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

Correct formula for height of a satellite from earths surface is:
  1. $\left(\frac{T^2 R^2}{4 \pi^2 g}\right)^{1 / 3}-R$
  2. $\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$
  3. $\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{-1 / 3}+R$
  4. $\left(\frac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R$

Solution


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

Asked in: JEE Main 2024 (04 Apr Shift 2)

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