JEE Main · 2004 · 24 Apr 2004 · Offline · Gravitation

A satellite of mass

m

revolves around the earth of radius

R

at a height

x

from its surface. If

g

is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

(A)gx
(B) $\frac{g R}{R - x}$
(C) $\frac{g R ^{2}}{R + x}$
(D) $(\frac{gR ^{2}}{R + x})^{1/2}$ ✓ Correct

Answer: (D)

Solution

For a satellite in a circular orbit gravity supplies the centripetal force,

\frac{G M _{e} m}{( R + x ) ^{2}} = \frac{m v _{0}^{2}}{R + x}

, so

v_{0}^{2} = \frac{G M _{e}}{R + x}

. Using

g = \frac{G M _{e}}{R ^{2}}

, that is

G M_{e} = g R^{2}

, and the orbital speed becomes

v_{0} = \frac{g R ^{2}}{R + x}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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