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Step-by-step Solution
Step 1: Recall the expression for escape velocity
The escape velocity $v_e$ from the surface of a planet of mass $M$ and radius $R$ is given by:
$$
v_e = \sqrt{\frac{2GM}{R}},
$$
where $G$ is the universal gravitational constant.
Step 2: Express mass in terms of density and radius
If the planet has a uniform mass density $\rho$, then its mass $M$ can be written as:
$$
M = \frac{4}{3} \pi R^3 \rho.
$$
Substituting this into the formula for escape velocity,
$$
v_e = \sqrt{\frac{2G \left(\frac{4}{3} \pi R^3 \rho\right)}{R}} \;=\; \sqrt{\frac{8 \pi G \rho}{3} R^2}.
$$
Notice that $v_e \propto R$.
Step 3: Compare the new radius to that of Earth
Let $v$ be the escape velocity from Earth (radius $R_\mathrm{E}$). For another planet whose radius is $4R_\mathrm{E}$ (with the same density), the escape velocity $v_e'$ from its surface will be:
$$
v_e' = \sqrt{\frac{8 \pi G \rho}{3} \,(4 R_\mathrm{E})^2} \;=\; 4 \,\sqrt{\frac{8 \pi G \rho}{3} R_\mathrm{E}^2}.
$$
Since $\sqrt{\frac{8 \pi G \rho}{3} R_\mathrm{E}^2}$ is just $v$, we get:
$$
v_e' = 4\,v.
$$
Step 4: Conclude the correct answer
Therefore, if the planet’s radius is four times that of Earth (with the same density), its escape velocity from its surface is four times that of Earth. Hence, the answer is
$$
4v.
$$
