Physics - Gravitation Question with Solution | TestHub

According to the Kepler's third law $$T^2\propto r^3$$ where, $T=$ time period of revolution $r=$ radius Now, ${\left(\frac{T_E}{T_P}\right)}^2={\left(\frac{r_E}{r_P}\right)}^3=\frac{r_E}{r_P}={\left(\frac{T_E}{T_P}\right)}^{2/3}$ $$\frac{r_E}{r_P}={\left(\frac{2\pi /{\omega }_E}{2\pi /{\omega }_P}\right)}^{2/3}\left[\because T=\frac{2\pi }{\omega }\right]$$ $$\frac{r_E}{r_P}={\left(\frac{{\omega }_P}{{\omega }_E}\right)}^{2/3}$$ According to the question, ${\omega }_P=2{\omega }_E$ and $r_E=R$ $\Rightarrow \frac{R}{r_P}={\left(\frac{2{\omega }_E}{{\omega }_E}\right)}^{2/3}$ $$\begin{array}{l}\frac{R}{r_P}=(2{)}^{2/3}\\ r_P=\frac{R}{(2{)}^{2/3}}=R(2{)}^{-2/3}\end{array}$$