Question 171 of 223

Correct formula for height of a satellite from earths surface is:

A$\left(\frac{T^2 R^2}{4 \pi^2 g}\right)^{1 / 3}-R$B$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$C$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{-1 / 3}+R$D$\left(\frac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R$

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Answer: (B) $\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$

$\Rightarrow \frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}$
$\Rightarrow \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\mathrm{v}^2$ ....(1)
$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \omega$
$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \frac{2 \pi}{\mathrm{T}}$ ....(2)
$\Rightarrow \frac{\mathrm{GM}}{\mathrm{R}^2}=\mathrm{g}$
$\Rightarrow \mathrm{GM}=\mathrm{gR}^2$ ....(3)
Put value from (2) \& (3) in eq. (1)
$\begin{aligned}
& \Rightarrow \frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}=(\mathrm{R}+\mathrm{h})^2\left(\frac{2 \pi}{\mathrm{T}}\right)^2 \\
& \Rightarrow \frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}=(\mathrm{R}+\mathrm{h})^3 \\
& \Rightarrow\left[\frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}\right]^{1 / 3}-\mathrm{R}=\mathrm{h}
\end{aligned}$

JEE Main Physics practice question on Gravitation (Motion of satellite), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.