If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth α h(h<<Rₑ) from the earth surface. The value of α will be _________.
(use Rₑ=6400 km)

Solution

Physics sir · 4 steps

Going up costs g twice as fast as going down, so the depth must be twice the height.

1. Gravity at height. For h ≪ Rₑ, the binomial approximation gives gₕ = g(1 - 2h/Rₑ). The factor 2 comes from the 1/r² fall-off outside the earth.
2. Gravity at depth. At depth d, only the mass inside radius Rₑ - d pulls you, so g_d = g(1 - d/Rₑ) exactly, with no factor 2.

   g(1 - 2h/Rₑ) = g(1 - (α h)/Rₑ)

3. Compare. Cancel g and the 1 on both sides. That leaves 2h = α h. The value of Rₑ cancels too, so 6400 km is not needed.
4. So. α = 2.

The common slip: 1. Using g(1 - h/Rₑ) for height, which drops the factor 2, gives 1.

The idea to keep. For small distances, g drops as 2h/R going up but only as d/R going down. Equal g means the depth is twice the height.

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