent to mass, according to the well-known Mass-Energy Equivalence.
The sun operates by crunching, by way of nuclear fusion, four hydrogen atoms into one helium atom. But the resulting helium atom has less mass then the four hydrogen atoms, and the energy of the remaining mass is radiated into space, in accord with the Mass-Energy Equivalence.
The orbital distance is an equilibrium between the gravitational energy between the two, the energy in the orbit, and the orbital distance. If the mass of the star were to decrease, but the orbital energy remain constant, the gravity between the two would decrease, and this would have to be compensated for by the orbital distance increasing.
Orbital dynamics are if there is three times the orbital energy, the object will orbit at nine times the distance as previously, but will orbit at only one-third the velocity. 9 / 3 = 3. An elliptical orbit, unlike a circular one, has a point in the orbit where it is closer to the star or planet that it is orbiting, the perigee, and a point in the orbit where it is furthest away, the apogee. But when the object is closer, it moves faster so that a line from the center of the object in orbit to the center of the object being orbited moves across equal areas of space in equal periods of time.
My explanation of the extremely elliptical orbits of comets is that the large star which preceded the sun exploded first in a nova, the outer layers being blasted away in an effort to attain stability, and then as a supernova, exploding from the center. Much of the mass of that star came back together, by gravity, to form the present sun and planets.
The initial blasting away of the outer layers of the star, in the nova, would have involved the lighter atoms in the outer reaches of the star, such as the hydrogen and oxygen which forms water. Some of the energy of the nova explosion would go into binding these atoms together into molecules, such as water. Since these outer atoms had a higher "starting point" than the mass that would later come back together by being thrown outward by the supernova, we can expect that the comets which are formed mostly by ices of these light molecules, particularly water, should today have more distant orbits around the sun than the planets, and that is just what we see.
But this means that the comets would have started out in orbit around a star that was much larger than the sun is now, before it exploded in the supernova and only some of the mass fell back together to form the sun. The comets were suddenly in orbit around a much less massive star, which decreased the orbital energy, and the orbits of the comets had to compensate by drastically "shrinking" the space within the orbit, which represents the orbital energy, and that is why comets are in such elliptical orbits today.
It requires less information to just have the orbit of the comet adjust at one point, than to have the orbital distance decrease all around. This is why an elliptical orbit has an apogee and a perigee. The momentum of the new orbit must remain. The new orbit can be described as a vector between the original mass at the orbit center, and the new and lesser mass there. The difference between the area of the elliptical orbit and the larger area of a circle around the center, with the apogee of the orbit as a point on it, represents the difference in the mass of the sun now, in comparison with the mass of the central star when the orbit first formed.
The fact that astronomical orbits are elliptical requires some special explanation because the information for such orbits ultimately comes from the information in electron orbitals, and there is no sign that those are elliptical at all. The eccentricity of the orbits of the planets around the sun cannot be due to any external factors because the points of apogee and perigee of the planets do not match up at all.
The distance of the moon in orbit around the earth is increasing, but it is not due to loss of mass in the earth. It is due to the moon raising a tidal bulge in the earth's oceans. The earth rotates faster then the moon revolves around it, and the gravitational connection exchanges energy between the two. The rotation of the earth is slowing, while the moon's orbital energy is increasing. Since the earth's mass is more or less constant, this means that the moon is moving further away from the earth, at the expense of the earth's rotation slowing.
11) THE RINGS OF SATURN
The principle that I noticed, and described in the posting "The Effective Center Of Gravity", on the physics and astronomy blog www.markmeekphysics.blogspot.com , solves the mystery of why there are gaps between the rings around Saturn.
"The Effective Center Of Gravity" is about the different between the center of mass and the center of gravity. Suppose that we are in a spacecraft in orbit around a planet. We are actually in orbit around the planet's center of gravity. It may seem logical that the center of gravity is the same thing as the center of mass, at the very center of the planet.
But there is a difference. The center of mass is constant, while the center of gravity is a matter of perspective.
For a given object, such as the planet, the force of gravity diminishes with distance, according to the Inverse Square Law. When our spacecraft is at a finite distance from the planet, the nearer side of the planet must have a greater gravitational effect on the spacecraft than the far side. This means that the center of gravity of the planet, from the perspective of the spacecraft, is closer to the spacecraft than the center of mass.
