Question

A rocket has to be launched from the Earth in such a way that it never returns. If $E$ is the minimum energy delivered by the rocket launcher, the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon is found to be $\frac{E}{N}$.
What is the value of $N$ ?
[Note: Assume $\rho_{ m }=\rho_{ e }$;
Volume of Earth $=64 \times$ Volume of moon]

Answer 1

The energy corresponding to escape speed is
$E =\frac{1}{2} m (2 gR )=\frac{1}{2} m \left(2 \frac{ GM }{ R ^{2}} R \right)$
$=\frac{1}{2} m \left(2 \frac{ GM }{ R }\right)=\frac{ GMm }{ R }$
$V _{ e } =64\, V _{ m }$
$\Rightarrow \frac{4}{3} \pi R _{ e }{ }^{3}=\frac{4}{3} \times \pi \times R _{ m }{ }^{3} \times 64$
$\Rightarrow R _{ e }=4\, R _{ m }$
$\because \rho_{ e }=\rho_{ m }$
$\frac{ M _{ e }}{ V _{ e }}=\frac{ M _{ m }}{ V _{ m }}$
$\Rightarrow \frac{ M _{ e }}{64\, V _{ m }}=\frac{ M _{ m }}{ V _{ m }}$
$\Rightarrow M _{ e }=64\, M _{ m }$
So $\frac{E_{m}}{E_{e}}=\frac{M_{m}}{R_{m}} \times \frac{R_{e}}{M_{c}}$
$=\frac{1}{64} \times 4=\frac{1}{16}...\left[\because R_{e}=4 R_{m}\right]$
$E_{m}=\frac{E}{16}$