Question

Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. A "near-Earth" orbit is at a height above the surface of the Earth that is very small compared to the radius of the Earth. [Hint: You may take the acceleration due to gravity as essentially the same as that on the surface.]

Answer 1

Answer:

$t=5063s$

Explanation:

The gravitational force a satellite of mass m orbiting the Earth of mass $M=5.97\times10^{24}Kg$ at a radius equal to the radius of the Earth $R=6371000m$ is:

$F=\frac{GMm}{R^2}$

Where $G=6.67\times10^{-11}Nm^2/Kg^2$ is the gravitational constant.

Newton's 2nd Law tells us that this satellite of mass m under the mentioned force F will experiment an acceleration given by F=ma.

This acceleration is a centripetal acceleration, so we can use the formula:

$a=\frac{v^2}{R}$

The velocity will be given by the relation between the circumference of the orbit and the time taken, that is, v=C/t, and we know that the circumference is $C=2\pi R$.

We can put everything together now:

$F=ma=\frac{GMm}{R^2}$

$a=\frac{GM}{R^2}$

$\frac{v^2}{R}=\frac{GM}{R^2}$

$v^2=\frac{GM}{R}$

$(\frac{2 \pi R}{t})^2=\frac{GM}{R}$

$\frac{4 \pi^2 R^3}{t^2}=GM$

$t^2=\frac{4 \pi^2}{GM}R^3$

$t=\sqrt{\frac{4 \pi^2}{GM}R^3}$

Substituting our values we have:

$t=\sqrt{\frac{4 \pi^2}{(6.67\times10^{-11}Nm^2/Kg^2)(5.97\times10^{24}Kg)}(6371000m)^3}=5063s$

Which are more than 84 minutes. For reference, the ISS has a period of almost 93 minutes, so we know the result makes sense (and the equation will give us this result considering it orbits 408Km above Earth's surface).