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Physics · GRAVITATION
हिन्दीA planet of mass \( m \) revolves around the Sun in an elliptical orbit. If the magnitude of its angular momentum about the Sun is \( L \), the areal velocity \( \frac{\Delta A}{\Delta t} \) swept out by the position vector of the planet is given by:
Correct answer · Option C
\( \frac{L}{2m} \)
Explanation
The swept area element is \( \Delta A = \frac{1}{2} |\mathbf{r} \times \mathbf{v}| \Delta t \), and since \( |\mathbf{r} \times \mathbf{p}| = L \), we have \( \frac{\Delta A}{\Delta t} = \frac{L}{2m} \).
