Two spherical planets P and Q have the same uniform density ,...

Here Planets $P$ and $Q$ have the same uniform density ' $\rho$ ' and surface areas $A$ and $4 A$ 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{4 A}{4 \pi}\right]^{3 / 2}=8 \mathrm{~m} \therefore \quad$The mass of Planet$R=8 \mathrm{~m}+\mathrm{m}=9 \mathrm{~m}$If the radius of$P=r$Then the radius of$Q=2 r \left[\because r_{Q}=\left(\frac{4 A}{4 \pi}\right)^{3 / 2}=2\left(\frac{A}{4 \pi}\right)^{3 / 2}\right] and radius of $R=9^{1 / 3} r$ \left[$\begin{array}{l} \because M_{R}=M_{P}+M_{Q} \\ r_{R}^{3}=r^{3}+(2 r)^{3}=9 r^{3} \end{array}$\right] As we know, escape velocity from the planet $$\begin{aligned} V_{e}=\sqrt{\frac{2 G M}{R}} & \therefore v_{P}=\sqrt{\frac{2 G M_{P}}{R_{p}}}=\sqrt{\frac{2 G m}{r}} \\ v_{Q} &=\sqrt{\frac{2 G M_{Q}}{R_{Q}}}=\sqrt{\frac{2 G(8 \mathrm{~m})}{2 r}}=2 v_{P} \\ v_{R} &=\sqrt{\frac{2 G(9 \mathrm{~m})}{9^{1 / 3} r}}=9^{1 / 3} v_{P} \end{aligned}$$