Question

Calculate the velocity with which a body must be thrown vertically upward from the surface of the earth so that it may reach a height of $10$R, where R is the radius of the earth and is equal to $6.4\times {10}^8$m(earth's mass $=6\times {10}^{24}$kg, gravitational constant $G=6.7\times {10}^{-11}$ $N{m}^{2}k{g}^{-2}$)

Answer 1

The gravitational potential energy of a body of mass m on earth's surface is
$U\left(R\right)=-\frac{GMm}{R}$
where M is the mass of the earth(supposed to be concentrated at its centre) and R is the radius of the earth(distance of the particle from the centre of the earth). The gravitational energy of the same body at a height $10$R from earth's surface, i.e., at a distance $11$R from earth's centre is
$U\left(11R\right)=-\frac{GMm}{R}$
Therefore, change in potential energy.
$U\left(11R\right)-U\left(R\right)=-\frac{GMm}{11R}-(-\frac{GMm}{R})=\frac{10}{11}\frac{GMm}{R}$
This difference must come from the initial kinetic energy given to the body in sending it to that height. Now, suppose the body is thrown up with a vertical speed v, so that its initial kinetic energy is $1/2$ $m{v}^2$. Then
$\frac{1}{2}m{v}^2=\frac{10}{11}\frac{GMm}{R}$ or $v=\sqrt{\frac{20}{11}\frac{Gmm}{R}}$
Putting the given values:
$v=\sqrt{\left(\frac{20\times (6.7\times {10}^{-11}N{m}^{2}k{g}^{-2})\times (6\times {10}^{24}kg)}{11(6.4\times {10}^{6}m)}\right)}$
$=1.07\times {10}^4$m $s^{-1}$.