Physics - Gravitation Question with Solution | TestHub

According to the principal of conservation of energy,

Initial kinetic energy + initial potential energy,

= final kinetic energy + final potential energy

$\Rightarrow \frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^{\prime 2}+0$

$\Rightarrow \frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$..... (i)

As consider,$v_e=$escape velocity

$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$........ (ii)

$\therefore$From equation (i) and (ii), we get

$\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_e^2$...... (iii)

$\Rightarrow v^{\prime 2}=v^2-v_e^2$

Now,$\left.\begin{array}{c}{v}_e=11.2 \mathrm{km} {\mathrm{s}}^{-1}\\ v=3v_e\end{array}\right\}$.... (iv)

From equation (iii) and (iv), we get

$v^{\prime 2}={\left(3v_e\right)}^2-v_e^2$

$v^{\prime 2}=9v_e^2-v_e^2=8v_e^2=8\times {\left(11.2\right)}^2$

$= 8\times {\left(11.2\right)}^2$

$\Rightarrow v^{\prime }=\sqrt{8\times {\left(11.2\right)}^2}=\sqrt{8}\times 11.2$

$v^{\prime }=2\times 1.414\times 11.2=31.68 \mathrm{km} {\mathrm{s}}^{-1}$

$\therefore$Speed of the body far away from the Earth,

$\mathrm{v}\text{'}=31.68 \mathrm{km} {\mathrm{s}}^{-1}=31.7 \mathrm{km} {\mathrm{s}}^{-1}$