Q157 / 180 · COMEDK 2025 (10 May Shift 1) · Physics
A planet is 121 times heavier than moon and has a diameter 9 times that of moon. If the escape velocity on the planet is v , then the escape velocity on the moon will be:
- A$\dfrac{33 v}{8}$
- Correct answerB$\dfrac{3 v}{11}$
- C$\dfrac{8 v}{33}$
- D$\dfrac{11 v}{3}$
Solution
The escape velocity $v_e$ of a body is given by the formula $v_e = \sqrt{\dfrac{2GM}{R}}$, where $M$ is the mass and $R$ is the radius of the body.
Let $M_p$ and $R_p$ be the mass and radius of the planet, and $M_m$ and $R_m$ be the mass and radius of the moon. Given $M_p = 121 M_m$ and the diameter $D_p = 9 D_m$, which implies $R_p = 9 R_m$.
The escape velocity on the planet is $v = \sqrt{\dfrac{2GM_p}{R_p}}$.
The escape velocity on the moon is $v_m = \sqrt{\dfrac{2GM_m}{R_m}}$.
Taking the ratio of the escape velocities:
$\dfrac{v_m}{v} = \sqrt{\dfrac{M_m}{R_m} \times \dfrac{R_p}{M_p}} = \sqrt{\dfrac{M_m}{R_m} \times \dfrac{9 R_m}{121 M_m}} = \sqrt{\dfrac{9}{121}} = \dfrac{3}{11}$.
Therefore, $v_m = \dfrac{3}{11} v$.
Answer: $\dfrac{3 v}{11}$
Let $M_p$ and $R_p$ be the mass and radius of the planet, and $M_m$ and $R_m$ be the mass and radius of the moon. Given $M_p = 121 M_m$ and the diameter $D_p = 9 D_m$, which implies $R_p = 9 R_m$.
The escape velocity on the planet is $v = \sqrt{\dfrac{2GM_p}{R_p}}$.
The escape velocity on the moon is $v_m = \sqrt{\dfrac{2GM_m}{R_m}}$.
Taking the ratio of the escape velocities:
$\dfrac{v_m}{v} = \sqrt{\dfrac{M_m}{R_m} \times \dfrac{R_p}{M_p}} = \sqrt{\dfrac{M_m}{R_m} \times \dfrac{9 R_m}{121 M_m}} = \sqrt{\dfrac{9}{121}} = \dfrac{3}{11}$.
Therefore, $v_m = \dfrac{3}{11} v$.
Answer: $\dfrac{3 v}{11}$
Hence, the correct answer is option: 'B'.