Q110 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
The density of a planet is three times that of earth and radius $2.5$ times that of earth. If $g_p$ and $g_e$ represents acceleration due to gravity on planet and earth respectively then ratio $g_p$ to $g_e$ is
- Correct answerA$7.5$
- B$3.5$
- C$6.5$
- D$9.5$
Solution
The acceleration due to gravity on the surface of a planet is given by
$g = \dfrac{GM}{R^2}$
Substituting the mass of the planet $M = \dfrac{4}{3}\pi R^3 \rho$, we get
$g = \dfrac{G}{R^2} \left(\dfrac{4}{3}\pi R^3 \rho\right) = \dfrac{4}{3}\pi G R \rho$
This implies that $g \propto R \rho$.
The ratio of acceleration due to gravity on the planet to that on Earth is
$\dfrac{g_p}{g_e} = \dfrac{R_p \rho_p}{R_e \rho_e}$
Given that $\rho_p = 3\rho_e$ and $R_p = 2.5R_e$, substituting these values yields
$\dfrac{g_p}{g_e} = (2.5) \times (3) = 7.5$
Answer: $7.5$
$g = \dfrac{GM}{R^2}$
Substituting the mass of the planet $M = \dfrac{4}{3}\pi R^3 \rho$, we get
$g = \dfrac{G}{R^2} \left(\dfrac{4}{3}\pi R^3 \rho\right) = \dfrac{4}{3}\pi G R \rho$
This implies that $g \propto R \rho$.
The ratio of acceleration due to gravity on the planet to that on Earth is
$\dfrac{g_p}{g_e} = \dfrac{R_p \rho_p}{R_e \rho_e}$
Given that $\rho_p = 3\rho_e$ and $R_p = 2.5R_e$, substituting these values yields
$\dfrac{g_p}{g_e} = (2.5) \times (3) = 7.5$
Answer: $7.5$
Hence, the correct answer is option: 'A'.