The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L. The magnitude of the areal velocity of the planet is: (a) 2L/M (b) L/2M (c) L/M (d) 4L/M

Answer: (b). Area swept in dt: dA = 12 r² dθ, so dA/dt = 12 r²ω. L = Mr²ω ⇒ dA/dt = L/2M So the answer is (b) L/2M.

The angular momentum of a planet of mass M moving around the sun in… | JEE Main 2021 (18 Mar, Shift 2)

Q15JEE Main 2021 · 18 MAR · SHIFT 2Easy

Kepler's Laws & Planetary Motion

The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is $\vec{L}$ . The magnitude of the areal velocity of the planet is:

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Answer: (b)

Area swept in $dt$ : $dA = \tfrac12 r^2\,d\theta$ , so $\dfrac{dA}{dt} = \tfrac12 r^2\omega$ .
$L = Mr^2\omega \Rightarrow \dfrac{dA}{dt} = \dfrac{L}{2M}$
So the answer is (b) $\dfrac{L}{2M}$ .

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