**Escape Velocity** ( Class 11 Physics Gravitation )

- Escape velocity is the minimum velocity that a body must attain to escape the gravitational field of the earth.
- Suppose if we throw a ball,it will fall back. This is happening due to the force of gravitation exerted on the ball by the surface of the earth due to which the ball is attracted towards the surface of the earth.
- If we increase the velocity to such an extent that the object which is thrown up will never fall back.This velocity is known as escape velocity.

The same ball is thrown with a velocity that itescapes the force of gravitation of earth and does not come back. This velocity is known as escape velocity.

- Mathematically:-
- Suppose we throw a ball and the initial velocity of the ball is equal to the escape velocity such that ball never comes back.
- Final Position will be infinity.
: At Infinity__At Final Position__- Total Energy (∞) = kinetic Energy (∞) + PotentialEnergy (∞)
- KineticEnergy (∞) = ½ mv
_{f}^{2}where v_{f}=final velocity - Potential Energy (∞) = -GMm/r + V
_{0} - where M=mass of the earth, m= mass of the ball,

V_{0}=potential energy at surface of earth, r=∞ r=distance from the centre of the earth.

- Therefore: – Potential Energy (∞) =0
- Total Energy (∞) =½ mv
_{f}^{2}(1) :-__At initial position__- E. = 1/2mv
_{i}^{2} - E= -GMm/ (R
_{e}+h) + V_{0} - Where h= height of the ball from the surface of the earth.
- Total Energy (initial) = 1/2mv
_{i}^{2}– GMm/ (R_{e}+h) (2) - According to law of conservation of energy
- Total Energy (∞) =Total Energy (initial)
- ½ mv
_{f}^{2}= 1/2mv_{i}^{2}– GMm/ (R_{e}+h) - As L.H.S = positive
- 1/2mv
_{i}^{2}– GMm/ (R_{e}+h) ≥ 0 - 1/2mv
_{i}^{2}= GMm/ (R_{e}+h) - By calculating
- v
_{i}^{2}= 2GM/ (R_{e}+h) - Assume Ball is thrown from earth surface h<<R
_{e} - This implies R
_{e}+h is same as Re as we can neglect h. - Therefore,v
_{i}^{2}= 2GM/ (R_{e}) - Or
**v**_{i}= √(2GM/R_{e}) - This is the initial velocity with which if the ball is thrown it will never fall back on the earth surface.

**In terms of ‘g’**

- g = GM/R
_{e}^{2} - Escape velocity can be written as

**V _{e}= √2gR_{e}**

__Example of Escape Velocity: No atmosphere on moon__

- Earth Escape velocity= V
_{e}= √2gR_{e} - Moon Escape velocity = V
_{e}= √2g_{m}R_{m}where R_{m}is the radius of the moon. - g
_{m}=1/6 g_{e}; R_{m}= 1/4 R_{e} - (V
_{e})_{moon}= √2 g_{m}R_{m}= √2xg/6x R_{e}/4 - After calculating we will get:

**(V _{e})_{moon} = 1/5 (V_{e})_{earth} = 2.3km/s**

- As this velocity is very less, the molecules cannot accumulate on the moon so there is no atmosphere on the moon.

** Problem**: – Does the escape speed of a body from the earth depend on

(a) the mass of the body,

(b) the location from where it is projected,

(c) the direction of projection,

(d) the height of the location from where the body is launched?

__Answer__

(a) No

(b) No

(c) No

(d) Yes

Escape velocity of a body from the Earth is given by the relation:

V_{e}= √2gR_{e}

g = Acceleration due to gravity

R = Radius of the Earth

It is clear from equation (i) that escape velocity is independent of the mass of the body and the direction of its projection. However, it depends on gravitational potential at the point from where the body is launched. Since this potential marginally depends on the height of the point, escape velocity also marginally depends on these factors.

** Problem**: A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

__Answer:-__

(a) No

(b) No

(c) Yes

(d) No

(e) No

(f) Yes

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic and potential energy varies from point to point in the orbit.

** Problem: –** The escape speed of a projectile on the earth’s surface is 11.2 km s

^{–1}. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

__Answer:-__

Escape velocity of a projectile from the Earth, v_{esc} = 11.2 km/s

Projection velocity of the projectile, vp = 3v_{esc}

Mass of the projectile = m

Velocity of the projectile far away from the Earth = v_{f}

Total energy of the projectile on the Earth=1/2m_{p}^{2} – 1/2mv_{esc}^{2}

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth =1/2mv_{f}^{2}

From the law of conservation of energy, we have,

1/2m_{p}^{2} – 1/2mv_{esc}^{2} = 1/2mv_{f}^{2}

v_{f} = √v_{p}^{2}-v_{esc}^{2}

=√(3 v_{esc})^{2} –(v_{esc})^{2}

=√8 v_{esc}

=31.68km/s

** Problem:-** ( Class 11 Physics Gravitation )

Calculate the escape velocity on the surface of the moon? Given that the radius of the moon is 1.7×10^{6}m and the mass of the moon is 10^{22}kg.

__Answer:-__

V_{e}= √2GM/R_{e}

R_{m} = 1.7×10^{6}m

M= 10^{22}kg

G=6.67×10^{-11}Nm^{2}/kg^{2}

V_{e }= √2×6.67×10^{-11}x10^{22}/1.7×10^{6}

=2.4×10^{3}m/s

The escape velocity on the surface of the moon is 2.4×10^{3}m/s.

** Problem:- **What is the escape velocity from Jupiter given that the mass is 300 times that of the Earth’s and its radius is 10times larger?

__Answer:-__

M_{J}=300M_{e}

R_{J}=10R_{e}

V_{e}= √2GM_{e}/R_{e} = V_{e}= √2GM_{J}/R_{J}

= √2Gx300M_{e}/10x R_{e}

After putting the values,

=√30×11.2m/s

=61.3km/s

The escape velocity on the surface of the Jupiter is 61.3km/s.

__Artificial Satellites:__

- Human built objects orbiting the earth for practical uses. There are several purposes which these satellites serve.
- Example:- Practical Uses of Artificial satellites
- Communication
- Television broadcasts
- Weather observation
- Military support
- Navigation
- Scientific research