One goes from the centre of the earth to a distance two third the radius of the earth. The acceleration due to gravity is highest at

Options

A. the centre of the earth
B. a distance half the radius of the earth
C. a distance one-third of the radius of the earth
D. a distance two-third of the radius of the earth

Answer

To solve the problem of where the acceleration due to gravity is highest when moving from the center of the Earth to a distance of two-thirds the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept of Gravitational Acceleration Inside the Earth**:
- The acceleration due to gravity (g) inside a spherical shell of uniform density varies linearly with the distance from the center of the Earth. At a distance \( r \) from the center, the gravitational acceleration \( g' \) can be expressed as:
\[ g' = \frac{G M(r)}{r^2} \]
where \( M(r) \) is the mass enclosed within radius \( r \).

2. **Calculating the Mass Enclosed**:
- The mass enclosed within a radius \( r \) can be calculated using the formula:
\[ M(r) = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \]
where \( \rho \) is the density of the Earth and \( V \) is the volume of the sphere.

3. **Substituting the Mass in the Gravitational Formula**:
- Substituting \( M(r) \) into the gravitational acceleration formula gives:
\[ g' = \frac{G \left( \rho \cdot \frac{4}{3} \pi r^3 \right)}{r^2} = \frac{4 \pi G \rho}{3} r \]
- This shows that \( g' \) is directly proportional to \( r \).

4. **Analyzing the Values of \( r \)**:
- The value of \( g' \) increases linearly as you move away from the center of the Earth. Therefore, the gravitational acceleration will be highest at the maximum distance from the center, which is at the surface of the Earth.

5. **Identifying the Maximum Point**:
- Since the question specifies moving to a distance of two-thirds the radius of the Earth (i.e., \( \frac{2}{3} R \)), we need to compare this with other distances:
- At the center of the Earth (\( r = 0 \)), \( g' = 0 \).
- At \( \frac{1}{3} R \) and \( \frac{1}{2} R \), \( g' \) will be less than at \( \frac{2}{3} R \).
- At the surface of the Earth (\( r = R \)), \( g' \) is at its maximum.

6. **Conclusion**:
- Therefore, the acceleration due to gravity is highest at the surface of the Earth (i.e., at a distance equal to the radius of the Earth, \( R \)). However, since the question is about the distance of two-thirds the radius, we conclude that the gravitational acceleration is highest at this point compared to any point closer to the center.

### Final Answer:
The acceleration due to gravity is highest at the surface of the Earth, but among the given options, it is highest at a distance of two-thirds the radius of the Earth.