Question

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

• A. $\frac{E}{16}$
• B. $\frac{E}{64}$
• C. $\frac{E}{4}$
• D. $\frac{E}{32}$

Answer 1

The correct option is A $\frac{E}{16}$

The energy given to an object, on the earth's surface, so that it escapes to infinity, is

$E=\frac{GMm}{R}=\frac{Gm}{R}\times \rho \times \frac{4}{3}\pi R^3$

$\Rightarrow E=\frac{4\pi Gm\rho R^2}{3}$

$\Rightarrow E\propto R^2$, as $m$ and $\rho$ are same for earth and moon.

$\Rightarrow \frac{E_e}{E_m}=\frac{R_e^2}{R_m^2}.......\left(i\right)$

Now, volume of earth is $64$ times volume of the moon.
$\Rightarrow \frac{4}{3}\pi R_e^3=64\times \frac{4}{3}\pi R_m^3$

$\Rightarrow \frac{R_e}{R_m}=\mathrm{4.......}\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get,

$\Rightarrow \frac{E_e}{E_m}=16$

Here, $E_e=E\Rightarrow E_m=\frac{E}{16}$

Hence, $\left(D\right)$ is the correct answer.

Alternate method: Escape velocity from earth's surface is given by, $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G\rho V}{R}}$ $=\sqrt{\frac{2G\rho \times 4\pi R^3}{3R}}=\sqrt{\frac{8}{3}\pi \rho G{R}^2}$ Similarly, for moon, $v_e^{\prime }=\sqrt{\frac{8}{3}\pi \rho G{R}_m^2}$ Now, volume of earth is $64$ times volume of the moon. $\Rightarrow \frac{4}{3}\pi R_e^3=64\times \frac{4}{3}\pi R_m^3$ $\Rightarrow \frac{R_e}{R_m}=4$ $\therefore v_e^{\prime }=\sqrt{\frac{8}{3}\pi \rho G{\left(\frac{R}{4}\right)}^2}=\frac{v_e}{4}$ Now, $\frac{E}{E^{\prime }}=\frac{\frac{1}{2}m{v}_e^2}{\frac{1}{2}m{v}_e^{\prime 2}}=\frac{v_e^2}{v_c^{\prime 2}}=\frac{v_e^2}{\left(\frac{v_e^2}{16}\right)}=16$ $\Rightarrow E^{\prime }=\frac{E}{16}$