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Step-by-Step Solution
Step 1: Understand the Problem
We are comparing the minimum energy required to launch a rocket such that it never returns (i.e., it escapes) from two different celestial bodies: the Earth and the Moon. We denote the energy needed from the Earth as $E$ and want to find the corresponding minimum energy from the Moon.
Step 2: Recall the Escape Energy Formula
The escape energy (which is essentially the escape potential energy magnitude) from a spherical body of mass $M$ and radius $R$ can be written as:
$ E_{\text{escape}} = \frac{GMm}{R} \,,$
where:
$G$ is the universal gravitational constant,
$M$ is the mass of the celestial body,
$m$ is the mass of the rocket,
$R$ is the radius of the celestial body.
Step 3: Use the Given Ratios for Earth and Moon
It is given that:
The Earth’s volume is 64 times that of the Moon.
The Earth and the Moon have the same density.
If $\rho$ is the common density, then the mass $M$ of a body is proportional to $\rho \times \text{volume}$. Thus:
$ \rho \cdot \frac{4}{3} \pi R_e^3 = 64 \left( \rho \cdot \frac{4}{3} \pi R_m^3 \right),
$
which simplifies to:
$ R_e^3 = 64\,R_m^3 \quad\Rightarrow\quad R_e = 4\,R_m.
$
Similarly, because they have the same density and the volume ratio is 64, the mass ratio is:
$ M_{\text{earth}} = 64 \, M_{\text{moon}}.
$
Step 4: Relate Earth’s and Moon’s Escape Energies
Let $E_{\text{earth}} = \frac{GM_{\text{earth}}m}{R_e}$ and $E_{\text{moon}} = \frac{GM_{\text{moon}}m}{R_m}.$ We want to find the ratio:
$ \frac{E_{\text{moon}}}{E_{\text{earth}}}
= \frac{\frac{GM_{\text{moon}}m}{R_m}}{\frac{GM_{\text{earth}}m}{R_e}}
= \frac{M_{\text{moon}}}{M_{\text{earth}}} \times \frac{R_e}{R_m}.
$
Substituting $M_{\text{moon}} = \frac{1}{64} \, M_{\text{earth}}$ and $R_e = 4\,R_m$, we get:
$ \frac{E_{\text{moon}}}{E_{\text{earth}}}
= \frac{1}{64} \times 4
= \frac{1}{16}.
$
Thus:
$ E_{\text{moon}} = \frac{E_{\text{earth}}}{16}.
$
Step 5: State the Final Answer
The minimum energy required to launch the rocket from the surface of the Moon, so that it never returns, is $\frac{E}{16}$. This corresponds to the correct answer given in the options.
