પૃથ્વીની સપાટીથી ઉપગ્રહની ઊંચાઈ માટેનું સાચું સૂત્ર. . . . . . . . છે.

Question

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

JEE MAIN 2024, Diffcult

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Solution

b
$\Rightarrow \frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})}$

$\Rightarrow \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\mathrm{v}^2 \ldots(1)$

$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \omega$

$\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \frac{2 \pi}{\mathrm{T}} .$

$\Rightarrow \frac{\mathrm{GM}}{\mathrm{R}^2}=\mathrm{g}$

$\Rightarrow \mathrm{GM}=\mathrm{gR}^2 .$

Put value from $(2) \& (3) in eq. (1)$

$\Rightarrow \frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}=(\mathrm{R}+\mathrm{h})^2\left(\frac{2 \pi}{\mathrm{T}}\right)^2$

$\Rightarrow \frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}=(\mathrm{R}+\mathrm{h})^3$

$\Rightarrow\left[\frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}\right]^{1 / 3}-\mathrm{R}=\mathrm{h}$

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