Answer to Question #219932 in Physics for Lgshwhh

Question

The ISS in the orbit 408 km away from the surface of the earth. Calculate the time taken to orbit the earth.

Answer 1

According to the third Kepler's law, the time taken to orbit the earth $T$ and the orbit radius $a$ are connected as follows (see https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law):

$$\dfrac{a^3}{T^2}\approx \dfrac{GM}{4\pi^2}$$

where $G = 6.674\times 10^{-11}m^3\cdot kg^{-1}\cdot s^{-2}$ is the gravitational constant, and $M = 5.974\times 10^{24} kg$ is the mass of the Earth.

The radius of the orbit is the elevation above the surface plus the radius of the Earth itself: $a = 408km +6371km =6.779\times 10^6m$.

Expressing $T$, obtain:

$$T = \sqrt{\dfrac{4\pi^2a^3}{GM}} = 2\pi\sqrt{\dfrac{a^3}{GM}}\\ T = 2\pi\sqrt{\dfrac{(6.779\times 10^6)^3}{6.674\times 10^{-11}\cdot 5.974\times 10^{24}}} \approx 5554s$$

Answer. 5554 s.