Assuming the earth to be a homogeneous sphere of radius `R`, its density in terms of `G` (constant of gravitation) and `g` (acceleration due to gravity on the surface of the earth)

Options

A. `3g//(4piRG)`
B. `4pig//(3RG)`
C. `4piRg//(3G)`
D. `4piRG//(3g)`

Explanation

To find the density of the Earth in terms of the gravitational constant \( G \) and the acceleration due to gravity \( g \), we can follow these steps:

### Step-by-Step Solution:

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

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

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

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

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

### Final Expression:
Thus, the density \( \rho \) of the Earth in terms of \( G \) and \( g \) is:
\[ \rho = \frac{3g}{4 \pi G R} \]