If `g` denotes the value of acceleration due to gravity at a point distance `r` from the centre of earth of radius `R`. If `r lt R`, then

Options

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

Answer

To solve the problem, we need to determine how the acceleration due to gravity \( g \) varies with the distance \( r \) from the center of the Earth when \( r < R \) (where \( R \) is the radius of the Earth).

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force \( F \) acting on a mass \( m \) at a distance \( r \) from the center of the Earth is given by Newton's law of gravitation:
\[ F = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Relating Force to Acceleration**:
The acceleration due to gravity \( g \) at distance \( r \) can be expressed as:
\[ g = \frac{F}{m} = \frac{G M}{r^2} \]
Here, \( g \) is the acceleration due to gravity at distance \( r \).

3. **Considering the Case When \( r < R \)**:
When \( r \) is less than \( R \), we need to consider only the mass of the Earth that is enclosed within the radius \( r \). According to the shell theorem, the gravitational attraction inside a uniform spherical shell of mass is zero. Thus, only the mass of the Earth within radius \( r \) contributes to the gravitational force.

4. **Finding the Mass Enclosed**:
The mass \( m_{\text{enclosed}} \) within radius \( r \) can be calculated using the density \( \rho \) of the Earth:
\[ m_{\text{enclosed}} = \rho \cdot \frac{4}{3} \pi r^3 \]
The density \( \rho \) can be expressed in terms of the total mass \( M \) of the Earth and its volume:
\[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \]

5. **Substituting Back into the Gravitational Formula**:
Now, substituting \( m_{\text{enclosed}} \) into the gravitational force equation:
\[ g = \frac{G m_{\text{enclosed}}}{r^2} = \frac{G \left(\rho \cdot \frac{4}{3} \pi r^3\right)}{r^2} \]
Simplifying this gives:
\[ g = \frac{G \cdot \frac{M}{\frac{4}{3} \pi R^3} \cdot \frac{4}{3} \pi r^3}{r^2} = \frac{G M}{R^3} r \]

6. **Conclusion**:
From the derived equation, we can see that:
\[ g \propto r \]
This means that the acceleration due to gravity \( g \) is directly proportional to the distance \( r \) from the center of the Earth when \( r < R \).

### Final Answer:
The relationship is \( g \propto r \) for \( r < R \).