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. (a)
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,
$$ M _Q=8 M \text { and } R _Q=2 r $$
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,
$$ \frac{v _P}{v _Q}=\frac{1}{2} \text { and } v _R>v _Q>v _P $$
