If `R` is the radius of the earth and `g` the acceleration due to gravity on the earth's surface, the mean density of the earth is

Options

A. `(4piG)/(3gR)`
B. `(3piR)/(4gG)`
C. `(3g)/(4piRG)`
D. `(piR)/(12G)`

Explanation

To find the mean density of the Earth given the radius \( R \) and the acceleration due to gravity \( g \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between gravity, mass, and radius**:
The acceleration due to gravity \( g \) at the surface of the Earth can be expressed using the formula:
\[ g = \frac{G M}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Express mass in terms of density**:
The mass \( M \) of the Earth can also be expressed in terms of its mean density \( \rho \) and volume. The volume \( V \) of the Earth (assuming it is a sphere) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass \( M \) can be written as:
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

3. **Substitute mass back into the gravity equation**:
Now, substitute the expression for \( M \) into the gravity equation:
\[ g = \frac{G \left(\rho \frac{4}{3} \pi R^3\right)}{R^2} \]

4. **Simplify the equation**:
Simplifying this gives:
\[ g = \frac{4}{3} \pi G \rho R \]

5. **Rearrange to find density**:
To find the mean density \( \rho \), rearrange the equation:
\[ \rho = \frac{3g}{4 \pi G R} \]

6. **Final expression for mean density**:
Thus, the mean density of the Earth is given by:
\[ \rho = \frac{3g}{4 \pi G R} \]

### Conclusion:
The mean density of the Earth in terms of the acceleration due to gravity \( g \), the radius \( R \), and the gravitational constant \( G \) is:
\[ \rho = \frac{3g}{4 \pi G R} \]