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Q.

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, than the orbital speed of the satellite is

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a

$\frac{{gR}^{2}}{R + x}$

b

$\frac{gR}{R - x}$

c

${(\frac{{gR}^{2}}{R + x})}^{1 / 2}$

d

$gx$

answer is B.

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Detailed Solution

$V_{0} = \sqrt{\frac{GM}{R + x}}; But g = \frac{Gm}{R^{2}} ∴ V_{0} = \sqrt{\frac{{gR}^{2}}{R + x}}$

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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, than the orbital speed of the satellite is