The acceleration due to gravity g and density of the earth `rho` are related by which of the following relations? (where G is the gravitational constant and `R_(E)` is the radius of the earth)

Options

A. `rho=(4piGR_(E))/(3g)`
B. `rho=(3g)/(4piGR_(E))`
C. `rho=(3G)/(4pigR_(E))`
D. `rho=(4pigR_(E))/(3G)`

Explanation

To find the relationship between the acceleration due to gravity \( g \) and the density of the Earth \( \rho \), we can follow these steps:

### Step 1: Understand the definitions
The density \( \rho \) of the Earth is defined as the mass \( M \) of the Earth divided by its volume \( V \). The formula for density is:
\[ \rho = \frac{M}{V} \]

### Step 2: Calculate the volume of the Earth
Assuming the Earth is a sphere, the volume \( V \) can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi R_E^3 \]
where \( R_E \) is the radius of the Earth.

### Step 3: Substitute the volume into the density formula
Substituting the volume of the Earth into the density formula gives us:
\[ \rho = \frac{M}{\frac{4}{3} \pi R_E^3} = \frac{3M}{4 \pi R_E^3} \]

### Step 4: Relate mass \( M \) to acceleration due to gravity \( g \)
From the law of universal gravitation, the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R_E^2} \]
where \( G \) is the gravitational constant.

### Step 5: Solve for mass \( M \)
Rearranging the equation for \( g \) gives us:
\[ M = \frac{g R_E^2}{G} \]

### Step 6: Substitute \( M \) back into the density equation
Now, substitute this expression for \( M \) back into the density equation:
\[ \rho = \frac{3}{4 \pi R_E^3} \cdot \frac{g R_E^2}{G} \]

### Step 7: Simplify the expression
This simplifies to:
\[ \rho = \frac{3g}{4 \pi G R_E} \]

### Conclusion
Thus, the relationship between the acceleration due to gravity \( g \) and the density of the Earth \( \rho \) is given by:
\[ \rho = \frac{3g}{4 \pi G R_E} \]