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 1: Write the formulas 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 gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

### Step 2: Express mass in terms of density
The mass \( M \) of a sphere can be expressed in terms of its density \( \rho \) and volume \( V \):

\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Substitute mass into the gravity formula
For the Earth, we have:

\[ g_e = \frac{G M_e}{R^2} = \frac{G \left(\rho_e \cdot \frac{4}{3} \pi R^3\right)}{R^2} \]

This simplifies to:

\[ g_e = \frac{4}{3} \pi G \rho_e R \]

For the Moon, we have:

\[ g_m = \frac{G M_m}{r^2} = \frac{G \left(\rho_m \cdot \frac{4}{3} \pi r^3\right)}{r^2} \]

This simplifies to:

\[ g_m = \frac{4}{3} \pi G \rho_m r \]

### Step 4: Find the ratio of the gravitational accelerations
Now, we can find the ratio of the acceleration due to gravity on the Moon to that on the Earth:

\[ \frac{g_m}{g_e} = \frac{\frac{4}{3} \pi G \rho_m r}{\frac{4}{3} \pi G \rho_e R} \]

### Step 5: Simplify the ratio
The terms \( \frac{4}{3} \pi G \) cancel out:

\[ \frac{g_m}{g_e} = \frac{\rho_m r}{\rho_e R} \]

### Final Result
Thus, the ratio of the acceleration due to gravity on the surface of the Moon to that on the surface of the Earth is:

\[ \frac{g_m}{g_e} = \frac{\rho_m r}{\rho_e R} \]