Q149 / 180 · COMEDK 2024 (Morning Shift) · Physics
If the earth has a mass nine times and radius four times that of planet X, the ratio of the maximum speed required by a rocket to pull out of the gravitational force of planet $\mathrm{X}$ to that of the earth is
- A$\dfrac{9}{4}$
- B$\dfrac{4}{9}$
- Correct answerC$\dfrac{2}{3}$
- D$\dfrac{3}{2}$
Solution
The escape velocity $v_{e}$ of a planet with mass $M$ and radius $R$ is given by the formula $v_{e} = \sqrt{\dfrac{2GM}{R}}$.
Let $M_{E}$ and $R_{E}$ be the mass and radius of the Earth, and $M_{X}$ and $R_{X}$ be the mass and radius of planet X. Given $M_{E} = 9M_{X}$ and $R_{E} = 4R_{X}$.
The escape velocity for Earth is $v_{E} = \sqrt{\dfrac{2GM_{E}}{R_{E}}} = \sqrt{\dfrac{2G(9M_{X})}{4R_{X}}} = \sqrt{\dfrac{9}{4}} \sqrt{\dfrac{2GM_{X}}{R_{X}}} = \dfrac{3}{2} v_{X}$, where $v_{X}$ is the escape velocity for planet X.
The ratio of the escape velocity of planet X to that of the Earth is $\dfrac{v_{X}}{v_{E}} = \dfrac{v_{X}}{\dfrac{3}{2} v_{X}} = \dfrac{2}{3}$.
Answer: $\dfrac{2}{3}$
Let $M_{E}$ and $R_{E}$ be the mass and radius of the Earth, and $M_{X}$ and $R_{X}$ be the mass and radius of planet X. Given $M_{E} = 9M_{X}$ and $R_{E} = 4R_{X}$.
The escape velocity for Earth is $v_{E} = \sqrt{\dfrac{2GM_{E}}{R_{E}}} = \sqrt{\dfrac{2G(9M_{X})}{4R_{X}}} = \sqrt{\dfrac{9}{4}} \sqrt{\dfrac{2GM_{X}}{R_{X}}} = \dfrac{3}{2} v_{X}$, where $v_{X}$ is the escape velocity for planet X.
The ratio of the escape velocity of planet X to that of the Earth is $\dfrac{v_{X}}{v_{E}} = \dfrac{v_{X}}{\dfrac{3}{2} v_{X}} = \dfrac{2}{3}$.
Answer: $\dfrac{2}{3}$
Hence, the correct answer is option: 'C'.