The force between two objects decreases by 36% when the distance between them is increasedby 4 m. The original distance between them in meter is

Options

A. 4
B. 16
C. 10
D. 12

Answer

To solve the problem, we need to use the concept of gravitational force and the relationship between force and distance. The gravitational force \( F \) between two objects is given by the formula:

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

where:
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses of the two objects,
- \( r \) is the distance between the centers of the two objects.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The force decreases by 36% when the distance is increased by 4 m.
- This means the new force \( F' \) is \( 64\% \) of the original force \( F \).

\[ F' = 0.64 F \]

2. **Setting Up the Equations**:
- The original force is given by:

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

- The new force after increasing the distance by 4 m is:

\[ F' = \frac{G m_1 m_2}{(r + 4)^2} \]

3. **Equating the Forces**:
- Since \( F' = 0.64 F \), we can write:

\[ \frac{G m_1 m_2}{(r + 4)^2} = 0.64 \cdot \frac{G m_1 m_2}{r^2} \]

- We can cancel \( G m_1 m_2 \) from both sides (assuming they are non-zero):

\[ \frac{1}{(r + 4)^2} = 0.64 \cdot \frac{1}{r^2} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives us:

\[ r^2 = 0.64 (r + 4)^2 \]

5. **Expanding the Right Side**:
- Expanding \( (r + 4)^2 \):

\[ (r + 4)^2 = r^2 + 8r + 16 \]

- So we have:

\[ r^2 = 0.64 (r^2 + 8r + 16) \]

6. **Distributing 0.64**:
- Distributing \( 0.64 \):

\[ r^2 = 0.64r^2 + 5.12r + 10.24 \]

7. **Rearranging the Equation**:
- Rearranging gives:

\[ r^2 - 0.64r^2 - 5.12r - 10.24 = 0 \]

- This simplifies to:

\[ 0.36r^2 - 5.12r - 10.24 = 0 \]

8. **Multiplying Through by 100**:
- To eliminate the decimal, multiply through by 100:

\[ 36r^2 - 512r - 1024 = 0 \]

9. **Using the Quadratic Formula**:
- We can apply the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 36 \), \( b = -512 \), and \( c = -1024 \).

\[ r = \frac{512 \pm \sqrt{(-512)^2 - 4 \cdot 36 \cdot (-1024)}}{2 \cdot 36} \]

\[ = \frac{512 \pm \sqrt{262144 + 147456}}{72} \]

\[ = \frac{512 \pm \sqrt{409600}}{72} \]

\[ = \frac{512 \pm 640}{72} \]

- This gives two possible solutions:

\[ r = \frac{1152}{72} = 16 \quad \text{(not valid as it should be less than 16)} \]
\[ r = \frac{-128}{72} \quad \text{(not valid as distance cannot be negative)} \]

- The valid solution is:

\[ r = 12 \text{ meters} \]

### Final Answer:
The original distance between the two objects is **12 meters**.