The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R form the surface of the earth is (g=acceleration due to gravity at the surface of the earth)

Options

A. `g/9`
B. `g/3`
C. `g/4`
D. g

Answer

To solve the problem of finding the acceleration of a body due to the attraction of the Earth at a distance of 2R from the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Distance from the Center of the Earth**:
- The radius of the Earth is denoted as \( R \).
- The distance from the surface of the Earth to the center is \( R \).
- Therefore, at a distance of \( 2R \) from the surface, the total distance from the center of the Earth is \( R + 2R = 3R \).

2. **Use the Formula for Gravitational Force**:
- The gravitational force \( F \) acting on a mass \( m \) at a distance \( d \) from the center of the Earth is given by:
\[ F = \frac{G M m}{d^2} \]
- Here, \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

3. **Substituting the Distance**:
- In our case, \( d = 3R \). Thus, the force becomes:
\[ F = \frac{G M m}{(3R)^2} = \frac{G M m}{9R^2} \]

4. **Find the Acceleration**:
- The acceleration \( a \) due to gravity at this distance can be found using Newton's second law, where \( a = \frac{F}{m} \):
\[ a = \frac{F}{m} = \frac{G M m}{9R^2 m} = \frac{G M}{9R^2} \]

5. **Relate to Surface Gravity \( g \)**:
- The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
- We can express the acceleration at \( 3R \) in terms of \( g \):
\[ a = \frac{G M}{9R^2} = \frac{1}{9} \cdot \frac{G M}{R^2} = \frac{g}{9} \]

6. **Conclusion**:
- Therefore, the acceleration of a body at a distance of \( 2R \) from the surface of the Earth is:
\[ a = \frac{g}{9} \]

### Final Answer:
The acceleration of a body due to the attraction of the Earth at a distance \( 2R \) from the surface is \( \frac{g}{9} \).

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