Assuming that the earth is a sphere of radius `R_E` with uniform density, the distance from its centre at which the acceleration due to gravity is equal to `g/3` (g is the acceleration due to gravity on the surface of earth) is

Options

A. `(R_E)/3`
B. `(2R_E)/3`
C. `(R_E)/2`
D. `(R_E)/4`

Explanation

To find the distance from the center of the Earth at which the acceleration due to gravity is equal to \( \frac{g}{3} \), we can follow these steps:

### Step 1: Understand the relationship of gravitational acceleration inside the Earth
The gravitational acceleration \( g' \) at a distance \( r \) from the center of the Earth can be expressed as:
\[ g' = g \left( \frac{r}{R_E} \right) \]
where \( g \) is the acceleration due to gravity at the surface of the Earth and \( R_E \) is the radius of the Earth.

### Step 2: Set up the equation for the given condition
We are given that \( g' = \frac{g}{3} \). Therefore, we can set up the equation:
\[ \frac{g}{3} = g \left( \frac{r}{R_E} \right) \]

### Step 3: Simplify the equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{3} = \frac{r}{R_E} \]

### Step 4: Solve for \( r \)
Now, we can solve for \( r \):
\[ r = \frac{R_E}{3} \]

### Conclusion
Thus, the distance from the center of the Earth at which the acceleration due to gravity is equal to \( \frac{g}{3} \) is:
\[ \boxed{\frac{R_E}{3}} \]