Q127 / 155 · WBJEE 2019 · Physics
Assume that the earth moves around the sun in a circular orbit of radius $R$ and there exists a planet which also move around the sun in a circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is
- Correct answerA$2^{-2 / 3} R$
- B$2^{2 / 3} R$
- C$ 2^{-1 / 3} R$
- D$\frac{R}{\sqrt{2}}$
Solution
According to the Kepler's third law
$$
T^{2} \propto r^{3}
$$
where, $T=$ time period of revolution
$r=$ radius
Now, $\left(\frac{T_{E}}{T_{P}}\right)^{2}=\left(\frac{r_{E}}{r_{P}}\right)^{3}=\frac{r_{E}}{r_{P}}=\left(\frac{T_{E}}{T_{P}}\right)^{2 / 3}$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{2 \pi / \omega_{E}}{2 \pi / \omega_{P}}\right)^{2 / 3} \quad\left[\because T=\frac{2 \pi}{\omega}\right]
$$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{\omega_{P}}{\omega_{E}}\right)^{2 / 3}
$$
According to the question, $\omega_{P}=2 \omega_{E}$ and $r_{E}=R$
$\Rightarrow \quad \frac{R}{r_{P}}=\left(\frac{2 \omega_{E}}{\omega_{E}}\right)^{2 / 3}$
$$
\begin{array}{l}
\frac{R}{r_{P}}=(2)^{2 / 3} \\
r_{P}=\frac{R}{(2)^{2 / 3}}=R(2)^{-2 / 3}
\end{array}
$$
Hence, the correct answer is option: 'A'.