Assuming the earth to be a sphere of uniform density, the acceleration due to gravity

Options

A. at a point outside the earth is inversely proportional to the square of its distance from the centre
B. at a point outside the earth is inversely proportional to its distance from the centre
C. at a point inside is zero
D. at a point inside is proportional to its distance from the centre

Explanation

To solve the problem regarding the acceleration due to gravity assuming the Earth to be a sphere of uniform density, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Gravitational Acceleration**:
The acceleration due to gravity (g) at a distance r from the center of a spherical mass (like Earth) can be expressed using Newton's law of gravitation.

2. **Gravitational Acceleration Outside the Earth**:
For a point outside the Earth (at a distance r from the center), the gravitational force can be calculated using the formula:
\[ g = \frac{GM}{r^2} \]
where G is the gravitational constant and M is the mass of the Earth.

3. **Mass of the Earth**:
The mass M of the Earth can be expressed in terms of its density (ρ) and volume (V). The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass of the Earth can be written as:
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]

4. **Substituting Mass in the Gravitational Formula**:
Substituting the expression for M into the formula for g gives:
\[ g = \frac{G \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{r^2} \]

5. **Acceleration Due to Gravity Inside the Earth**:
For a point inside the Earth at a distance r from the center, the gravitational acceleration is given by:
\[ g' = \frac{GM'}{r^2} \]
where M' is the mass enclosed within radius r. The mass M' can be expressed as:
\[ M' = \rho \cdot \frac{4}{3} \pi r^3 \]
Thus, the gravitational acceleration inside the Earth becomes:
\[ g' = \frac{G \left(\rho \cdot \frac{4}{3} \pi r^3\right)}{r^2} = \frac{4}{3} \pi G \rho r \]
This shows that g' is directly proportional to r.

6. **Conclusion**:
- Outside the Earth: \( g \propto \frac{1}{r^2} \)
- Inside the Earth: \( g' \propto r \)

### Final Answer:
The acceleration due to gravity outside the Earth is inversely proportional to the square of the distance from the center of the Earth, while inside the Earth, it is directly proportional to the distance from the center.