Question

A satellite is located 2.75 * 10 ^ 5 m above the surface of the Earth. What is the period of the satellite's orbit?

Answer 1

We can find the period of the satellite's orbit from the Newton's form of Kepler's third law:

$$\dfrac{T^2}{(R_E+h)^3}=\dfrac{4\pi^2}{GM_E},$$ $$T=\sqrt{\dfrac{4\pi^2(R_E+h)^3}{GM_E}},$$ $$T=\sqrt{\dfrac{4\pi^2\cdot(2.75\cdot10^5\ m+6.37\cdot10^6\ m)^3}{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.98\cdot10^{24}\ kg}}=5389\ s=1.49\ h.$$