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How much time a satellite in an orbit at height 35780 km above earth’s surface would take, if the mass of the earth would have been four times its original mass?
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Given:
Height of the satellite, h = 35780 km = 35.78×106 m
Let the original mass of Earth be M. Then its new mass M’ will be 4M.

∴ M’= 4×6×1024kg

Radius of the Earth (R)= 6.4x106 m

G= Gravitational constant 6.67x10-11 N-m2/kg2 (Ref-Text Book data)

Time taken by the satellite to revolved around the Earth's orbit is given as
T=2π(R+h)/vc
Now, vc is given as

\(v_c = \sqrt{\frac{GM'}{R+h}}\)

\(T = \frac{2π(R+h)}{\sqrt{\frac{GM'}{R+h}}}\)

\(T = \frac{2π(R+h)^{3/2}}{\sqrt{GM'}}\)

\(T = \frac{2π(6.4×10^6+35.78× 10^6 )^{3/2}}{\sqrt{6.67×10^{-11}×4× 6×10^{24}}}s\)

\(T = 4.303×10^4 s Approx\)

\(T = 4.303×10^4 /3600 h Approx\)

\(T = 11.95 h Approx\)

T= 11 hours 57 minutes approx

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