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Correct formula for height of a satellite from earths surface is :
$\left(\frac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R$
$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$
$\left(\frac{\mathrm{T}^2 \mathrm{R}^2}{4 \pi^2 \mathrm{~g}}\right)^{1 / 3}-\mathrm{R}$
$\left(\frac{T^2 R^2}{4 \pi^2}\right)^{-1 / 3}+R$
Solution
$\Rightarrow \frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}$
$\Rightarrow \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\mathrm{v}^2 \ldots(1)$
$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \omega$
$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \frac{2 \pi}{\mathrm{T}} .$
$\Rightarrow \frac{\mathrm{GM}}{\mathrm{R}^2}=\mathrm{g}$
$\Rightarrow \mathrm{GM}=\mathrm{gR}^2 .$
Put value from $(2) \& (3) in eq. (1)$
$\Rightarrow \frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}=(\mathrm{R}+\mathrm{h})^2\left(\frac{2 \pi}{\mathrm{T}}\right)^2$
$\Rightarrow \frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}=(\mathrm{R}+\mathrm{h})^3$
$\Rightarrow\left[\frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}\right]^{1 / 3}-\mathrm{R}=\mathrm{h}$
