The acceleration due to gravity is `g` at a point distant `r` from the centre of earth of radius `R`. If `r lt R`, then

Options

A. `gpropr`
B. `g prop r^(2)`
C. `g prop r^(-1)`
D. `g prop r^(-2)`

Explanation

To solve the problem regarding the acceleration due to gravity \( g \) at a point distant \( r \) from the center of the Earth, where \( r < R \) (inside the Earth), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the scenario**:
We are given that the distance \( r \) is less than the radius \( R \) of the Earth. This means we are considering a point inside the Earth.

2. **Using the formula for gravitational force**:
The gravitational force \( F \) acting on an object of mass \( m \) at a distance \( r \) from the center of the Earth is given by:
\[ F = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

3. **Acceleration due to gravity**:
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth can be expressed as:
\[ g = \frac{F}{m} = \frac{G M}{r^2} \]

4. **Considering the mass of the Earth inside a sphere**:
When we are inside the Earth, only the mass of the Earth that is at a distance less than \( r \) contributes to the gravitational force. According to the shell theorem, the gravitational effect of a spherical shell of uniform density is zero inside the shell. Therefore, only the mass \( M' \) of the Earth that is within radius \( r \) contributes to the gravitational force:
\[ M' = \frac{M}{R^3} r^3 \]
where \( M \) is the total mass of the Earth.

5. **Substituting \( M' \) into the gravity formula**:
Now, substituting \( M' \) into the equation for \( g \):
\[ g = \frac{G M'}{r^2} = \frac{G \left(\frac{M}{R^3} r^3\right)}{r^2} \]

6. **Simplifying the expression**:
Simplifying this gives:
\[ g = \frac{G M}{R^3} r \]
This shows that the acceleration due to gravity \( g \) is directly proportional to the distance \( r \) from the center of the Earth when \( r < R \).

7. **Conclusion**:
Hence, we conclude that:
\[ g \propto r \quad \text{(for } r < R\text{)} \]

### Final Answer:
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth (where \( r < R \)) is directly proportional to \( r \).