WBJEE · Physics · Gravitation
A satellite of mass m rotates round the earth in a circular orbit of radius R. If the angular momentum of the satellite
is J, then its kinetic energy (K) and the total energy (E) of the satellite are
- A \(\mathrm{K}=\frac{\mathrm{J}^2}{m R^2}, \mathrm{E}=-\frac{\mathrm{J}^2}{2 m R^2}\)
- B \(\mathrm{K}=\frac{\mathrm{J}^2}{2 m R^2}, \mathrm{E}=-\frac{\mathrm{J}^2}{2 m R^2}\)
- C \(\mathrm{K}=\frac{J^2}{2 m R^2}, \mathrm{E}=-\frac{J^2}{m R^2}\)
- D \(K=\frac{J^2}{2 m R^2}, E=\frac{J^2}{m R^2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{K}=\frac{\mathrm{J}^2}{2 m R^2}, \mathrm{E}=-\frac{\mathrm{J}^2}{2 m R^2}\)
Step-by-step Solution
Detailed explanation
where "I, is moment of inertia" Hint: \(K E=\frac{J^2}{2 I}=\frac{J^2}{2 m R^2}, \quad E=-K E\)
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