QJEEM23Ph8e45: JEE Main 2023 Physics - Class 11 JEEM23Ph8e45 Solution: The ratio of escape velocity of a planet to the escape velocity …

The escape velocity of a planet or a celestial body is given by:

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

where $$G$$ is the gravitational constant, $$M$$ is the mass of the planet, and $$r$$ is the radius of the planet.

Let the subscripts "p" and "e" denote the planet and earth, respectively. Then, we have:

$$\frac{v_{e,p}}{v_{e,e}}=\frac{\sqrt{\frac{2G{M}_p}{r_p}}}{\sqrt{\frac{2G{M}_e}{r_e}}}=\sqrt{\frac{M_p{r}_e}{M_e{r}_p}}$$

Substituting the given values, we get:

$$\frac{v_{e,p}}{v_{e,e}}=\sqrt{\frac{16\cdot 1}{1\cdot 4}}=\sqrt{4}=2$$

Therefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.