The acceleration due to gravity on the moon is one-sixth that on earth. If the average density of moon is three-fifth that of earth, the moon's radius in terms of earth's radius `R_(e)` is -

Options

A. `0.16 R_e`
B. `0.27 R_e`
C. `0.32 R_e`
D. `0.36R_e`

Answer

To solve the problem, we need to find the radius of the moon (R_m) in terms of the radius of the Earth (R_e). We are given two key pieces of information:

1. The acceleration due to gravity on the moon (g_m) is one-sixth that on Earth (g_e).
2. The average density of the moon (ρ_m) is three-fifths that of the Earth (ρ_e).

### Step-by-Step Solution:

**Step 1: Write the formulas for gravitational acceleration.**

The gravitational acceleration (g) at the surface of a celestial body is given by the formula:

\[ g = \frac{G \cdot M}{R^2} \]

where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the body,
- \( R \) is the radius of the body.

**Step 2: Express the mass of the Earth and the Moon using density.**

The mass can be expressed in terms of density and volume:

\[ M = \rho \cdot V \]

The volume \( V \) of a sphere is given by:

\[ V = \frac{4}{3} \pi R^3 \]

Thus, the mass of the Earth (M_e) and the Moon (M_m) can be expressed as:

\[ M_e = \rho_e \cdot \frac{4}{3} \pi R_e^3 \]
\[ M_m = \rho_m \cdot \frac{4}{3} \pi R_m^3 \]

Given that \( \rho_m = \frac{3}{5} \rho_e \), we can write:

\[ M_m = \left(\frac{3}{5} \rho_e\right) \cdot \frac{4}{3} \pi R_m^3 \]

**Step 3: Set up the equations for gravitational acceleration.**

From the problem, we have:

\[ g_m = \frac{1}{6} g_e \]

Substituting the expressions for gravitational acceleration:

\[ \frac{G \cdot M_m}{R_m^2} = \frac{1}{6} \cdot \frac{G \cdot M_e}{R_e^2} \]

**Step 4: Substitute the masses into the equation.**

Substituting the expressions for \( M_e \) and \( M_m \):

\[ \frac{G \cdot \left(\frac{3}{5} \rho_e \cdot \frac{4}{3} \pi R_m^3\right)}{R_m^2} = \frac{1}{6} \cdot \frac{G \cdot \left(\rho_e \cdot \frac{4}{3} \pi R_e^3\right)}{R_e^2} \]

**Step 5: Simplify the equation.**

Cancel \( G \), \( \frac{4}{3} \pi \), and \( \rho_e \) from both sides:

\[ \frac{\frac{3}{5} R_m^3}{R_m^2} = \frac{1}{6} \cdot \frac{R_e^3}{R_e^2} \]

This simplifies to:

\[ \frac{3}{5} R_m = \frac{1}{6} R_e \]

**Step 6: Solve for \( R_m \).**

Rearranging gives:

\[ R_m = \frac{1}{6} \cdot \frac{5}{3} R_e \]

Calculating this gives:

\[ R_m = \frac{5}{18} R_e \]

### Final Answer:

The radius of the moon in terms of the radius of the Earth is:

\[ R_m = \frac{5}{18} R_e \]