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Q.

A satellite is moving in circular orbit around the earth with orbital speed v. It radius of earth is R and acceleration due to gravity on its surface is g, then height of the satellite above the surface of earth is

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a

2gR2V2-R2

b

gR2V2-R

c

2gR2V2-R

d

gR2V2

answer is B.

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Detailed Solution

(2)Orbital  speed   V= GMr       Now  g= GMR2  ⇒   GM=gR2Also,   r=R+h∴    V= gR2R+h  ⇒   h=  gR2V2−R
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