Escape Velocity

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Escape velocity from spherical planet (mass M radius R) is:

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√(2GM/R)

√(GM/R)

√(3GM/R)

2GM/R

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Subject: physics
Chapter: gravitation
Topic: escape velocity
Difficulty: Easy
Year: 2025

Solution

Correct Answer:

√(2GM/R)

Escape velocity ( $v_{e}$ ) is the minimum speed needed for an object to escape the gravitational influence of a planet.

Using conservation of energy: Total energy at surface = Total energy at infinity ( $0$ ). $\frac{1}{2} m v_{e}^{2} - \frac{GM m}{R} = 0$ $v_{e} = \frac{2 GM}{R}$
For Earth, $v_{e} \approx 11.2 km/s$ .

This easy difficulty physics question is from the chapter gravitation, covering the topic of escape velocity. It appeared in the 2025 exam.

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Escape Velocity

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More escape velocity Practice Questions

Escape velocity from Earth surface (R, g)?

What is the escape velocity from Earth's surface? (g = 10 m/s², R = 6.4 × 10⁶ m)

Escape velocity from a moon of half Earth's radius and quarter its mass is (Earth v_e = v):

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