JEE Main 2023 · 8 April · evening · Single correct · +4 / −1
The acceleration due to gravity at height h above the earth if h << R (Radius of earth) is given by
- (a) g′=g(1-(2 h)/R)
- (b) g′=g(1-(2 h²)/R²)
- (c) g′=g(1-h²/(2 R²))
- (d) g′=g(1-h/(2 R))
Answer and solution
Answer (a) g′=g(1-(2 h)/R)
Solution
Physics sir · 3 steps
Write g at height h as an inverse square, then use the binomial approximation because h is tiny compared to R.
- Inverse square law. At the surface g = GM/R². At height h the distance from the centre is R + h, so g' = GM/(R+h)².
g' = GM/R²(1+h/R)⁻² = g(1+h/R)⁻²
- Binomial approximation. For small x, (1+x)ⁿ ≈ 1 + nx. Here x = h/R and n = -2, so the bracket becomes 1 - 2h/R. Terms in h²/R² are far smaller and are dropped.
- So. g' = g(1 - 2h/R), which is option (a).
The idea to keep. Above the surface, g falls as 1/r², so for small heights the fractional drop is 2h/R: the power 2 becomes the factor 2. Inside the earth at depth d it is 1 - d/R, with no 2.
Answer from the NTA final answer key. Solution by PrepWiser.