MHT CET · Physics · Rotational Motion
A satellite of mass 'm', revolving round the earth of radius 'r' has kinetic energy (E). Its angular momentum is
- 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 \(\left(2 \mathrm{mEr}^{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(2 \mathrm{~m} \mathrm{Er}^{2}\right)^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}=\frac{1}{2} \mathrm{mV}^{2} \therefore \mathrm{V} =\sqrt{\frac{2 \mathrm{E}}{\mathrm{m}}} \)
Angular momentum \(\mathrm{L} =\mathrm{mvr} \quad=\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}} \cdot \mathrm{r}} \)
\( =\sqrt{2 \mathrm{mEr}^{2}}=\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}\)
Angular momentum \(\mathrm{L} =\mathrm{mvr} \quad=\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}} \cdot \mathrm{r}} \)
\( =\sqrt{2 \mathrm{mEr}^{2}}=\left(2 \mathrm{mEr}^{2}\right)^{\frac{1}{2}}\)
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