A satellite is to be placed in equatorial geostationary orbit around earth for communication. The height (in meter) of such a satellite is 3.57×10pm. Find p[ME=6×1024 kg,RE=6400 km,T=24 h,G=6.67×10−11 N m2 kg−2]

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Detailed Solution

T2=4π2(RE+h)3GME⇒h=(GMET24π2)1/3−RE⇒h=(6.67×10−11×6×1024×(86400)24×(3.14)2)1/3−6.4×106=3.57×107 m

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