The closer the spacecraft is to the planet, the greater is the proportional difference in the distance to the near side of the planet in relation to the distance to it's far side. This means that when the spacecraft is closer to the planet, the center of gravity from it's perspective is even further from the center of mass of the planet. In fact, it is only when the spacecraft is at an infinite distance from the planet that the two are identical.
I find that this concept of "The Effective Center Of Gravity" provides a simple explanation of what we see in the rings around Saturn. The rings are composed mostly of pieces of water ice. The rotation of Saturn aligns the particles into a finely-defined plane.
https://en.wikipedia.org/wiki/Rings_of_Saturn#/media/File:Saturn_Ring_Material.jpg
The reason that Saturn has it's famous and spectacular ring system is the planet's low density. The low density of the planet means that the center of gravity can be even further from the center of mass.
Saturn is the least dense of all the planets. It is actually less dense than water. All of the large planets, Jupiter, Saturn, Uranus and, Neptune, have ring systems. But all except Saturn's are faint, as seen from earth. The dense inner planets of the Solar System, including the earth, do not have ring systems.
Debris in orbit around planets can combine together to form moons. But my concept shows that the closer the two centers of gravity around which two pieces of debris are in orbit, the more likely they are to be able to join together by their mutual gravity and continue in orbit around the planet.
Saturn's low density means that objects can be in stable orbits close to the planet, but this also means that pieces of debris that are in orbits that are not at exactly the same altitude have the centers of gravity around which they orbit as further apart. When a planet is of low density, the difference between the centers of gravity of objects in two orbits is greater, relative to the scale of the orbits. The great equatorial bulge on Saturn, caused by it's rapid rotation, brings about the possibility of even more distance between the centers of gravity of objects in orbit.
The result is that in less-dense planets, the zone in which objects in orbit around the planet will find it more difficult to merge by their mutual gravity is broader. This is the zone in which planetary rings can form, high enough to orbit but with too different centers of gravity within the planet to merge into moons.
Put another way, we could describe tidal forces as what keeps moons from forming close to the planet. Just as, the closer we get to the planet, the greater the difference between the center of mass and the real center of gravity becomes, the closer a would-be moon is to the planet, the greater the difference between the gravity from the planet on one side of the moon, relative to the other side of the moon. This difference in gravity is referred to as tidal force, and it would pull the moon apart if it could form. This is another way of expressing what I am pointing out by way of the effective centers of gravity.
As we move higher above a planet, the real center of gravity of objects in orbit gets closer to the actual center of mass of the planet, and thus the real centers of gravity of objects in different orbits gets closer together. We reach an altitude above which objects that are close enough for there to be significant mutual gravity between them can merge together into moons. This is known on Saturn as the Roche Limit.
Look at the following diagram of the interior of Saturn, with the rings outside. The rings are labeled A-F, not with regard for altitude from the planet, but in the order that they were discovered.
https://en.wikipedia.org/wiki/Saturn#/media/File:Saturn_diagram.svg
Saturn has a rocky core, above which are ices, then metallic hydrogen and helium, then liquid hydrogen, and then the atmosphere. The center of the planet is the most dense, and each layer above would be of lesser density then the one below.
We can see why the rings of Saturn exist, but why would there be the gaps that there are between the rings? The ring system of Saturn consists of several sharply-defined rings with wide gaps between. The widest, and best-known, gap is the Cassini Division between the two most prominent rings, A and B.
There have been various attempts to explain the rings, such as small moons forming and "sweeping up" the debris within the gap, and "resonances" with the gravity of Saturn's moon, but the moons are distant from the planet, relative to the rings.
https://en.wikipedia.org/wiki/Moons_of_Saturn#/media/File:Saturn%27s_Rings_PIA03550.jpg
My explanation for the gaps in Saturn's rings, using this concept of the effective center of gravity being different from the center of mass, is simply that if an object orbits a real center of gravity, in a higher-density area, it's orbital energy will be higher. Since greater orbital energy means a higher orbit, this means that a sudden change in density, as we move from one layer to another inside the planet, would have to be matched by a corresponding gap in orbiting debris due to the sudden increase in orbital energies that corresponded to the real center of gravity within the planet from that orbit suddenly moving to a zone of higher density.
It is easy to see, in the cross-section diagram of the planet, how the changes from one zone to another going into the planet is matched by a gap in the rings in a mirror image of moving away from the planet. There is no better way to explain the sharply-defined gaps between the rings of Saturn.
