Answer

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**Hint:**The minimum energy of the launcher corresponds to the gravitational potential energy of the rocket at the surface of the earth or at the surface of the moon. As density of both the moon and the earth is assumed to be equal and as the volume of the earth is given to be 64 times that of the earth, a relation between the mass of the earth and the mass of the moon can be obtained which will help us in determining the gravitational potential energy at of the launcher at each surface.

**Formulas used:**

The gravitational potential energy of an object at the surface of a more massive body is given by, ${U_G} = - \dfrac{{GMm}}{R}$ where $G$ is the gravitational constant, $M$ is the mass of the more massive body, $m$ is the mass of the object and $R$ is the distance from the centre of the massive body to the object.

The mass of an object is given by, $m = \rho V$ where $\rho $ is the density of the object and $V$ is the volume of the object.

**Complete step by step solution:**

Step 1: Sketch a rough figure of the rocket launching from the surface of the earth or the moon.

It is given that the densities of the moon and that of the earth are equal.

i.e., ${\rho _{earth}} = {\rho _{moon}} = \rho $ --------- (1)

It is also given that the volume of the earth is 64 times that of the moon.

i.e., ${V_{earth}} = 64{V_{moon}}$ -------- (2).

Let $E$ be the minimum energy of the launcher at the surface of the earth and $E$’ be the minimum energy of the launcher at the surface of the moon.

Let $m$ be the mass of the launcher.

Step 2: Obtain the relation between the mass of the earth and that of the moon using equations (1) and (2).

Multiply the L.H.S of equations (1) and (2) and the R.H.S of equations (1) and (2).

Then we have $\rho {V_{earth}} = 64\rho {V_{moon}}$ --------- (3)

Since the mass of any object is the product of its density and its volume, equation (3) gives us ${M_{earth}} = 64{M_{moon}}$ -------- (4)

Step 3: Using equation (2) obtain a relation for the radius of the earth and that of the moon.

From equation (2) we have ${V_{earth}} = 64{V_{moon}}$

$ \Rightarrow \dfrac{4}{3}\pi {R_{earth}}^3 = 64 \times \dfrac{4}{3}\pi {R_{moon}}^3$

The above equation then gives us ${R_{earth}}^3 = 64{R_{moon}}^3$

$ \Rightarrow {R_{earth}} = 4{R_{moon}}$ --------- (5)

Step 4: Express the minimum energy of the launcher at the surface of the moon and that of the earth.

The minimum energy of the launcher is given by the gravitational potential energy of the launcher at the surface of the earth and that of the moon i.e., $E = - {U_G}$

Now the minimum energy of the launcher at the surface of the moon will be

$E' = \dfrac{{G{M_{moon}}m}}{{{R_{moon}}}}$ ------- (6)

Then the minimum energy of the launcher at the surface of the earth is $E = \dfrac{{G{M_{earth}}m}}{{{R_{earth}}}}$ ------ (7)

On substituting equations (4) and (5) in equation (7) we get, $E = \dfrac{{64G{M_{moon}}m}}{{4{R_{moon}}}}$

$ \Rightarrow E = \dfrac{{16G{M_{moon}}m}}{{{R_{moon}}}} = 16E'$

$ \Rightarrow E' = \dfrac{E}{{16}}$

Thus the minimum energy of the launcher at the surface of the moon is $\dfrac{1}{{16}}$ times that at the surface of the earth.

**So the correct option is (B).**

**Note:**Here both the earth and the moon are assumed to have a spherical shape and so the volume of the earth and the moon is given as ${V_{earth}} = \dfrac{4}{3}\pi {R_{earth}}^3$ and ${V_{moon}} = \dfrac{4}{3}\pi {R_{moon}}^3$ respectively. The distance between the rocket launcher and the surface of the earth or the moon is taken as the radius of the earth sphere or the moon sphere.

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