Satellite orbit — altitude, speed, period

What just happened?

You raised the orbit, and the satellite slowed down and needs more time for one revolution. The farther from Earth, the lower the speed and the longer the period.

Where does this come from?

On a circular orbit, gravity acts as the centripetal force — it keeps the satellite moving in a circle:

The satellite's mass $m$ cancels out. What remains is the dependence of speed and period on the orbital radius alone:

— the larger $r$, the smaller $v$ and the larger $T$. This is exactly Kepler's third law: $T^2 \propto r^3$.

Geostationary orbit

When $T = 24\text{ h}$ ($h \approx 35786\text{ km}$ above the equator), the satellite orbits Earth at the same rate as the planet rotates — from the ground it looks as if it hangs motionless. This is how TV and weather satellites work.

Why doesn't the satellite fall?

It is falling the whole time — gravity constantly curves its path toward Earth. But its horizontal speed is so large that Earth's surface "falls away" beneath it just as fast. This continuous falling "past" the planet is precisely the orbit.

Why this matters for the exam:

Three formulas — $r = R+h$, $v = \sqrt{GM/r}$, $T = 2\pi\sqrt{r^3/GM}$ — solve most satellite problems. The trick: first compute the orbital radius $r$ (add Earth's radius to the altitude!), then substitute into $v$ or $T$.

Remember: a higher orbit means lower speed and a longer period — $v = \sqrt{GM/r}$, $T^2 \propto r^3$. Gravity plays the role of the centripetal force here.