Answer to Question #254179 in Physics for sabby

Question

What is the speed of a satellite moving in a circular orbit at a height of 3800km above

the surface of the earth? (b) what is the period of the satellite in hours? The mass of the

earth is 5.97x10^24 kg. The radius of the earth is 6.38x10^6 m.

Answer 1

Given:

$G=6.67*10^{-11}\:\rm N\cdot m^2/kg^2$

$R_E=6.38*10^{6}\:\rm m$

$h=3.80*10^{6}\:\rm m$

$M_E=5.97*10^{24}\:\rm kg$

(a) the speed of a satellite moving in a circular orbit 

$$v=\sqrt{\frac{GM_E}{R_E+h}}$$

$$v=\sqrt{\frac{6.67*10^{-11}*5.97*10^{24}}{6.38*10^{6}+3.80*10^{6}}}=6250\:\rm m/s$$

(b) what is the period of the satellite in hours

$$T=\frac{2\pi(R_E+h)}{v}$$

$$T=\frac{2\pi(6.38*10^{6}+3.80*10^{6})}{6250}=10200\:\rm s=2.84\: hr$$