JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Rotational Mechanics

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If the angular momentum of a planet of mass m , moving a round the Sun in a circular orbit its L , about the center of the Sun, its areal velocity is:

(A) $\frac{L}{m}$
(B) $\frac{4 L}{m}$
(C) $\frac{L}{2 m}$ ✓ Correct
(D) $\frac{2 L}{m}$

Answer: (C)

Solution

Areal velocity is the rate at which the radius vector sweeps out area,

\frac{d A}{d t} = \frac{1}{2} r^{2} \frac{d θ}{d t}

. The planet's angular momentum about the Sun is

L = m r^{2} \frac{d θ}{d t}

, so

r^{2} \frac{d θ}{d t} = \frac{L}{m}

. Substituting this straight in gives

\frac{d A}{d t} = \frac{L}{2 m}

, which is simply Kepler's second law in disguise: a constant

L

means a constant areal velocity.

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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