A hypothetical planet has an average mass density 9 times that of...

The escape velocity from the surface of a planet is given by $v_e = \sqrt{\dfrac{2GM}{R}}$.

The mass of the planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \dfrac{4}{3}\pi R^3 \rho$.

Substituting this into the escape velocity formula, we get:

$v_e = \sqrt{\dfrac{2G}{R} \left(\dfrac{4}{3}\pi R^3 \rho\right)} = R\sqrt{\dfrac{8\pi G \rho}{3}}$

This implies that $v_e \propto R\sqrt{\rho}$.

Given that the escape velocity of the hypothetical planet is equal to that of Earth ($v_{ep} = v_{eE}$):

$R_p\sqrt{\rho_p} = R_E\sqrt{\rho_E}$

We are given $\rho_p = 9\rho_E$. Substituting this value:

$R_p\sqrt{9\rho_E} = R_E\sqrt{\rho_E}$

$3R_p = R_E$

$R_p = \dfrac{R_E}{3}$

Answer: $\dfrac{R_E}{3}$