Physics - Gravitation Question with Solution | TestHub

Here Planets $P$ and $Q$ have the same uniform density ' $\rho$ ' and surface areas $A$ and $4A$ respectively. Let the mass of $P,M_P$ be $m$. Then $m=\rho \times \frac{4}{3}\pi r^3=\rho \times \frac{4}{3}\pi {\left[\frac{A}{4\pi }\right]}^{3/2}$ The mass of $M_Q=\rho \times \frac{4}{3}\pi {\left[\frac{4A}{4\pi }\right]}^{3/2}=8\mathrm{m}$ $\therefore$ The mass of Planet $R=8\mathrm{m}+\mathrm{m}=9\mathrm{m}$ If the radius of $P=r$ Then the radius of $Q=2r$ $\left[\because r_Q={\left(\frac{4A}{4\pi }\right)}^{3/2}=2{\left(\frac{A}{4\pi }\right)}^{3/2}\right]$ and radius of $R=9^{1/3}r$ $\left[\begin{array}{l}\because M_R=M_P+M_Q\\ r_R^3=r^3+(2r{)}^3=9r^3\end{array}\right]$ As we know, escape velocity from the planet $\begin{array}{rl}{V}_e=\sqrt{\frac{2GM}{R}}& \therefore v_P=\sqrt{\frac{2G{M}_P}{R_p}}=\sqrt{\frac{2Gm}{r}}\\ v_Q& =\sqrt{\frac{2G{M}_Q}{R_Q}}=\sqrt{\frac{2G(8\mathrm{m})}{2r}}=2v_P\\ v_R& =\sqrt{\frac{2G(9\mathrm{m})}{9^{1/3}r}}=9^{1/3}{v}_P\end{array}$