Question

Assuming the orbit of mars around the sun to be circular, the revolving red planet's angular momentum is proportional to the $n^{t h}$ power of its radius. Here $n$ is

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. 1.5

Answer 1

Answer: (C)

The C.P. force $\frac{m v^{2}}{R}$ is provided by the gravitational force $\frac{G M m}{R^{2}}$ .
$∴$ $\frac{m v^{2}}{R}=\frac{G M m}{R^{2}}$ $∴v=\sqrt{\frac{G M}{R}}$
But the angular momentum $L=Ιω=mR^{2}\times \frac{v}{R}=mvR$
$∴$ $L=mR\sqrt{\frac{G M}{R}}=m\sqrt{G M}\times \frac{R}{\sqrt{R}}$ $=m\sqrt{G M}⋅R^{\frac{1}{2}}=KR^{\frac{1}{2}}$
$∴$ $L∝R^{\frac{1}{2}}$ as $K=m\sqrt{G M}$ is a constant.
But $L∝R^{n}$ $∴$ $n=\frac{1}{2}$