Gravitation 2 Question 6
9. Two spherical planets $P$ and $Q$ have the same uniform density $\rho$, masses $M_{P}$ and $M_{Q}$, and surface areas $A$ and $4 A$, 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
(a) $v_{Q}>v_{R}>v_{P}$
(b) $v_{R}>v_{Q}>v_{P}$
(c) $v_{R} / v_{P}=3$
(d) $v_{P} / v_{Q}=1 / 2$
(2012)
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Answer:
Correct Answer: 9. (b,d)
Solution:
- Surface area of $Q$ is four times. Therefore, radius of $Q$ is two times. Volume is eight times. Therefore, mass of $Q$ is also eight times.
So, let
$M_{P}=M$ and $R_{P}=r$
Then,
$$ M_{Q}=8 \mathrm{M} \text { and } R_{Q}=2 r $$
Now, mass of $R$ is $\left(M_{P}+M_{Q}\right)$ or $9 M$. Therefore, radius of $R$ is $(9)^{1 / 3} r$. Now, escape velocity from the surface of a planet is given by
$$ \begin{aligned} v & =\sqrt{\frac{2 G M}{r}}(r=\text { radius of that planet }) \\ \therefore \quad v_{P} & =\sqrt{\frac{2 G M}{r}} \\ v_{Q} & =\sqrt{\frac{2 G(8 M)}{(2 r)}} \Rightarrow v_{R}=\sqrt{\frac{2 G(9 M)}{(9)^{1 / 3} r}} \end{aligned} $$
From here we can see that,
$$ \frac{v_{P}}{v_{Q}}=\frac{1}{2} \text { and } v_{R}>v_{Q}>v_{P} $$
