The acceleration due to gravity on moon is only one sixth that on earth. Ratio of moon radius `(R_m)` to earth.s radius `(R_e)` should be:
a) `6/1` if both are assumed to have same density
b) `1/6` if both are assumed to have same density
c) `5//18` if `rho_e/rho_m` is given as 5/3
d) `9//5` if `rho_m/rho_e` is given as `3//5`

Options

A. only a & b are true
B. only b & c are true
C. only a, b & d are true
D. all are true

Answer

To solve the problem, we need to find the ratio of the radius of the Moon `(R_m)` to the radius of the Earth `(R_e)` based on the given conditions regarding their densities and the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Acceleration Due to Gravity**:
The acceleration due to gravity `g` at a planet's surface can be expressed as:
\[ g = \frac{G \cdot M}{R^2} \]
where `G` is the gravitational constant, `M` is the mass of the planet, and `R` is the radius of the planet.

2. **Expressing Mass in Terms of Density**:
The mass `M` can be expressed in terms of density `ρ` and volume `V`:
\[ M = \rho \cdot V \]
For a sphere, the volume `V` is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass becomes:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Substituting Mass into the Gravity Equation**:
Substituting the expression for mass into the gravity equation gives:
\[ g = \frac{G \cdot \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4}{3} \pi G \cdot \rho \cdot R \]
This shows that the acceleration due to gravity is directly proportional to the density and the radius of the planet.

4. **Setting Up the Ratio of Gravity on Earth and Moon**:
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth:
\[ g_m = \frac{1}{6} g_e \]
Using the expressions for gravity:
\[ \frac{g_e}{g_m} = 6 = \frac{\left( \frac{4}{3} \pi G \cdot \rho_e \cdot R_e \right)}{\left( \frac{4}{3} \pi G \cdot \rho_m \cdot R_m \right)} \]
Simplifying this gives:
\[ 6 = \frac{\rho_e \cdot R_e}{\rho_m \cdot R_m} \]

5. **Rearranging for the Radius Ratio**:
Rearranging the equation gives:
\[ \frac{R_m}{R_e} = \frac{1}{6} \cdot \frac{\rho_e}{\rho_m} \]

6. **Considering Different Density Ratios**:
- If both have the same density (`ρ_e = ρ_m`), then:
\[ \frac{R_m}{R_e} = \frac{1}{6} \]
- If `\frac{\rho_e}{\rho_m} = \frac{5}{3}`, then:
\[ \frac{R_m}{R_e} = \frac{1}{6} \cdot \frac{5}{3} = \frac{5}{18} \]
- If `\frac{\rho_m}{\rho_e} = \frac{3}{5}`, then:
\[ \frac{R_m}{R_e} = \frac{1}{6} \cdot \frac{3}{5} = \frac{1}{10} \]

### Final Answers:
- The ratio of the Moon's radius to the Earth's radius is:
- `1/6` if both are assumed to have the same density.
- `5/18` if `ρ_e/ρ_m` is given as `5/3`.
- `1/10` if `ρ_m/ρ_e` is given as `3/5`.