If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
$\frac{M}{A}$
$2MA$
$A^2 M$
$A M^2$
Solution
Areal velocity $A=\frac{1}{2} R^2 \omega$
$\therefore \quad$ Multiplying by $M$ on both sides
$
\begin{aligned}
M A & =\frac{1}{2} M R^2 \omega \\
M A & =\frac{1}{2} I \omega
\end{aligned}
$
$I=$ moment of inertia.
$
M A=\frac{1}{2} L
$
where, $L=$ angular momentum
$
\begin{array}{rlrl}
& =I \omega \\
\therefore & L & =2 M A
\end{array}
$