If the distance between two particles is reduced to half, the gravitational attraction between them will be

Options

A. halved
B. quadrupled
C. doubled
D. reduced to a quarter

Answer

To solve the problem, we need to understand how gravitational force changes with distance. The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( d \) is given by Newton's law of gravitation:

\[ F = \frac{G \cdot m_1 \cdot m_2}{d^2} \]

where \( G \) is the gravitational constant.

### Step-by-Step Solution:

1. **Identify the initial gravitational force**:
The initial gravitational force \( F \) when the distance between the two masses is \( d \) is given by:
\[ F = \frac{G \cdot m_1 \cdot m_2}{d^2} \]

2. **Change the distance**:
According to the problem, the distance between the two particles is reduced to half. Therefore, the new distance \( d' \) is:
\[ d' = \frac{d}{2} \]

3. **Calculate the new gravitational force**:
We need to find the new gravitational force \( F' \) when the distance is \( d' \):
\[ F' = \frac{G \cdot m_1 \cdot m_2}{(d')^2} \]
Substituting \( d' = \frac{d}{2} \):
\[ F' = \frac{G \cdot m_1 \cdot m_2}{\left(\frac{d}{2}\right)^2} \]

4. **Simplify the expression**:
The expression for \( F' \) becomes:
\[ F' = \frac{G \cdot m_1 \cdot m_2}{\frac{d^2}{4}} = \frac{G \cdot m_1 \cdot m_2 \cdot 4}{d^2} \]

5. **Relate \( F' \) to \( F \)**:
Now we can relate \( F' \) to the original force \( F \):
\[ F' = 4 \cdot \frac{G \cdot m_1 \cdot m_2}{d^2} = 4F \]

6. **Conclusion**:
Thus, when the distance between the two particles is reduced to half, the gravitational attraction between them becomes four times greater. Therefore, the answer is that the gravitational attraction will be quadrupled.

### Final Answer:
The gravitational attraction between them will be **4 times** greater.

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