Question

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 X to that of the earth is:

• A. \( \frac{2}{3} \)
• B. \( \frac{9}{4} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{4}{9} \)

Hint:

Escape velocity depends on the mass and radius of the planet. The formula \( v_e = \sqrt{\frac{2GM}{R}} \) shows that a larger mass or smaller radius results in a higher escape velocity.

Answer 1

The Correct Option is A

The escape velocity \(v_e\) is given by the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \] where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet. For Earth: \[ v_{e, \text{Earth}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} \] For Planet X: \[ v_{e, \text{X}} = \sqrt{\frac{2GM_{\text{X}}}{R_{\text{X}}}} \] Given that the mass of Earth is 9 times the mass of Planet X and the radius of Earth is 4 times the radius of Planet X, we can write the ratio of the escape velocities as: \[ \frac{v_{e, \text{X}}}{v_{e, \text{Earth}}} = \sqrt{\frac{M_{\text{X}}}{M_{\text{Earth}}} \times \frac{R_{\text{Earth}}}{R_{\text{X}}}} \] Substituting the given values: \[ \frac{v_{e, \text{X}}}{v_{e, \text{Earth}}} = \sqrt{\frac{1}{9} \times \frac{4}{1}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \] Thus, the required ratio is \( \frac{2}{3} \), which corresponds to option (A).