The average density of the earth

Options

A. Is directly proportional to g
B. Is inversely proportional to g
C. Does not depend on g
D. Is a complex function of g

Answer

To solve the problem regarding the relationship between the average density of the Earth and the acceleration due to gravity (G), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( M \) be the mass of the Earth.
- Let \( R \) be the radius of the Earth.
- Let \( \rho \) be the average density of the Earth.
- Let \( g \) be the acceleration due to gravity at the surface of the Earth.

2. **Using the Formula for Gravitational Acceleration**:
- The formula for gravitational acceleration at the surface of the Earth is given by:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the universal gravitational constant.

3. **Expressing Mass in Terms of Density**:
- The mass \( M \) of the Earth can also be expressed in terms of its density and volume. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
- Therefore, the mass can be expressed as:
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]

4. **Substituting Mass into the Gravitational Formula**:
- Substituting the expression for mass \( M \) into the gravitational acceleration formula gives:
\[ g = \frac{G \cdot \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} \]

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

6. **Analyzing the Relationship**:
- From the final equation, we can see that the average density \( \rho \) is directly proportional to \( g \) (acceleration due to gravity) when \( R \) (radius of the Earth) and \( G \) (universal gravitational constant) are constants.

### Conclusion:
The average density of the Earth is directly proportional to the acceleration due to gravity \( g \).