If R E be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < R E )

Solution

Physics sir · 4 steps

Use the exact formulas for depth and height, take the ratio, then expand the brackets carefully.

1. At depth r. Only the mass inside radius R_E - r pulls on you, so g falls linearly: g_d = g(1 - r/R_E).
2. At height r. The distance from the centre becomes R_E + r, so gₕ = (g R_E²)/(R_E + r)² = g/(1 + r/R_E)². Keep this exact form. Do not use the approximation 1 - 2r/R_E, because the options carry terms up to r³.

   g_d/gₕ = (1 - x)(1 + x)², x = r/R_E

3. Expand. (1 + x)² = 1 + 2x + x². Multiplying by (1 - x) gives 1 + 2x + x² - x - 2x² - x³ = 1 + x - x² - x³.
4. So. The ratio is 1 + r/R_E - r²/R_E² - r³/R_E³, which is option (d).

The idea to keep. Depth gives g(1 - d/R) exactly, height gives g R²/(R + h)² exactly. When the options have higher powers, keep the exact forms and expand.

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