JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 11)
If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the center of the Sun, its areal velocity is :
$${L \over m}$$
$${4L \over m}$$
$${L \over 2m}$$
$${2L \over m}$$
Explanation
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dA = $${1 \over 2}$$ r2d$$\theta $$
$$ \therefore $$ $${{dA} \over {dt}} = {1 \over 2}{r^2}{{d\theta } \over {dt}}$$
$$ \Rightarrow $$ $${{dA} \over {dt}} = {1 \over 2}{r^2}\omega $$ . . . . . (1)
We know,
angular momentum,
L = $$mvr$$
= $$m\left( {\omega r} \right)r$$
= mr2$$\omega $$
$$ \therefore $$ $$\omega $$ = $${L \over {m{r^2}}}$$ . . . . . (2)
Put value of $$\omega $$ in equation(1),
$${{dA} \over {dt}}$$ = $${1 \over 2}{r^2}$$ ($${L \over {m{r^2}}}$$)
= $${L \over {2m}}$$
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