Question

The escape velocity of a projectile on the Earth's surface is $11.2\text{km/s}$. A body is projected out with thrice this speed. What is the speed of the body $\left(\text{in km/s}\right)$ far away from the Earth? Ignore the presence of the Sun and other planets.

Answer 1

Let $v$ and $v^{\prime }$ be the initial and final speed of the body.
The Kinetic energies of the body will be $\frac{1}{2}m{v}^2\text{and}\frac{1}{2}m{v}^{\prime 2}$.
Potential Energies on the earth surface = $=\frac{GMm}{R}$ and
final potential energy $\left(at\infty \right)=0$
Applying the principle of conservation of energy i.e.,
Initial KE + Initial PE = Final KE + Final PE
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^{\prime 2}+0$
But $\frac{GMm}{R}=\frac{1}{2}m{v}_e^2\text{where}{v}_e$ is the escape speed and $v=3v_e$
$\therefore \frac{1}{2}m{v}^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}^{\prime 2}$
$\Rightarrow \frac{1}{2}m(3v_e{)}^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}^{\prime 2}$
$\Rightarrow v^{\prime }=\sqrt{8v_e}\Rightarrow v^{\prime }=\sqrt{8}\times 11.2=31.68\text{km/s}.$
The speed of the body far away from the earth is $31.68\text{km/s}$.