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Question

# A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes 5.26×103s to complete one revolution with a centripetal acceleration equal to 9.32 m/s2. The height of satellite orbiting above the earth is(Earth's radius =6.37×106m)

A
220 km
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B
160 km
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C
70 km
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D
120 km
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Solution

## The correct option is C 160 kmGiven : T=5.26×103s, a=9.32m/s2Centripetal acceleration a=v2r=9.32or v2=9.32ror v=√9.32(Re+h)Time period of revolution is given by T=2π(Re+h)v=2π(Re+h)√9.32(Re+h)or T=2π√Re+h√9.32∴ 5.26×103=2×3.14√Re+h3.05√Re+h=2.55×103Re+h=6.53×106h=6.53×106−6.37×106 (∵Re=6.37×106 m)h=0.16×106mh=160×103m=160km

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