At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3 R above earth surface. Where R is Radius of earth (Take R=6400 km ). The depth d is equal to

Solution

Physics sir · 3 steps

At height 3R use the exact inverse-square formula, then match that value with the depth formula.

1. Gravity at height. Here h = 3R is not small compared with R, so the approximate formula g(1 - 2h/R) does not work. Use the exact one: gₕ = g/(1 + h/R)² = g/(1+3)² = g/16.
2. Gravity at depth. Below the surface, g_d = g(1 - d/R). This value must be four times gₕ, so g_d = 4 × g/16 = g/4.

   g(1 - d/R) = g/4 ⇒ d/R = 3/4

3. So. d = 3/4 × 6400 km = 4800 km, which is option (d).

The idea to keep. Above the surface, g falls as 1/r². Below the surface, it falls linearly with depth. For large heights, always use g/(1+h/R)², never the small-h approximation.

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