JEE Mains · Physics · STD 11 - 7. gravitation
A satellite of mass \(\frac{M}{2}\) is revolving around earth in a circular orbit at a height of \(\frac{R}{3}\) from earth surface. The angular momentum of the satellite is \(M \sqrt{\frac{G M R}{x}}\). The value of \(x\) is ______ , where \(M\) and \(R\) are the mass and radius of earth, respectively. ( G is the gravitational constant)
- A 7
- B 8
- C 9
- D 3
Answer & Solution
Correct Answer
(D) 3
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { orbital velocity } v_0=\sqrt{\frac{\mathrm{GM}}{4 \mathrm{R} / 3}}=\sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \\ & \text { Angular momentum of satellite }=\frac{\mathrm{M}}{2} \mathrm{v}_0 \frac{4 R}{3} \\ & =\frac{M}{2} \cdot \sqrt{\frac{3…
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