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Name:  Proctor G.
Status: other
Age: 30s
Location: N/A
Country: N/A
Date: 2000-2001


Question:
The following question and answer were found by a web search for escape velocity. Steve L. has asked the same question which I am interested in, and have found various ambiguous answers to.

Is escape velocity the velocity necessary at any given altitude (distance from the earth's (or other body's) centre) to propel it out of the earth's gravitational field WITHOUT the use of rockets or other machinery? In other words, if there was an explosion on the earth large enough to propel an object at a velocity of 25 000 mph straight outward from the earth, that object would escape the earth's gravitational field. I am assuming that this is true, in which case we come back to Steve L.'s question. IF that object has instead some form of propulsion other than the large explosion, why must it reach 25 000 mph to escape the earth's gravitational field. If it had enough fuel why could it not propel itself at a constant 5 mph for instance, until it was far enough from the earth so as not to be affected by its gravity? And if this is true, then why does a rocket have to reach an "escape velocity" of 25 000 mph when by the time it has attained that speed it would be considerably further from the earth's centre than where it started (at the surface)? Isn't the escape velocity less the further you are from the centre of mass? Therefore isn't it possible that by the point at which the rocket reached a velocity of 15 000 mph it would be distant enough from the earth to have attained escape velocity FROM THAT POINT?


Replies:
You are absolutely correct in your analysis. If you are interested in the mathematics, the equation is:

E = -GMm/r + mvv/2

where E is the total energy, G is Newton's gravitational constant (6.67 x 10E-11 Nmm/(kg kg), M is the mass of the earth (6 x 10E+24 kg), m is the mass of the satellite, r is the distance of the satellite from the center of the earth, and v is the speed of the satellite. The first term is the gravitational potential energy and the second is the kinetic energy of the satellite.

When E=0, the satellite has reached escape velocity. Then vv = 2GM/r. Notice that when r is infinite, v=0 is the escape velocity. This makes sense, since the satellite has escaped at that point.

On the other hand when r is the radius of the earth (6.5x10E+6 m), the escape velocity is 11,000 m/s or about 25,000 mi/hr. Once the rocket achieved that velocity (and was outside the atmosphere so there was no drag), it could shut down its engine and it would continue to infinity (and arrive there at zero velocity).

You could, however, keep the rocket engine running so the rocket speed was always 5 mi/hr and still get to infinity. It would, however, take a long time and an enormous amount of fuel. If you slowed down just 5 mi/hr so the rocket was at rest, you would burn ALL your fuel without moving a bit. Going 5 mi/hr is a little better, but not much!

Best, Dick Plano

Note: 10E+24 is 1 followed by 24 zeros and vv is v squared (v times v).


What you are suggesting is true. The further from the earth's center the lower the escape velocity. Thus, if you moved upward at a constant velocity of 5 mph you would eventually escape the earth's gravity. Initially it would require a large rate of fuel consumption but by the time you escaped earth's gravity you would not be using any fuel at all.

The physics and economics of the situation are such that it is easier and cheaper to build a rocket that accelerates rapidly to the escape velocity early in the flight than it would be to build a rocket that accelerates to a lower velocity and holds that speed for a LONG time to escape earth orbit. Control issues are more critical for low velocities as well.

Greg Bradburn



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