Worked example 12.3: Circular Earth orbit

Question: A satellite moves in a circular orbit around the Earth with speed $v=6000 {\rm m/s}$. Determine the satellite's altitude above the Earth's surface. Determine the period of the satellite's orbit. The Earth's mass and radius are $M_\oplus =5.97\times 10^{24} {\rm kg}$ and $R_\oplus = 6.378\times 10^6 {\rm m}$, respectively.

Answer: The acceleration of the satellite towards the centre of the Earth is $v^2/r$, where $r$ is its orbital radius. This acceleration must be provided by the acceleration $G M_\oplus/r^2$ due to the Earth's gravitational attraction. Hence,

$$\frac{v^2}{r} = \frac{G M_\oplus}{r^2}.$$

The above expression can be rearranged to give

$$r = \frac{G M_\oplus}{v^2} = \frac{(6.673\times 10^{-11})\times(5.97\times 10^{24})}{(6000)^2} =1.107\times 10^7 {\rm m}.$$

Thus, the satellite's altitude above the Earth's surface is

$$h = r - R_\oplus = 1.107\times 10^7 - 6.378\times 10^6 = 4.69\times 10^6 {\rm m}.$$

The satellite's orbital period is simply

$$T = \frac{2 \pi r}{v} = \frac{2\times\pi\times (1.107\times 10^7)}{6000} = 3.22 {\rm hours}.$$