# A rocket has to be launched from earth in such a way that it never returns.

Question:

A rocket has to be launched from earth in such a way that it never returns. If $\mathrm{E}$ is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.

1. (1) $\frac{\mathrm{E}}{64}$

2. (2) $\frac{E}{32}$

3. (3) $\frac{\mathrm{E}}{4}$

4. (4) $\frac{\mathrm{E}}{16}$

Correct Option: , 4

Solution:

(4) Escape velocity,

$v_{c}=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G \rho V}{R}}$

$=\sqrt{\frac{2 G S \times 4 \pi R^{3}}{R}}=\sqrt{\frac{8}{3} \pi \rho G R^{2}}$

For moon, $v_{c}^{\prime}=\sqrt{\frac{8}{3} \pi \rho G R_{m}^{2}}$

Given, $\frac{4}{3} \pi R^{3}=64 \times \frac{4}{3} \pi R_{m}^{3}$ or $R_{m}=\frac{R}{4}$

$\therefore \quad v_{e}^{\prime}=\sqrt{\frac{8}{3} \pi \rho G\left(\frac{R}{4}\right)^{2}}=\frac{v_{c}}{4}$

$\frac{E}{E^{\prime}}=\frac{\frac{1}{2} m v_{e}^{2}}{\frac{1}{2} m v_{e}^{\prime 2}}=\frac{v_{e}^{2}}{v_{c}^{2}}=\frac{v_{e}}{\left(\frac{v_{e}}{4}\right)}=16$

or $E^{\prime}=\frac{E}{16}$