Question

For a satellite to be in a circular orbit 760kmkm above the surface of the earth. What is the period of the orbit (in hours)?

Answer 1

The equation you'll to consider is the following:\[\frac{ T ^{2} }{ r ^{3} }=\frac{ 4\pi ^{2} }{ GM _{E} }\]where T is the orbital period of the orbiting satellite; r is the mean orbital radius; G is the gravitational constant; and ME is the mass of the Earth.

Note two things:
1. The period will be in seconds, so you'll need to convert.
2. The mean orbital radius is measured from the center of the Earth.

Answer 2

Just for your info @tornadosiren - the formula given above can be derived from some basic principles:

Gravitational force between two bodies =

\[F= \frac{ GMm }{ r ^{2} }\]

And for a circular path the centripetal force required is

\[mr \omega ^{2}\]

If you equate those two forces:

\[\omega ^{2}=\frac{ GM }{ r ^{3} }\]

Since
\[T=\frac{ 2\pi }{ \omega } \]
This gives the equation above