The period of a satellite against its height

The higher a satellite orbits, the longer it takes to go round. You will make the program draw the period against the height, find the height of the orbit that takes exactly $3$ hours, and count how many times a satellite $700\ \text{km}$ up goes round in a day.

The idea. For a circular orbit, $T = 2\pi\sqrt{\dfrac{r^3}{GM}}$ with $r = R + h$. A day has $86\,400\ \text{s}$.

Your tasks.

1. Work out the period T, in hours, for every height in h.
2. Find the height at which the period is closest to $3$ hours.
3. Work out how many orbits a day a satellite $700\ \text{km}$ up makes.