Question

The escape velocity on earth is 11.2 km/s. If the body is projected out with twice this velocity, then the speed of the body far away from the earth, ignoring the presence of any other object in-universe, will be:
1. 12.2 km/s
2. 22.4 km/s
3. 19.4 km/s
4. 15.2 km/s

Answer 1

Correct Answer - Option 3 : 19.4 km/s

CONCEPT:

• Escape Velocity: Escape velocity is defined as the speed at which an object travels to break free from either the planet's or moon's gravity and leave without any development of propulsion.

Body	Escape Velocity in km/s
sun	618 km/s
moon	2.38 km/s
Earth	11.2 km/s
Jupiter	59.5 km/s

 

• Law of conservation of energy: The law of conservation of energy states that energy can neither be created nor be destroyed. Although, it may be transformed from one form to another. 

 

EXPLANATIONS:

• The escape velocity of a projectile from the Earth,

V_sec = 11.2 km/s

• Projection velocity of the projectile,

V_p = 2V_esc

• Mass of the Projectile = m
• Let the velocity of the projectile far away from the Earth = V_f
• The total energy of the projectile on the Earth = 

\(=\frac{1}{2}mv^{2}_p - \frac{1}{2}mv^{2}_(esc)\)

• The gravitational potential energy of the projectile far away from the Earth is zero.
• The Total energy of the projectile far away from the Earth = 

\(\frac{1}{2}mv^{2}_f\)

• From the law of conservation of energy, we have

\(\frac{1}{2}mv^{2}_p - \frac{1}{2}mv^{2}_(esc) = \frac{1}{2}mv^{2}_f\)

\(v_f = \sqrt{v^{2}_p - v^{2}_(esc))}\)

\(= \sqrt{3}v_(esc)\)

\(=\sqrt{3}\times 11.2\)

\(= 19.376 km/s\)