If masses of two point objects are tripled and distance between them is doubled , then gravitational force of attraction between them will

Options

A. Increases by 225 %
B. Decrease by 56%
C. Increases by 125%
D. Decreases by 144%

Answer

To solve the problem, we will use the formula for gravitational force and analyze the changes in mass and distance step by step.

### Step-by-Step Solution:

1. **Understand the Gravitational Force Formula**:
The gravitational force \( F \) between two point masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
where \( G \) is the gravitational constant.

2. **Identify the Changes**:
According to the problem:
- The masses are tripled: \( m_1 \) becomes \( 3m_1 \) and \( m_2 \) becomes \( 3m_2 \).
- The distance is doubled: \( r \) becomes \( 2r \).

3. **Substitute the New Values into the Formula**:
The new gravitational force \( F' \) can be expressed as:
\[ F' = \frac{G \cdot (3m_1) \cdot (3m_2)}{(2r)^2} \]

4. **Simplify the Expression**:
Calculating \( (2r)^2 \):
\[ (2r)^2 = 4r^2 \]
Now substituting this back into the equation for \( F' \):
\[ F' = \frac{G \cdot 9m_1 \cdot m_2}{4r^2} \]

5. **Relate \( F' \) to the Original Force \( F \)**:
We can express \( F' \) in terms of the original force \( F \):
\[ F' = \frac{9}{4} \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} = \frac{9}{4} F \]

6. **Calculate the Percentage Increase**:
To find the percentage increase in the gravitational force:
\[ \text{Percentage Increase} = \left( \frac{F' - F}{F} \right) \times 100 \]
Substituting \( F' = \frac{9}{4} F \):
\[ \text{Percentage Increase} = \left( \frac{\frac{9}{4} F - F}{F} \right) \times 100 \]
Simplifying this gives:
\[ = \left( \frac{\frac{9}{4} F - \frac{4}{4} F}{F} \right) \times 100 = \left( \frac{5}{4} \right) \times 100 = 125\% \]

### Conclusion:
The gravitational force of attraction between the two masses increases by **125%**.