Question

A particle is projected from point A at a distance 4R from the centre of earth with speed v as shown. What should be the maximum speed of particle if minimum distance happens to be the radius of earth

Answer 1

$\Large\boxed{\sf{\quad(3)\quad\!\sqrt{\dfrac{2GM}{R}}\quad}}$

Let the mass of the particle be $\sf{m.}$ The gravitational force of attraction of the particle due earth when it is at a distance $\sf{r}$ from the center of the earth is,

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

Work done by this force to move the particle through a small distance $\sf{dr}$ towards earth is,

$\longrightarrow\sf{dW=\dfrac{GMm}{r^2}\ dr}$

Hence the total work done to move it from the point A $\sf{(r=4R)}$ towards the surface of the earth $\sf{(r=R),}$ where the minimum distance becomes the radius of the earth is,

$\displaystyle\longrightarrow\sf{W=\int\limits_{4R}^{R}\dfrac{GMm}{r^2}\ dr}$

$\displaystyle\longrightarrow\sf{W=GMm\int\limits_{4R}^{R}\dfrac{1}{r^2}\ dr}$

$\displaystyle\longrightarrow\sf{W=-GMm\left[\dfrac{1}{r}\right]_{4R}^{R}}$

$\displaystyle\longrightarrow\sf{W=-GMm\left(\dfrac{1}{R}-\dfrac{1}{4R}\right)}$

$\displaystyle\longrightarrow\sf{W=-\cdot\dfrac{3GMm}{4R}}$

For maximum $\sf{v_0}$ such that the minimum distance becomes the radius of the earth, the particle should come to rest at the earth's surface. Hence by work - energy theorem,

$\displaystyle\longrightarrow\sf{-\dfrac{1}{2}m\left(v_0\right)^2\cos^230^o=-\dfrac{3GMm}{4R}}$

[$\sf{\left(v_0\right)\cos30^o}$ is the velocity of the particle in the considered direction, i.e., towards the center of the earth.]

$\displaystyle\longrightarrow\sf{\dfrac{3\left(v_0\right)^2}{4}=\dfrac{3GM}{2R}}$

$\displaystyle\longrightarrow\sf{\left(v_0\right)^2=\dfrac{2GM}{R}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{v_0=\sqrt{\dfrac{2GM}{R}}}}}$