# Geostationary Satellite

I am unable to understand one particular thing about geostationary satellite. Let us assume that I am on earth at a particular place and remain there (moving with the place during rotation) such that the satellite is just above me all the time.

Now my orbital velocity will be

$$v_o = \sqrt{GM_e/R_e}$$

Talking about the geostationary satellite at a height $h$ we can say that the orbital velocity is

$$v_{geo} = \sqrt{GM_e/(R_e+h) }$$

This showa that the velocity of the satellite has decreased and simultaneously its radius has increased.

But it must have my angular velocity to have the same time period.

$\omega =v/r$

For the satellite $v$ has decreased and $r$ has increased. How can the angular velocity of satellite be equal to that of mine.

## 1 Answer

You wrote $v_o = \sqrt{G M_e / R_e}$ for your orbital velocity. But you're not in orbit! Rather, you're going in a circle with the angular frequency of the Earth, $\Omega \approx \frac{2 \pi}{1\;\text{day}}$. This makes it possible to set your angular frequency equal to that of the satellite.