Question

Class 11th !

Derive an expression for escape velocity .​

Answer 1

Explanation:

Derive an expression for escape velocity. ... Escape velocity is the minimum velocity with which a body must be projected vertically upward so that it may just escape the surface of the Earth. Expression for escape velocity: Let a body of mass m be escaped from the gravitational field of the earth.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the mass of the planet be "M".
• Radius of the planet be "R".
• And, "m" be the mass of the body.

$\huge{\bold{\underline{Explanation:-}}}$

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From Law of conservation of energy,

$\large{\boxed{\tt E = K.E + P.E}}$

Where :-

• E = Total Energy,
• K.E = Kinetic energy,
• P.E = Potential Energy.

If the mass "m" is placed on the earth then it's Kinetic energy will be Zero.

Therefore,

${\large{\tt E = 0 + P.E}}$

But P.E = ${\tt \dfrac{- GMm}{R}}$

Substituting in the above equation,

$\large{\tt E = 0 + \dfrac{- GMm}{R}}$

Now,

$\large{\boxed{\tt E = - \dfrac{GMm}{R}}}$

The negative sign here indicates that the body is bound to the earth.

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But The Modulus of the Total energy equals the binding energy,

Therefore,

$\large{\boxed{\tt Binding \: Energy = |E|}}$

Now,

$\large{\tt Binding \: Energy = |E| = \left| \dfrac{- GMm}{R}\right|}$

Therefore,

$\large{\boxed{\tt Binding \: Energy = |E| = \dfrac{ GMm}{R}}}$

Binding energy:-

It is the energy when given to a body it escapes from a particular Planet's gravitational field.

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To escape the gravitational field of a planet this energy (Binding energy) is supplied,

This energy is supplied in The Form of Kinetic energy.

Therefore,

$\large{\boxed{\tt K.E = |E|}}$

Substituting the Formulas,

$\large{\tt \dfrac{1}{2}mv_e^2 = \dfrac{GMm}{R}}$

(Here ${\tt v_e}$ is Escape velocity of the body)

$\large{\tt \dfrac{1}{2}\cancel{m}v_e^2 = \dfrac{GM\cancel{m}}{R}}$

$\large{\tt \dfrac{1}{2}v_e^2 = \dfrac{GM}{R}}$

Now,

$\large{\tt v_e^2 = \dfrac{2GMm}{R}}$

$\large{\tt v_e = \sqrt{\dfrac{2GMm}{R}}}$

But we know,

$\large{\tt GM = gR^2}$

Substituting it,

$\large{\tt v_e = \sqrt{\dfrac{2gR^2}{R}}}$

$\large{\tt v_e = \sqrt{\dfrac{2gR\cancel{^2}}{\cancel{R}}}}$

This becomes,

$\huge{\boxed{\boxed{\tt v_e = \sqrt{2gR}}}}$

Hence derived.

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Some Important Values:-

• Escape velocity of Earth is 11.2 Km/sec.
• Escape velocity of Moon is 2.4 Km/sec

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Additional information:-

• Escape velocity is independent of mass "m" and angle of projection .

• Escape velocity depends on

1. Mass of the planet
2. Radius of planet.

• H₂gas escape from the earth's atmosphere as, RMS Velocity of H₂gas is greater than the escape velocity of the earth.

• For the same reason there is no atmosphere on the moon

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