Answer to Question #292323 in Mechanics | Relativity for Ashter

Question

a satellite orbits earth at an altitude 356 kilometers above the planets surface. Compute for its (a) radius (b) speed (c) orbital period. (mE=5.97x10^24 kg)

Answer 1

Radius

$R=R_e+h$

$$R=6400000+356000=6756000m$$

We know that

In circular motion

$\frac{GmM}{R}=\frac{mv^2}{R}$

$V=\sqrt{\frac{GM}{R}}$

Put (R)=R_e+h,M=M_e

Speed $V=\sqrt{\frac{Gm_e}{R_e+h}}$

$V=\sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6400\times10^3+356000}}$

$$V=\sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6756000}}=7.6772km/sec$$

V=7.677km/sec

Time period

$T=\frac{2\pi R}{V}$

$T=\frac{2\times3.14\times6756000}{7677}=5526.59sec$

$T=\frac{5526.59}{60\times60}=1.53hr$