An object is allowed to fall from a height R above the earth, where R is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be

Solution

Physics sir · 4 steps

Over a drop as big as R, g does not stay constant, so use energy conservation with -GMm/r.

1. Why not kinematics. The object starts at distance 2R from the earth's centre and ends at R. Over that range g changes by a factor of 4, so v² = 2gh does not apply.
2. Energy conservation. The object starts from rest. The kinetic energy it gains equals the potential energy it loses. Potential energy at distance r from the centre is -GMm/r.

   1/2mv² = (-GMm/2R) - (-GMm/R) = GMm/2R

3. Bring in g. At the surface, g = GM/R², so GM = gR². Then 1/2v² = gR²/2R = gR/2, which gives v² = gR.
4. So. The object strikes the surface with v = √gR, which is option (b).

The idea to keep. When the height is comparable to R, use U = -GMm/r and conserve energy. Then replace GM with gR² to write the answer in terms of g.

Answer from the NTA final answer key. Solution by PrepWiser.