Assuming the earth to be a homogencous 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 )is

Options

A. `3 g // (4 pi RG)`
B. ` 4 pi g// (3 RG)`
C. ` 4 pi Rg (3G)`
D. ` 4 pi RG // (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 1: Use the formula for acceleration due to gravity
The acceleration due to gravity \( g \) at the surface of the Earth is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( M \) is the mass of the Earth and \( R \) is the radius of the Earth.

### Step 2: Rearrange the formula to find the mass \( M \)
From the equation above, we can rearrange it to express the mass \( M \) of the Earth:
\[ M = \frac{gR^2}{G} \]

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

### Step 4: Set the two expressions for mass equal to each other
Now we can set the two expressions for mass equal to each other:
\[ \frac{gR^2}{G} = \rho \left(\frac{4}{3} \pi R^3\right) \]

### Step 5: Solve for density \( \rho \)
To find the density \( \rho \), we can rearrange the equation:
\[ \rho = \frac{gR^2}{G} \cdot \frac{3}{4\pi R^3} \]
This simplifies to:
\[ \rho = \frac{3g}{4\pi G R} \]

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

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