Escape Velocity

## Question:

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 v^{2} /
(X+R)^{2}, R=rad of earth, X = height off ground, would
the escape velocity be (mgR^{2})^{.5} ?? Thanks-
## Answer:

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.