An artificial satellite is revolving around the earth in a circular orbit. Its velocity is one third of the escape velocity. Its height from the earth’s surface is (in km)
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a
22400
b
12800
c
3200
d
1600
answer is A.
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Detailed Solution
v=ve3 GMR+h=132GMR Squaring on both sides we get ⇒GMR+h=192GMR ⇒9R=2R+2h ⇒7R=2h ⇒h=7R2=7×6400km2=22400km