JEE Main 2023 — Gravitation Question with Solution

The escape velocity of a planet is given by the formula:

$${\mathrm{v}}_{\text{escape}}=\sqrt{\frac{2GM}{R}}$$

where G is the gravitational constant, M is the mass of the planet, and R is its radius.

If the escape velocity of planet A is v_A and the escape velocity of planet B is v_B, then we can write the following relationship:

$$\frac{v_B}{v_A}=\frac{\sqrt{\frac{2G{M}_B}{R_B}}}{\sqrt{\frac{2G{M}_A}{R_A}}}=\sqrt{\frac{R_A{M}_B}{M_A{R}_B}}$$

$$\Rightarrow$$ $$\sqrt{\frac{R_A{M}_B}{M_A{R}_B}}=2$$

$$\Rightarrow$$ $$\frac{M_B}{M_A}\times \frac{1}{3}=4$$

$$\Rightarrow$$ $$\frac{M_B}{M_B}=12$$

We know, The acceleration due to gravity on a planet can be calculated using the formula:

$$\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^2}$$

where G is the gravitational constant, M is the mass of the planet, and R is its radius.

$$\frac{g_A}{g_B}=\frac{M_A{R}_B^2}{M_B{R}_A^2}=\frac{1}{12}\times {\left(\frac{3}{1}\right)}^2=\frac{9}{12}=\frac{3}{4}$$