JEE Main 2023 — Gravitation Question with Solution

The acceleration due to gravity at the surface of a planet is given by the formula:

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

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

In this problem, we are given two planets A and B, with radii $$R_A$$ and $$R_B=1.5R_A$$, and densities $${\rho }_A$$ and $${\rho }_B=\frac{{\rho }_A}{2}$$, respectively. We can use the formula above to calculate the acceleration due to gravity at the surface of each planet:

$$g_A=\frac{G{M}_A}{R_A^2}=\frac{4}{3}\pi G{\rho }_A{R}_A$$

and $$g_B=\frac{G{M}_B}{R_B^2}=\frac{4}{3}\pi G{\rho }_B{R}_B=\frac{4}{3}\pi G\frac{{\rho }_A}{2}1.5R_A=\frac{3}{2}{g}_A$$

To find the ratio of the accelerations due to gravity at the surfaces of planets A and B, we can divide the expression for $$g_B$$ by the expression for $$g_A$$:

$$\frac{g_B}{g_A}=\frac{\frac{3}{2}{g}_A}{g_A}=\frac{3}{2}$$

So, we can see that the acceleration due to gravity at the surface of planet B is $$\frac{3}{2}$$ times larger than that at the surface of planet A.

However, the question asks for the ratio of the gravitational accelerations at the surfaces of the planets, which is equal to the ratio of the densities times the ratio of the radii:

$$\frac{g_B}{g_A}=\frac{{\rho }_B}{{\rho }_A}\cdot \frac{R_B}{R_A}=\frac{1/2}{1}\cdot \frac{1.5R_A}{R_A}=\frac{3}{4}$$

Therefore, the ratio of the gravitational accelerations at the surfaces of the planets is $$3/4$$, which is our final answer.