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?

Given:

Height of the satellite, *h* = 35780 km = 35.78×10^{6 }m

Let the original mass of Earth be *M*. Then its new mass M’ will be 4*M*.

∴ M’= 4×6×10^{24}kg

Radius of the Earth (*R*)= 6.4x10^{6 }m

G= Gravitational constant 6.67x10^{-11 }N-m^{2}/kg^{2 }(Ref-Text Book data)

Time taken by the satellite to revolved around the Earth's orbit is given as

T=2π(R+h)/v_{c}

Now, *v*_{c }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**