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Question: Answered & Verified by Expert
A satellite of mass 'm', revolving round the earth of radius 'r' has kinetic energy (E). Its angular momentum is
PhysicsRotational MotionMHT CETMHT CET 2020 (14 Oct Shift 1)
Options:
  • A $\left(\mathrm{mEr}^{2}\right)^{\frac{1}{2}}$
  • B $\left(\mathrm{mEr}^{2}\right)$
  • C $\left(2 \mathrm{~m} \mathrm{Er}^{2}\right)^{\frac{1}{2}}$
  • D $\left(2 \mathrm{mEr}^{2}\right)$
Solution:
1595 Upvotes Verified Answer
The correct answer is: $\left(2 \mathrm{~m} \mathrm{Er}^{2}\right)^{\frac{1}{2}}$
$\begin{aligned} \mathrm{E}=\frac{1}{2} \mathrm{mV}^{2} \quad \therefore \mathrm{V} &=\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \\ \text { Angular momentum } \quad \mathrm{L} &=\mathrm{mvr} \quad=\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}} \cdot \mathrm{r}} \\ &=\sqrt{2 \mathrm{mEr}^{2}}=\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}} \end{aligned}$

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