1. Escape velocity formula

For a planet of mass $M$ and radius $R$, the escape velocity is

$v_e = \sqrt{\frac{2GM}{R}}$

Since mass of a planet is

$M = \frac{4}{3}\pi R^3 \rho$

we substitute into the escape velocity formula:

$v_e = \sqrt{\frac{2G}{R}\left(\frac{4}{3}\pi R^3 \rho\right)}$

$v_e = \sqrt{\frac{8}{3}\pi G R^2 \rho}$

Thus,

$v_e \propto R\sqrt{\rho}$

2. Compare planet A and planet B

Given:

• Escape velocity on planet A: $v_A = 12\ \text{km s}^{-1}$
• Density of planet B: $\rho_B = 4\rho_A$
• Radius of planet B: $R_B = \frac{1}{2}R_A$

Using

$v \propto R\sqrt{\rho}$

we get

$\frac{v_B}{v_A} = \frac{R_B\sqrt{\rho_B}}{R_A\sqrt{\rho_A}}$

Substitute the given ratios:

$\frac{v_B}{v_A} = \frac{\frac{1}{2}R_A \cdot \sqrt{4\rho_A}}{R_A\sqrt{\rho_A}}$

$\frac{v_B}{v_A} = \frac{1}{2} \cdot \frac{2\sqrt{\rho_A}}{\sqrt{\rho_A}} = 1$

So,

$v_B = v_A = 12\ \text{km s}^{-1}$

3. Check options

• A: $12\ \text{km s}^{-1}$ ✅
• B: $24\ \text{km s}^{-1}$ ❌
• C: $36\ \text{km s}^{-1}$ ❌
• D: $6\ \text{km s}^{-1}$ ❌

4. Final answer

The correct option is

$\boxed{\text{A: } 12\ \text{km s}^{-1}}$

5. Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

So, the answer agrees with the stored correct answer.