Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

Solution

Physics sir · 3 steps

Inside a uniform earth, g falls linearly with depth, so a quarter of the way down costs a quarter of g.

1. Only inner mass pulls. At depth d, the pulls from the shell of earth above the body cancel out. Only the sphere of radius R - d below it attracts the body. That mass grows as r³ while g goes as M/r², so g is proportional to r.

   g_d = g(1 - d/R)

2. Put in numbers. With d = R/4, g_d = g(1 - 1/4) = 3/4g. The mass of the body does not change, so its weight falls by the same factor.
3. So. Weight at depth = 3/4 × 300 N = 225 N, which is option (d).

The idea to keep. Going down into a uniform earth, g drops linearly: g_d = g(1 - d/R). Going up, it drops as 1/r². Do not mix the two formulas.

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