An object is released from rest at a distance h above the surface of a planet (mass = M,radius = R

Question

An object is released from rest at a distance h above the surface of a planet (mass = M,radius = R < h).With what speed will the object strike the surface of the planet? Disregard any dissipative effects of the atmosphere of the planet.
A) $$\left[ \frac { 2 G M h } { R ( R + h ) } \right] ^ { 1 / 2 }$$
B) $$\left[ \frac { 2 G M } { R } \right] ^ { 1 / 2 }$$
C) $\left[ \frac { 2 G M ( h - R ) } { R h } \right] ^ { 1 / 2 }$
D) $\left[ \frac { 2 G M } { R + h } \right] ^ { 1 / 2 }$
E) $\left[ \frac { 2 G M } { R - h } \right] ^ { 1 / 2 }$

Quizplus · Accepted Answer

The correct answer is A because the object is released from rest at a distance h above the surface of a planet, so its initial velocity is zero. The only force acting on the object is the gravitational force of the planet, which is conservative. Therefore, the total mechanical energy of the object is conserved. The initial mechanical energy of the object is zero, and its final mechanical energy is equal to its kinetic energy just before it strikes the surface of the planet. Therefore, we can use the conservation of mechanical energy to find the speed of the object just before it strikes the surface of the planet.

An Object Is Released from Rest at a Distance H [2GMhR(R+h)]1/2\left[ \frac { 2 G M h } { R ( R + h ) } \right] ^ { 1 / 2 }[R(R+h)2GMh​]1/2

An Object Is Released from Rest at a Distance H $\left[ \frac { 2 G M h } { R ( R + h ) } \right] ^ { 1 / 2 }$