I want to emphasize on this blog how much there still is to be discovered, and that anyone can discover. If you think you have noticed something new you can easily look online to see if anyone else has ever noticed it.
There is a lot to be discovered just by messing around with numbers. I noticed another way to express the escape velocity of the earth or another planet.
The escape velocity is the speed that an object would have to be moving to escape the earth's gravity, starting from the earth's surface. The escape velocity does not apply to rockets, because the rocket has an engine, it only applies to ballistic objects such as artillery. If an object is moving at less than the escape velocity it will reach a certain height and then fall back down. The concept of escape velocity is somewhat theoretical because, as far as is known, no ballistic object has ever escaped from the earth's gravity.
Aside from escape velocity there is also orbital velocity, the velocity required to go into orbit around the earth or other planet starting from the surface. As we might expect a lower velocity is required to go into orbit, compared to escaping the planet's gravity altogether. An object can first be placed in orbit, and then given an additional burst of energy to escape the planet's gravity altogether.
The formula for the escape velocity of a planet is: EV = square root of ( 2GM / r ).
Where EV is escape velocity, G is the universal gravitational constant, M is the mass of the planet and, r is the radius or distance to the center of the planet.
I noticed another way of calculating and expressing escape velocity, using the surface gravity acceleration, rather than the mass of the planet or the universal gravitational constant.
There is a fixed acceleration due to gravity near the earth's surface, neglecting air resistance. It is 32 feet per second squared, or 9.8 meters per second squared. This means that an object dropped from a height will have a downward velocity of 32 feet per second, or 9.8 meters per second, at the end of the first second of fall. At the end of the second second of fall the velocity will be 64 feet per second, or 19.6 meters per second, and so on. We can easily calculate how far an object has fallen by halving it's terminal velocity and multiplying it by the fall time. If an object starts falling at zero velocity and it's terminal velocity is 20 meters per second, that means it has fallen 10 meters.
One day I got to wondering what altitude an object would have to fall from, based on gravitational acceleration near the surface, to reach escape velocity by the time it impacted the surface of the earth. I was amazed to find that the altitude is equal to the radius of the earth, about 4,000 miles or 6,000 km. Again this must be based on the acceleration due to gravity at the surface, it ignores air resistance and the fact that gravity gets weaker as we move further from the earth, according to the Inverse Square Law.
This means that, if we could shoot a ballistic object into the sky so that it reached an altitude equal to the earth's radius it would not fall back down, it would keep going into space.
If you want to see the numbers by which I arrived at this: The terminal velocity is 7 miles per second which, multiplied by 5,280 = 36,960 feet per second. 36,960 / 32 means that it must have fallen for 1,155 seconds to reach this velocity. If the falling object started at a velocity of zero, and steadily increased to a terminal velocity of 7 miles per second, that means it's average velocity was 3.5 miles per second. If it's average velocity was 3.5 miles per second, and it fell for 1,155 seconds, that means it fell from an altitude of 3.5 x 1,155 = 4,042.5 miles, which is just about exactly the radius of the earth.
The conventional formula for escape velocity is, again, EV = square root of ( 2GM / r ).
My method also includes the radius of the planet, r, but uses it as the altitude. The conventional formula must use the square root because gravity decreases as we move away from the earth, by the Inverse Square Law. But I think it is easier to measure and deal with the acceleration due to gravity, which can be measured in a science class experiment, than the mass of the earth and the universal gravitational constant.
Aside from the escape velocity there is also the orbital velocity, the velocity required for an object to go into orbit. As we might expect the orbital velocity is less than the escape velocity, because the object doesn't have to escape the earth's gravity altogether. Again neither velocity applies to rockets, only to ballistic objects without their own engines. While the escape velocity is 7 miles per second the orbital velocity is only 5 miles per second.
But what do you notice here, considering that gravity in space operates by the Inverse Square Law? Also considering that we might consider these velocities as having two elements, the first element is reaching orbit and the second is going on from there into space?
Considering that these are rounded figures the square of the first is just about exactly twice the square of the second. We know that the force in a moving object is related to squares, an object moving at twice the velocity has four times the force. What this means is it requires twice as much energy to move an object at escape velocity as it does at orbital velocity.
Remember that all around you, every day, there are relatively simple things that have never been pointed out.
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