JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Gravitation

Correct formula for height of a satellite from earths surface is :

(A) $(\frac{T ^{2} R ^{2} g}{4 π})^{1/2} - R$
(B) $(\frac{T ^{2} R ^{2} g}{4 π ^{2}})^{1/3} - R$ ✓ Correct
(C) $(\frac{T ^{2} R ^{2}}{4 π ^{2} g})^{1/3} - R$
(D) $(\frac{T ^{2} R ^{2}}{4 π ^{2}})^{- 1/3} + R$

Answer: (B)

Solution

For a satellite of period

T

at height

h

above a planet of radius

R

, orbital speed and gravity give (using

GM = g R^{2}

T = \frac{2 π ( R + h ) ^{3/2}}{g R ^{2}} ⟹ T^{2} = \frac{4 π ^{2} ( R + h ) ^{3}}{g R ^{2}}

Solving for

h

h = (\frac{T ^{2} R ^{2} g}{4 π ^{2}})^{1/3} - R

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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