Question

The escape velocity of a body from the earth’s surface, Vesc = ________.

A) √(GM/R)
B) 2√(GM/R)
C) √(2GM/R)
D) √(GM/2R)​

Answer 1

Answer:

From the Earth's surface escape velocity is $\sqrt{\frac{2GM}{R}}$. Option C) is correct.

Explanation:

• To escape the gravitational field of the Earth, the minimum velocity a body should possess is called escape velocity.
• On the surface of the Earth, the value of escape velocity is $11.2$  $km/s$.

Step 1:

To calculate escape velocity on the Earth's surface:

When a body is thrown upwards, the law of conservation of energy takes place.

Loss in kinetic energy $=$ Gain in gravitational potential energy

Step 2:

If the body has a mass $m$ and is thrown upwards with a velocity $V_{esc}$.

$\frac{1}{2} mV_{esc} ^{2} =\frac{GMm}{R}$

where $M$ is the Earth's mass, $R$ is the Earth's radius, and $G$ is the gravitational constant.

$\frac{1}{2} V_{esc} ^{2} =\frac{GM}{R}$

$V_{esc} ^{2} =\frac{2GM}{R}$

$V_{esc} =\sqrt{\frac{2GM}{R}}$

So, on the Earth's surface escape velocity is $\sqrt{\frac{2GM}{R}}$.

Option C) is correct.