Question

derive an expression for the escape velocity of an object from the Earth surface

Answer 1

Answer:

Refer the attachment for diagram.

Let us consider the earth to be a sphere with centre O, radius R and mass M. Let the mass of the object projected be having a mass 'm' which is projected from Point A from the sphere.

Let us extend the path of OA and mark points P and Q such that, OP = x and PQ = dx in length.

Let the minimum velocity required for escaping be v₀, then the kinetic energy of the object is:

⇒ K.E = 1/2 mv₀²

When the object projected is at point P, the gravitational force between object and the earth is given by:

$\bf{ F = \dfrac{GMm}{x^2} }$

Work done by body in moving against gravitational force is given by:

$\bf{ dW = F.dx}\\\\\\\bf{dW = \dfrac{GMm}{x^2}.dx}$

The total work done in taking the object from radius of the earth to infinity can be found by integrating the equation.

$\bf{W = \int\limits^\infty_R{dW} = \int\limits^\infty_R {\frac{GMm}{x^2}} .\, dx$

On simplifying the equation we get,

$\bf{ W = -GMm [\:\dfrac{1}{\infty} - \dfrac{1}{R} \:]$

$\bf{\dfrac{1}{\infty}\: \textbf{can be approximated to 0 and we get,}$

$\implies \bf{ W = \dfrac{GMm}{R} }$

To escape the gravitational pull, the work done must be equal to Kinetic energy of the body.

$\implies \bf{ \dfrac{1}{2} mv_0\:^{2} = \dfrac{GMm}{R} }$

'm' gets cancelled on both sides and we get,

$\implies \bf{ \dfrac{v_0\:^{2} }{2} = \dfrac{GM}{R}$

$\implies \bf{ v_0 = \sqrt{ \dfrac{2GM}{R} }$

$\textbf{We know that,} \\\\\implies \bf{ g = \dfrac{GM}{R}$

Hence substituting we get,

$\implies \bf{ v_o = \sqrt{2gR} \implies \sqrt{ 2 \times 9.8 \times 6.4 \times 10^6} \implies 11.2 \times 10^3 m/s$

Which can be simplified as 11.2 km/sec

Hence Derived !