Acceleration due to gravity on moon is `1//6` of the acceleration due to gravity on earth. If the ratio of densities of earth `(rho_(e))` and moon `(rho_(m))` is `((rho_(e))/(rho_(m)))=5/3` then radius of moon `(R_(m))` in terms of `R_(e)` will be

Options

A. `5/18 R_(e)`
B. `1/6 R_(e)`
C. `3/18 R_(e)`
D. `/(2sqrt(3)) R_(e)`

Explanation

To find the radius of the moon \( R_m \) in terms of the radius of the Earth \( R_e \), we can follow these steps:

### Step 1: Understand the relationship between acceleration due to gravity and density
The acceleration due to gravity \( g \) on a celestial body can be expressed as:
\[ g = \frac{GM}{R^2} \]
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 celestial body can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass can be written as:
\[ M = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 3: Substitute mass into the gravity equation
Substituting the expression for mass into the gravity equation gives:
\[ g = \frac{G \left(\rho \frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} \pi G \rho R \]
This shows that \( g \) is directly proportional to the product of density \( \rho \) and radius \( R \).

### Step 4: Set up the ratio of gravitational accelerations
Given that the acceleration due to gravity on the moon \( g_m \) is \( \frac{1}{6} \) of that on Earth \( g_e \):
\[ \frac{g_m}{g_e} = \frac{1}{6} \]
Using the proportionality established, we can write:
\[ \frac{\rho_m R_m}{\rho_e R_e} = \frac{1}{6} \]

### Step 5: Substitute the ratio of densities
We are given that the ratio of densities is:
\[ \frac{\rho_e}{\rho_m} = \frac{5}{3} \]
This implies:
\[ \frac{\rho_m}{\rho_e} = \frac{3}{5} \]
Substituting this into the ratio of gravitational accelerations gives:
\[ \frac{3}{5} \cdot \frac{R_m}{R_e} = \frac{1}{6} \]

### Step 6: Solve for \( R_m \)
Now we can solve for \( R_m \):
\[ R_m = \frac{1}{6} \cdot \frac{5}{3} R_e \]
\[ R_m = \frac{5}{18} R_e \]

### Final Answer
Thus, the radius of the moon \( R_m \) in terms of the radius of the Earth \( R_e \) is:
\[ R_m = \frac{5}{18} R_e \]

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