Question

Find minimum velocity of escape if the body is carried by a rocket and it projected from the rocket at a hight of 500 miles from the earth's surface.

Answer 1

Earth's radius = R miles, Mass of earth = M kg, mass body = m kg

Energy of the body at height 500mi = 800 km when projected: $P.E. + K.E. = - \frac{GMm}{(R+800)} + \frac{1}{2} m v1^2$

Suppose it goes away from Earth and revolves in an Orbit around the Earth at radius X > R, then

Energy of the body at height X : $P.E. + K.E. = - \frac{GMm}{(R+X)} + \frac{1}{2} m v2^2.$ Let us say the body has very near zero velocity as it nears infinite distance away from Earth.  If the body nears infinite distance, then its energy will be  zero as R+X ⇒ Infinity.

So, Energy of the body at height 500mi to go to infinity $= 0 = - \fracGMm}{R+800} + \frac{1}{2} m v1^2 \\ v1 = \sqrt{\frac {2GM}{R+800}} \\ \\ = \sqrt{\frac{2*6.67*10^{-11}*6*10^{24}}{7170*10^3}} = 10.565 km/sec$