Suppose you are playing cricket and you hit the ball with some velocity, the ball will again come down on the surface of the ground. But in case you hit the ball with greater velocity, the ball will escape out of the gravitational field. This is what we call escape velocity. Let us learn more about this.

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## Escape Velocity

To understand the term Escape Velocity let us carry out a small activity.

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Suppose you are having a ball in your hand. You throw a ball in the air and you see that it comes down. We know that it comes back because of the force of gravitation. Now you throw the ball with greater velocity, in this case, the balls reach a greaterÂ height but eventually, it comes down and falls on the surface of the ground.

Because it still experiences the force of attraction by the surface of the earth. Now suppose you throw the ball with such a high velocity that it never comes back on the ground. This is where escape velocity comes into the picture. Escape velocity is the velocity that a body must attain to escape a gravitational field.

So if you throw the ball with the velocity which is at leastÂ equal to the escape velocity, in that case, the ball will go out of the gravitational field.

### Mathematical Expression

Suppose the ball is initially in your hand. So that is the initialÂ position of the ball. Now throw the ball at a greater velocity, that it never comes back. As we don’t know where did the ball go, so its final velocity isÂ **âˆž**. So with this assumption let us derive the expression.

#### At initial position,

Total energy = Kinetic energy + Potential energy

Or, T.E = K.E + P.E

Kinetic Energy =Â \( \frac{1}{2} \) mvÂ²

Potential Energy =Â \( \frac{- GMm}{R_e + h}\)

Here M is the mass of the earth and m is the mass of the ball.Â At the earth’s surface, P.E (0) = 0.

Hence, T. E (0) =Â Â \( \frac{1}{2} \) mvÂ²

#### At final position (âˆž),

K.E = \( \frac{1}{2} \) mv_{f}Â²

P.E =Â \( \frac{-GM_em}{R_e + h} \)Â = 0 {h =âˆž}

Now by the law of conservation of energy, the total energy at the initial position should be equal to the final position.

T.E (âˆž)Â = T.E (0)

Or, \( \frac{1}{2} \) mv_{f}Â² =Â \( \frac{1}{2} \) m v_{i}Â² –Â \( \frac{GM_em}{R_e + h} \)

L.H.S has to be alawys +ve, which implies

\( \frac{1}{2} \) m v_{i}Â² –Â \( \frac{GM_em}{R_e + h} \)Â Â â‰¥ 0

â‡’Â \( \frac{1}{2} \) m v_{i}Â² =Â \( \frac{GM_em}{R_e + h} \)

â‡’ v_{i}Â² =Â \( \frac{2GM_e}{R_e + h} \)

Assume the ball is thrown from earth’s surface.

h << R_{eÂ }â‡’R_{e}+h ~Â R_{e}

â‡’Â v_{i}Â² =Â \( \frac{2GM_e}{R_e } \)

â‡’ v_{i} =Â Â \( \sqrt{\frac{2GM_e}{R_e }} \)

This is the velocity in which the objects never comes back. In terms of ‘ g ‘

g =Â \( \frac{GM}{RÂ²} \)

â‡’ gR_{e }=Â Â \( \frac{GM_e}{R_e} \)

v_{eÂ Â }=Â âˆš(2gR_{e})

## Solved Questions For You

Q1. The escape velocity of a particle depends on its mass m as:

- mass m as mÂ²
- as m
^{-1} - mass m as m
^{0} - as m
^{1}

Ans: C. Escape velocity, v_{e}= âˆš2gR. It is independent of the mass of the particle. Thus, it will depend on m^{0}

Q2.Â The earth retains its atmosphere. This is due to:

- the special shape of the earth
- the escape velocity which is greater than the mean speed of the atmospheric molecules.
- the escape velocity which is lessÂ than the mean speed of the atmospheric molecules.
- the suns gravitational effect.

Ans: B. The earth retains its atmosphere. This is due to the escape velocity is been greater than the mean speed of theÂ atmospheric molecules.

When earth is near know it move faster some gravity of earth act on it and it produce restriction so speed may be slownear sun

When earth is near the sun how it move faster some gravity of sun act on it and it produce restriction and speed may be slow down