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(i) Derive escape velocity. 

(ii) Derive expression for the escape velocity of an object from the surface of planet. 

(iii) Does it depend on location from where it is projected?

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(i) Escape velocity on earth (or any other planet) is defined as the minimum velocity with which the body has to be projected vertically upwards from the surface of earth (or any other planet) so that it just crosses the gravitational field of earth (or of that planet) and never returns on its own.

(ii) Let earth be perfect sphere of mass M, radius R with centre at O. Let a body of mass m to be projected from a point A on the surface of earth (planet). Join OA and produce it further. Take two points P and Q at a distance x and (x + dx) from the centre of the earth.

Gravitational force of attraction on the body at P is F = \(\frac{GMm}{\mathrm x^2}\)

This much force has to be applied on the body in the upward direction.

Work done in taking the body against gravitational attraction from P to Q is.

dW = Fdx = \(\frac{GMm}{\mathrm x^2}\)dx

Total work done in taking the body against gravitational attraction from surface of earth (i.e., x = R) to a region beyond the gravitational field of earth (i.e., x = \(\infty\)) can be calculated by integrating the above expression with the limit, x = R to x = \(\infty\)

Thus total work done is

W = \(\int^{\infty}_R \frac{GMm}{\mathrm x^2}\)dx

= GMm\(\int^{\infty}_R\)x2 dx

= GMm\(\big[\frac{\mathrm x^{-2+1}}{-2+1}\big]^{\infty}_R\)

= -GMm\(\big[\frac{1}{\mathrm x}\big]^{\infty}_R\)

= -GMm\(\big[\frac{1}{\infty}-\frac{1}{R}\big]\)

\(\frac{GMm}{R}\)

This work is done at the cost of kinetic energy given to the body at the surface of the earth. If v1 is the escape velocity of the body projected from the surface of earth, then

Kinetic energy of the body,

\(\frac{1}{2}\)\(mv^2_e\)  = \(\frac{GMm}{R}\)

\(v^2_e=\sqrt{\frac{2GM}{R}}\)

\(v_e\) = \(\sqrt{\frac{2GM}{R}}\)..................(i)

But, g = \(\frac{GM}{R^2}\)

or, GM = gR2

Putting this value in (1),

\(v_e\) = \(\sqrt{\frac{2gR^2}{R}}\) ...........(2)

\(\sqrt{2gR}\)

(iii) The value of escape velocity depends upon the mass and radius of the planet of the surface from which the body is to be projected. Clearly, the value of escape velocity of a body will be different for different planets.

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