Q.State the formula for acceleration due to gravity at depth 'd' and at altitude 'h'. Hence show that their ratio is equal to $\dfrac{g_d}{g_h}=\dfrac{R-d}{R-2h}$, by assuming that the altitude is very small as compared to the radius of the Earth.

[!TLDR] $g_d=g(1-d/R)$ (depth) and $g_h\approx g(1-2h/R)$ (small altitude); their ratio, for small h, simplifies to $g_d/g_h=(R-d)/(R-2h)$. [!ANSWER] $g_d=g(1-d/R) …

The acceleration due to gravity at depth d below the Earth's surface is (exactly, for a uniform-density Earth) $g_d=g\left(1-\dfrac{d}{R}\right)=g\cdot\dfrac{R-d}{R}.$ The acceleration due to gravity at altitude h above the surface, for h SMALL compared to R (the condition given in the question), uses the binomial-approximated form $g_h\approx g\left(1-\dfrac{2h}{R}\right)=g\cdot\dfrac{R-2h}{R}.$ Dividing the two expressions, the common factor g/R cancels exactly: $\dfrac{g_d}{g_h}=\dfrac{g(R-d)/R}{g(R-2h)/R}=\dfrac{R-d}{R-2h}.$ This shows that, under the stated small-altitude assumption, the ratio of gravity at a given depth to gravity at a given (small) altitude reduces to a simple ratio of two length differences, $(R-d)$ and $(R-2h)$ -- a direct consequence of the depth relation being exactly linear …

State both the exact depth formula and the small-altitude approximation for g, then divide them, cancelli …

Using the EXACT (inverse-square) altitude formula instead of the small-h LINEAR approximation the question specifically asks to assum …