Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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

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

Solution

<p>$$\begin{aligned} & T=2 \pi \sqrt{\frac{r^3}{G M}} \\ & T^2=\frac{4 \pi^2}{G M}(R+h)^3 \\ & h=\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{G M \cdot R}{R^2} \cdot \frac{T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{\frac{1}{3}}-R \end{aligned}$$</p>

About this question

Subject: Physics · Chapter: Gravitation · Topic: Satellites and Orbital Motion

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