The gravitational force between two bodies is F. If the mass of each body is doubled and the distance between them is halved, then the force between them will be

Options

A. `(4F)/9`
B. `(9F)/4`
C. 16F
D. `F//16`

Answer

To solve the problem, we need to use the formula for gravitational force between two bodies, which is given by:

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

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

### Step 1: Write the initial gravitational force
The initial gravitational force between the two bodies is given as:

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

### Step 2: Modify the masses and distance
According to the problem, the mass of each body is doubled, and the distance between them is halved. Therefore, the new masses and distance can be represented as:

- New mass \( m_1' = 2m_1 \)
- New mass \( m_2' = 2m_2 \)
- New distance \( r' = \frac{r}{2} \)

### Step 3: Substitute the new values into the gravitational force formula
Now, we will substitute these new values into the gravitational force formula:

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

Substituting the new values:

\[ F' = \frac{G \cdot (2m_1) \cdot (2m_2)}{\left(\frac{r}{2}\right)^2} \]

### Step 4: Simplify the expression
Now, let's simplify the expression:

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

This can be rewritten as:

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

### Step 5: Relate it back to the original force
We can relate this back to the original gravitational force \( F \):

\[ F' = 16 \cdot \frac{G \cdot m_1 \cdot m_2}{r^2} = 16F \]

### Conclusion
Thus, the new gravitational force \( F' \) is:

\[ F' = 16F \]

### Final Answer
The gravitational force between the two bodies after doubling the masses and halving the distance will be \( 16F \).

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