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If $A$ is the areal velocity of a planet of mass $M$, its angular momentum is
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1753 Upvotes
Verified Answer
The correct answer is:
$2MA$
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
$$
\begin{array}{rlrl}
& =I \omega \\
\therefore & L & =2 M A
\end{array}
$$
$\therefore \quad$ 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
$$
\begin{array}{rlrl}
& =I \omega \\
\therefore & L & =2 M A
\end{array}
$$
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