Escape Velocity


Hey there. I was wondering- if one were to try to determine the escape velocity of a body off the center of the earth, where the air resistance is proportional to v2 / (X+R)2, R=rad of earth, X = height off ground, would the escape velocity be (mgR2).5 ?? Thanks-


Escape velocity is a concept that depends on the conservation of energy. It is that velocity where the kinetic energy equals the negative gravitational potential energy, giving the body a total mechanical energy of zero. Then as that kinetic energy is traded for potential as distance from the planet increases the object slows down but does not stop short of infinite height.

Conservation of energy only applies in the absence of air resistance so the concept of escape velocity only applies once the atmosphere has been left behind. The escape velocity remains (2GM/(R+H)).5 where G is the universal gravitation constant 6.672e-11, M is the mass of the Earth 5.98e24kg and R is the radius of the Earth 6.37e6m, and H is the height above the surface in meters. In practice we finesse the air resistance by taking H to be 3.6e3m or so.

Imagine a rocket that accelerated so quickly that it achieved escape velocity while still in significant air. It would be destroyed by the forces and temperatures involved.