The acceleration due to gravity on the moon is only one sixth that of earth.It the earth and moon are assume to have the same density, the ratio of the radii of moon and earth will be

Options

A. `(1)/(6)`
B. `(1)/((6)^(1//3))`
C. `(1)/(36)`
D. `(1)/((6)^(2//3))`

Answer

To find the ratio of the radii of the moon and the earth given that the acceleration due to gravity on the moon is one sixth that of the earth, and assuming both celestial bodies have the same density, we can follow these steps:

### Step 1: Understand the relationship between gravity and radius
The acceleration due to gravity (g) is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius.

### Step 2: Set up the equations for the moon and the earth
Let:
- \( g_E \) = acceleration due to gravity on Earth
- \( g_M \) = acceleration due to gravity on Moon

According to the problem:
\[ g_M = \frac{1}{6} g_E \]

Using the formula for gravity, we can express this as:
\[ \frac{g_M}{g_E} = \frac{GM}{R_M^2} \div \frac{GE}{R_E^2} = \frac{GM}{GE} \cdot \frac{R_E^2}{R_M^2} \]

### Step 3: Relate the masses of the moon and the earth
Since the density (\( \rho \)) is the same for both the moon and the earth, we can express the mass in terms of density and volume:
\[ M = \rho V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the mass can be expressed as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

For the moon and earth:
- \( M_M = \rho \cdot \frac{4}{3} \pi R_M^3 \)
- \( M_E = \rho \cdot \frac{4}{3} \pi R_E^3 \)

### Step 4: Substitute the masses into the gravity ratio
Now substituting the masses into the ratio:
\[ \frac{M_M}{M_E} = \frac{R_M^3}{R_E^3} \]

### Step 5: Substitute back into the gravity ratio equation
Substituting this back into the gravity ratio gives:
\[ \frac{g_M}{g_E} = \frac{R_E^2}{R_M^2} \cdot \frac{R_M^3}{R_E^3} = \frac{R_M}{R_E} \]

### Step 6: Set the ratio equal to the known value
From the problem, we know:
\[ \frac{g_M}{g_E} = \frac{1}{6} \]
Thus:
\[ \frac{R_M}{R_E} = \frac{1}{6} \]

### Final Result
The ratio of the radius of the moon to the radius of the earth is:
\[ \frac{R_M}{R_E} = \frac{1}{6} \]