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 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.