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, we can start from the formula for the acceleration due to gravity at the surface of the Earth, which is given by:

1. **Formula for acceleration due to gravity**:
\[ g = \frac{GM}{R^2} \]
where:
- \( g \) is the acceleration due to gravity,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

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

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

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

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

7. **Identifying the correct option**:
Looking at the options provided, we see that option 3 states:
\[ \text{Option 3: } \frac{3g}{4 \pi R G} \]
This matches our derived expression for the mean density of the Earth.

**Conclusion**: The correct answer is option 3: \( \frac{3g}{4 \pi R G} \).

---