Question

Derive an explain escape velocity . v = √2gr​ plzz answer correctly.... don't spam...
answer in long....

Answer 1

Escape Velocity

It is the minimum velocity required for a particle to escape from the gravitational field of the earth, and it is expressed as,

$\displaystyle\longrightarrow\sf {\underline {\underline {v=\sqrt{2gR}}}}$

Proof for the Expression of the Escape Velocity:-

Consider a particle of mass $\displaystyle\sf {m}$ which has to be projected from the surface of earth of mass $\displaystyle\sf {M}$ and radius $\displaystyle\sf {R.}$

The force required to escape the particle from the gravitational field of the earth is equal to the gravitational force of attraction of the particle due to the earth, i.e.,

$\displaystyle\longrightarrow\sf {F=\dfrac {GMm}{R^2}}$

The work done to project the particle outside the earth by a small distance $\displaystyle\sf {dR}$ is,

$\displaystyle\longrightarrow\sf {dW=\dfrac {GMm}{R^2}\ dR}$

Then the total work done to project the particle from the surface of the earth to infinity is,

$\displaystyle\longrightarrow\sf {W=\int\limits_R^{\infty}\dfrac {GMm}{R^2} dR}$

$\displaystyle\longrightarrow\sf {W=GMm\int\limits_R^{\infty}\dfrac {1}{R^2} dR}$

$\displaystyle\longrightarrow\sf {W=-GMm\left [\dfrac {1}{R}\right]_R^{\infty}}$

$\displaystyle\longrightarrow\sf {W=\dfrac {GMm}{R}}$

This work done is stored in the particle as its kinetic energy. Therefore,

$\displaystyle\longrightarrow\sf {\dfrac {1}{2}mv^2=\dfrac {GMm}{R}}$

$\displaystyle\longrightarrow\sf {v=\sqrt{\dfrac {2GM}{R}}\quad\quad\dots (1)}$

But,

$\displaystyle\longrightarrow\sf {GM=gR^2}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf {\underline {\underline {v=\sqrt{2gR}}}}$