An artificial satellite is revolving round the earth in a circular orbit. Its velocity is half the escape velocity. Its height from earth's surface is
A
6400km
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B
12800km
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C
3200km
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D
1600km
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Solution
The correct option is A6400km Orbital velocity of the artificial satellite v0=R√gR+h Escape velocity vc=√2gR Given, v0=12vc 12√2gR=R√gR+h 1√2=√RR+h or R+h=2R h=R=6400km.