1. Use surface area to relate radii

For a sphere, $A = 4\pi R^2$

Let the radii of planets $P,Q,R$ be $R_P,R_Q,R_R$.

Given:

• Planet $P$ has surface area $A$
• Planet $Q$ has surface area $4A$

So, $4\pi R_Q^2 = 4(4\pi R_P^2)$ $R_Q^2 = 4R_P^2$ $R_Q = 2R_P$

2. Use same density to relate masses

Since all planets have the same uniform density $\rho$, $M = \frac{4}{3}\pi R^3 \rho$ So mass is proportional to $R^3$: $M \propto R^3$

Hence, $\frac{M_Q}{M_P} = \left(\frac{R_Q}{R_P}\right)^3 = 2^3 = 8$ So, $M_Q = 8M_P$

Given for planet $R$: $M_R = M_P + M_Q = 9M_P$

Again, since density is same, $\frac{M_R}{M_P} = \left(\frac{R_R}{R_P}\right)^3$ Thus, $9 = \left(\frac{R_R}{R_P}\right)^3$ $R_R = 9^{1/3} R_P$

3. Escape velocity formula

Escape velocity from a planet is $v_e = \sqrt{\frac{2GM}{R}}$

For same density, since $M \propto R^3$, $v_e \propto \sqrt{\frac{R^3}{R}} = R$ So for planets of equal density, $v_e \propto R$

Therefore, $V_P : V_Q : V_R = R_P : R_Q : R_R = 1 : 2 : 9^{1/3}$

Now, $9^{1/3} \approx 2.08$ So, $V_R > V_Q > V_P$

Thus Option B is correct and Option A is false.

4. Check option D

Since $\frac{V_P}{V_Q} = \frac{R_P}{R_Q} = \frac{1}{2}$ So Option D is correct.

5. Check option C

$\frac{V_R}{V_P} = \frac{R_R}{R_P} = 9^{1/3} \neq 3$ So Option C is false.

6. Final correct options

The correct options are: $\boxed{B, D}$