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Question

# A satellite is to be placed in equatorial geostationary orbit around earth for communication. The height of such a satellite is [ME=6×1024kg,RE=6400km,T=24h,G=6.67×10−11Nm2kg−2]

A
3.57×105m
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B
3.57×106m
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C
3.57×107m
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D
3.57×108m
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Solution

## The correct option is B 3.57×107mTime period of satellite, T=2π(RE+h)√GME(RE+h)=2π(RE+h)3/2√GMESquaring both sides, we get T2=4π(RE+h)3GME(RE+h)3=GMET24π2(RE+h)=(GMET24π2)1/3 or h=(GMET24π2)1/3−REHere, ME=6×1024kgRE=6400km=6400×103m=6.4×106mT=24h=24×60×60s=86400sG=6.67×10−11Nm2kg−2On substituting the given values, we geth=(6.67×10−11×6×1024×(86400)24×(3.14)2)−6.4×106=4.21×107−6.4×106=3.57×107m

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