The gravitational force of attraction between two celestial bodies is F. If these two bodies are moving away from each other and the new distance between them is twice that of the original distance after 5 years, then change in gravitational force between them is ____

Options

A. 2F
B. `(F)/(2)`
C. `(3F)/(4)`
D. `(F)/(4)`

Explanation

To solve the problem, we need to understand how gravitational force changes with distance. The formula for gravitational force \( F \) between two masses \( M_1 \) and \( M_2 \) separated by a distance \( D \) is given by:

\[ F = \frac{G \cdot M_1 \cdot M_2}{D^2} \]

Where \( G \) is the gravitational constant.

### Step-by-Step Solution:

1. **Identify the Initial Force**:
- Let's denote the initial gravitational force as \( F_1 \).
- According to the problem, the initial distance between the two bodies is \( D \).
- Therefore, the initial force can be expressed as:
\[ F_1 = \frac{G \cdot M_1 \cdot M_2}{D^2} \]

2. **Determine the New Distance**:
- After 5 years, the distance between the two bodies becomes \( 2D \).

3. **Calculate the New Gravitational Force**:
- Let's denote the new gravitational force as \( F_2 \).
- The new force can be expressed as:
\[ F_2 = \frac{G \cdot M_1 \cdot M_2}{(2D)^2} \]
- Simplifying this gives:
\[ F_2 = \frac{G \cdot M_1 \cdot M_2}{4D^2} \]

4. **Relate the New Force to the Initial Force**:
- We can express \( F_2 \) in terms of \( F_1 \):
\[ F_2 = \frac{1}{4} \cdot \frac{G \cdot M_1 \cdot M_2}{D^2} = \frac{F_1}{4} \]

5. **Calculate the Change in Gravitational Force**:
- The change in gravitational force can be calculated as:
\[ \text{Change in Force} = F_1 - F_2 \]
- Substituting the values we have:
\[ \text{Change in Force} = F - \frac{F}{4} \]
- This simplifies to:
\[ \text{Change in Force} = F - 0.25F = 0.75F = \frac{3F}{4} \]

### Final Answer:
The change in gravitational force between the two celestial bodies is \( \frac{3F}{4} \).

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