Question

If the acceleration due to gravity on the surface of a planet is 2.5 times that on Earth and radius, 10 times that of the Earth, then the ratio of the escape velocity on the surface of a planet to that on Earth is:

• A. 1:1
• B. 1:2
• C. 2:1
• D. 1:5
• E. 5:1

Hint:

Escape velocity depends on both the gravitational acceleration and the planet’s radius. If both increase, the escape velocity increases.

Answer 1

Answer: (E)

Escape velocity (\( v_e \)) is given by the formula: \[ v_e = \sqrt{2 g R} \] where \( g \) is the acceleration due to gravity and \( R \) is the radius of the planet. 
For the given planet: \[ g_{{planet}} = 2.5 g_{{Earth}}, \quad R_{{planet}} = 10 R_{{Earth}} \] Now, computing the escape velocity ratio: \[ \frac{v_{e,{planet}}}{v_{e,{Earth}}} = \sqrt{\frac{2.5 g_{{Earth}} \times 10 R_{{Earth}}}{2 g_{{Earth}} R_{{Earth}}}} \] \[ = \sqrt{\frac{25}{5}} = \sqrt{5} \approx 5 \] Thus, the ratio of escape velocity on the planet to that on Earth is 5:1.