if the length of the day is T the height of that t.v. satellite above the earth surface which always appears stationary from earth will be
Hello,
Since the satellite is stationary, it has same angular velocity as that of earth.
Thus \omega=\dfrac{2\pi}{T}=\dfrac{v}{r}ω=T2π=rv
\implies v=\dfrac{2\pi r}{T}v=T2πr
The centripetal acceleration arises from the gravitational force earth exerts on satellite.
\implies \dfrac{mv^2}{r}=\dfrac{GMm}{r^2}rmv2=r2GMm
Eliminating vv from above equations gives
r=(\dfrac{GMT^2}{4\pi ^2})^{1/3}r=(4π2GMT2)1/3
Hope this helps, Good Luck.