The acceleration due to gravity at height h above the earth if $\mathrm{h} \ll \mathrm{R}$ (radius of earth) is given by

(A) $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$

(B) $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}^{2}}{\mathrm{R}^{2}}\right)$

(C) $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{h}}{2 \mathrm{R}}\right)$

(D) $g^{\prime}=g\left(1-\frac{h^{2}}{2 R^{2}}\right)$

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Gravitation

The acceleration due to gravity at height h above the earth if

h ≪ R

(radius of earth) is given by

(A) $g^{'} = g (1 - \frac{2 h}{R})$ ✓ Correct
(B) $g^{'} = g (1 - \frac{2 h ^{2}}{R ^{2}})$
(C) $g^{'} = g (1 - \frac{h}{2 R})$
(D) $g^{'} = g (1 - \frac{h ^{2}}{2 R ^{2}})$

Answer: (A)

Solution

At height h above the surface,

g (h) = \frac{GM}{( R + h ) ^{2}} = \frac{GM}{R ^{2}} (1 + \frac{h}{R})^{- 2}

. For

h ≪ R

, using binomial approximation:

(1 + \frac{h}{R})^{- 2} \approx 1 - \frac{2 h}{R}

. Therefore

g (h) = g (1 - \frac{2 h}{R})

, where

g = \frac{GM}{R ^{2}}

is the surface gravity.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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