Two celestial bodies are separated by some distance. If the mass of any one of the bodies is doubled while the mass of other is halved then how far should they be taken so that the gravitational force between them becomes one-fourth ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Gravitational Force Formula
The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[ F = \frac{G m_1 m_2}{r^2} \]
where \( G \) is the gravitational constant.

### Step 2: Define the New Masses
According to the problem, one mass is doubled and the other mass is halved:
- New mass \( m_1' = 2m_1 \)
- New mass \( m_2' = \frac{m_2}{2} \)

### Step 3: Set Up the New Gravitational Force
The new gravitational force \( F' \) can be expressed as:
\[ F' = \frac{G m_1' m_2'}{r'^2} \]
Substituting the new masses:
\[ F' = \frac{G (2m_1) \left(\frac{m_2}{2}\right)}{r'^2} = \frac{G m_1 m_2}{r'^2} \]

### Step 4: Relate the New Force to the Original Force
According to the problem, the new gravitational force \( F' \) should be one-fourth of the original force \( F \):
\[ F' = \frac{F}{4} \]
Substituting the expressions for \( F \) and \( F' \):
\[ \frac{G m_1 m_2}{r'^2} = \frac{1}{4} \cdot \frac{G m_1 m_2}{r^2} \]

### Step 5: Simplify the Equation
Since \( G m_1 m_2 \) is common on both sides, we can cancel it out:
\[ \frac{1}{r'^2} = \frac{1}{4} \cdot \frac{1}{r^2} \]
This simplifies to:
\[ r'^2 = 4r^2 \]

### Step 6: Solve for the New Distance
Taking the square root of both sides gives us:
\[ r' = 2r \]
This means the distance should be doubled.

### Conclusion
To achieve a gravitational force that is one-fourth of the original force, the distance between the two celestial bodies should be doubled.

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