An object is taken to a height above the surface of earth at a distance 5/4 R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :

Solution

Physics sir · 3 steps

Weight follows inverse square of distance from the centre, so square the ratio 4/5 and see what is lost.

1. Inverse square law. Outside the earth, g at distance r from the centre is g' = GM/r². Comparing with the surface value g = GM/R² gives the ratio below.

   g'/g = (R/r)² = (R/(5/4R))² = (4/5)² = 16/25

2. Weight left. Mass does not change, so weight scales the same way as g. The object keeps 16/25 = 0.64, that is 64% of its surface weight.
3. So. The decrease is 100% - 64% = 36%, which is option (a). The value R = 6400 km is not needed because only the ratio r/R matters.

The idea to keep. Above the surface, use g' = g(R/r)² exactly. The shortcut (Δ g)/g = 2h/R works only when h is much smaller than R.

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