Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_{P}$ and $M_{Q}$ and…
Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_{P}$ and $M_{Q}$ and surface areas $A$ and 4A respectively. A spherical planet $R$ also has uniform density $\rho$ and its mass is $\left(M_{P}+M_{Q}\right)$. The escape velocities from the planets $P, Q$ and $R$ are $V_{P}, V_{Q}$ and $V_{R}$, respectively. Then
$V_{Q}>V_{R}>V_{P}$
$V_{R}>V_{Q}>V_{P}$
$V_{R} / V_{P}=3$
$V_{P} / V_{Q}=\frac{1}{2}$
Solution
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
$\begin{aligned} V_{e}=\sqrt{\frac{2 G M}{R}} & \therefore v_{P}=\sqrt{\frac{2 G M_{P}}{R_{p}}}=\sqrt{\frac{2 G m}{r}} \\ v_{Q} &=\sqrt{\frac{2 G M_{Q}}{R_{Q}}}=\sqrt{\frac{2 G(8 \mathrm{~m})}{2 r}}=2 v_{P} \\ v_{R} &=\sqrt{\frac{2 G(9 \mathrm{~m})}{9^{1 / 3} r}}=9^{1 / 3} v_{P} \end{aligned}$
!