Answer to Question #119543 in Physics for rose

Question

international space station makes 15.65 revolutions per day in its orbit around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

Answer 1

The radius of ofbit is given by the expresson (see https://en.wikipedia.org/wiki/Geostationary_orbit#Derivation_of_geostationary_altitude):

$$R = \sqrt[3]{\dfrac{\mu T^2}{4\pi^2}}$$

where $\mu = 398.6\cdot 10^{3} km^3/s^2$ is the the geocentric gravitational constant and $T = (15.65 rpd)^{-1} = (1.81\cdot 10^{-4} rps)^{-1} \approx5521s$ is the period of revolution. Substituting values, obtain:

$$R = \sqrt[3]{\dfrac{398.6\cdot 10^{3}\cdot 5521^2}{4\pi^2}} \approx 6752km$$

Subtructing the Earth's equatorial radius, 6,378 kilometres, obtain the height of this satellite above the surface of the Earth:

$$h = 6752-6378 = 374 km$$

Answer. h = 374 km.