The weight of a body on the surface of the earth is 100 N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

Solution

Physics sir · 3 steps

Measure distance from the centre of the earth, not from the surface, and apply the inverse square law.

1. Distance from centre. Gravity depends on the distance from the earth's centre. At height h = R/4, that distance is r = R + R/4 = 5R/4.
2. Inverse square law. The force on the same mass falls as 1/r². So compare it with the surface value, where r = R.

   Fₕ/F₀ = (R/(R+h))² = (R/(5R/4))² = 16/25

3. So. Fₕ = 100 × 16/25 = 64 N, which is option (b).

The idea to keep. For any height, write gₕ = g(R/(R+h))². The shortcut g(1 - 2h/R) works only when h is much smaller than R.

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