The ratio of escape velocity of a planet to the escape velocity of earth will be:-

Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.

The escape velocity of a planet or a celestial body is given by:<br/><br/> $v_e = \sqrt{\frac{2GM}{r}}$<br/><br/> where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. <br/><br/> Let the subscripts "p" and "e" denote the planet and earth, respectively. Then, we have:<br/><br/> $$\frac{v_{e,p}}{v_{e,e}} = \frac{\sqrt{\frac{2G M_p}{r_p}}}{\sqrt{\frac{2G M_e}{r_e}}} = \sqrt{\frac{M_p r_e}{M_e r_p}}$$ <br/><br/> Substituting the given values, we get:<br/><br/> $\frac{v_{e,p}}{v_{e,e}} = \sqrt{\frac{16 \cdot 1}{1 \cdot 4}} = \sqrt{4} = 2$ <br/><br/> Therefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.