Question

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 $(M_p+M_Q)$. The escape velocities from the planets $P,Q$ and $R$, are $V_p,V_Q$ and $V_R$, respectively. Then

• A. $\frac{V_p}{V_Q}=\frac{1}{2}$
• B. $V_R>V_Q>V_P$
• C. $\frac{V_R}{V_P}=3$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{V_p}{V_Q}=\frac{1}{2}$
B $V_R>V_Q>V_P$

let mass of planet P $=$ M and radius of planet P $=$ R

$A_p=A$

$A_q=4A$

$\therefore R_p=R$

And

$R_q=R/2$

Also,

$density=\frac{mass}{volume}$

mass of planet Q$=$ 8M and radius of planet Q$=$ 2R

mass of planet R$=$ 9M and radius of planet R $=9^{\frac{1}{3}}R$

so escape velocity of P $=V_p=\sqrt{\frac{2GM}{R}}$

$V_Q=\sqrt{\frac{2G8M}{2R}}=2V_P$

$V_R=\sqrt{\frac{2G8M}{9^{\frac{1}{3}}R}}=9^{\frac{1}{3}}{V}_P$

so $V_R>V_Q>V_P$

and $\frac{V_Q}{V_P}=2$