R and r are the radii of the earth and moon respectively , p_e and P_m are densities of earth and mon respectively . The ratio of the acceleration due to geavity on the surfaces of moon and earth is

Options

A. `R/r(rho_e)/rho_m`
B. `r/R(rho_e)/rho_m`
C. `r/R(rho_m)/rho_e`
D. `r/R(rho_m)/rho_e`

Answer

To find the ratio of the acceleration due to gravity on the surfaces of the Moon and the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( R \) be the radius of the Earth.
- Let \( r \) be the radius of the Moon.
- Let \( \rho_E \) be the density of the Earth.
- Let \( \rho_m \) be the density of the Moon.
- Let \( M \) be the mass of the Earth.
- Let \( m \) be the mass of the Moon.

2. **Formula for Acceleration due to Gravity**:
The acceleration due to gravity \( g \) on the surface of a celestial body is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant.

3. **Calculate the Acceleration due to Gravity for Earth**:
For Earth, the acceleration due to gravity \( g_E \) can be expressed as:
\[ g_E = \frac{GM}{R^2} \]

4. **Calculate the Mass using Density**:
The mass \( M \) of the Earth can be expressed in terms of its density and volume:
\[ M = \rho_E \times V \]
The volume \( V \) of a sphere is given by \( V = \frac{4}{3} \pi R^3 \). Therefore,
\[ M = \rho_E \times \frac{4}{3} \pi R^3 \]

5. **Substituting Mass into the Gravity Formula**:
Substituting the expression for \( M \) into the formula for \( g_E \):
\[ g_E = \frac{G \left( \rho_E \times \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4}{3} \pi G \rho_E R \]

6. **Calculate the Acceleration due to Gravity for Moon**:
Similarly, for the Moon, the acceleration due to gravity \( g_m \) is:
\[ g_m = \frac{Gm}{r^2} \]
where \( m \) can be expressed as:
\[ m = \rho_m \times \frac{4}{3} \pi r^3 \]
Thus,
\[ g_m = \frac{G \left( \rho_m \times \frac{4}{3} \pi r^3 \right)}{r^2} = \frac{4}{3} \pi G \rho_m r \]

7. **Finding the Ratio of Acceleration due to Gravity**:
Now, we can find the ratio of the acceleration due to gravity on the Moon to that on the Earth:
\[ \frac{g_m}{g_E} = \frac{\frac{4}{3} \pi G \rho_m r}{\frac{4}{3} \pi G \rho_E R} \]
The \( \frac{4}{3} \pi G \) terms cancel out:
\[ \frac{g_m}{g_E} = \frac{\rho_m r}{\rho_E R} \]

8. **Final Result**:
Therefore, the ratio of the acceleration due to gravity on the surfaces of the Moon and Earth is:
\[ \frac{g_m}{g_E} = \frac{r}{R} \cdot \frac{\rho_m}{\rho_E} \]