At which height can a satellite be in a geostationary orbit? To remain geostationary,obtain the equation for its height from the surface of the Earth.

(N/A) geostationary satellite must have an orbital period $T$ equal to the Earth's rotation period,which is $T = 24 \text{ hours} = 86400 \text{ s}$.
From Kepler's third law,the square of the orbital period is proportional to the cube of the orbital radius $r$: $T^2 = \frac{4\pi^2 r^3}{GM}$.
Here,$r = R_e + h$,where $R_e$ is the Earth's radius and $h$ is the height above the surface.
Rearranging for $r$: $r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$.
Substituting $GM = gR_e^2$,we get $r = \left( \frac{gR_e^2 T^2}{4\pi^2} \right)^{1/3}$.
The height $h$ is given by $h = r - R_e = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3} - R_e$.
Substituting standard values ($M \approx 5.97 \times 10^{24} \text{ kg}$,$G \approx 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$),the calculated height $h$ is approximately $35,800 \text{ km}$ above the Earth's surface.