The time period of a satellite, revolving above earth's surface at a height equal to $R$ will be (Given $\mathrm{g}=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=$ radius of earth)

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Gravitation

The time period of a satellite, revolving above earth's surface at a height equal to

R

will be (Given

g = π^{2} m / s^{2}, R =

radius of earth)

(A) $4 R$
(B) $8 R$
(C) $32 R$ ✓ Correct
(D) $2 R$

Answer: (C)

Solution

At height

h = R

, orbital radius

r = R + h = 2 R

. For circular orbit,

\frac{m v ^{2}}{r} = \frac{GM m}{r ^{2}}

, giving

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

. Time period

T = \frac{2 π r}{v} = \frac{2 π ( 2 R )}{g R /2} = \frac{4 π R 2}{g R} = 4 π \frac{2 R}{g}

. With

g = π^{2}

T = 4 π \frac{2 R}{π ^{2}} = 4 2 R = 32 R

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option

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