QJEEM23Phca88: JEE Main 2023 Physics - Class 11 JEEM23Phca88 Solution: The escape velocities of two planets and are in the ratio . If…

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:

${{{v_B}} \over {{v_A}}} = {{\sqrt {{{2G{M_B}} \over {{R_B}}}} } \over {\sqrt {{{2G{M_A}} \over {{R_A}}}} }} = \sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}}$

$\Rightarrow$ $\sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}} = 2$

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

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

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

$\mathrm{g} = \frac{\mathrm{G} \mathrm{M}}{\mathrm{R}^2}$

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

${{{g_A}} \over {{g_B}}} = {{{M_A}R_B^2} \over {{M_B}R_A^2}} = {1 \over {12}} \times {\left( {{3 \over 1}} \right)^2} = {9 \over {12}} = {3 \over 4}$