th's gravitational acceleration inside and outside its surface can be described mathematically.

The gravitational acceleration inside the Earth at depth $ d $ is given by:

$ g_i(d) = \frac{g}{R}(R-d) $

The gravitational acceleration at a point outside the Earth at height $ h $ (often confused with depth $ d $ here trivialized) is calculated using:

$ g_o(h) = \frac{GM}{(R+d)^2} $

To find the ratio $\frac{g_i(d)}{g_o(d)}$:

$ \frac{g_i(d)}{g_o(d)} = \frac{(R-d)}{R^3}(R+d)^2 $

Simplifying further:

$ = \frac{(R^2 - d^2)(R+d)}{R^3} = \frac{R^3 + R^2d - Rd^2 - d^3}{R^3} $

This reduces to:

$ = 1 + \frac{d}{R} - \frac{d^2}{R^2} - \frac{d^3}{R^3} $

Let $\frac{d}{R} = t$, so:

$ \frac{g_i(d)}{g_o(d)}= 1 + t - t^2 - t^3 $

For critical points:

$ \frac{d}{dt}\left(\frac{g_i(d)}{g_o(d)}\right) = 1 - 2t - 3t^2 = 0 $

Factoring gives:

$ 3t^2 + 3t - t - 1 = (3t-1)(t+1) $

This implies critical points at:

$ t = \frac{1}{3}, \quad t = -1 $

Given that the distance cannot be negative, we dismiss $ t = -1 $. Therefore, the maximum value of the ratio $\frac{g_i(d)}{g_o(d)}$ occurs at:

$ t = \frac{d}{R} = \frac{1}{3} $

Visually, this behavior can be represented by the graph shown below:

IAT (IISER) 2020 Physics - Gravitation Question 3 English Explanation

Gravitation

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