Question

• Derive an approximate formula for the speed of a satellite in a circular orbit at an altitude of 500 km from the surface of the Earth. How long does it take for the satellite to complete one orbit?

Answer 1

According to Newton's law of gravitation, satellite moves around the earth due to centripetal force by earth on satellite.

Let, speed of satellite is $v$ ,mass is $m$ and height from surface of the earth is $h=500km=5\times10^{5}m$

Thus,

$$G\frac{M_em}{(R_e+h)^2}=\frac{mv^2}{R_e+h}\\ \implies \frac{GM_e}{R_e+h}=v^2$$

But, Radius of earth $R_e$ is very larger than $h$ , so we can apply the binomial approximation and thus

$$v^2=\frac{GM_e}{R_e+h}=\frac{GM_e}{R_e(1+\frac{h}{R_e})}=\frac{GM_e}{R_e}(1+\frac{h}{R_e})^{-1}\\ \implies v^2\approx \frac{GM_e}{R_e}(1-\frac{h}{R_e})\approx g(R_e-h)\\ \implies v=\sqrt{g(R_e-h)}$$

Thus, on plugin the value we get

$$v=\sqrt{9.8(6400000-500000)}\approx7604m/s$$

Since, for one complete revolution satellite covers

$$d=2\pi R_e$$

Thus, time takes to complete one revolution is

$$T=\frac{d}{v}=\frac{2\pi R_e}{v}$$

On plug int he data we get,

$$T=5288.32s$$