JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 16)
Correct formula for height of a satellite from earths surface is :
$$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$$
$$\left(\frac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R$$
$$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{-1 / 3}+R$$
$$\left(\frac{T^2 R^2}{4 \pi^2 g}\right)^{1 / 3}-R$$
Explanation
$$\begin{aligned}
& T=2 \pi \sqrt{\frac{r^3}{G M}} \\
& T^2=\frac{4 \pi^2}{G M}(R+h)^3 \\
& h=\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\
& =\left(\frac{G M \cdot R}{R^2} \cdot \frac{T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\
& =\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{\frac{1}{3}}-R
\end{aligned}$$
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