Two point objects of mass 2x and 3x are separated by a distance r. keeping the distance fixed, how much mass should be transferred from 3x to 2x , so that gravitational force between them becomes maximum ?

Options

A. `x/4`
B. `x/3`
C. `x/2`
D. `(2x)/(3)`

Answer

To solve the problem of how much mass should be transferred from the object with mass \(3x\) to the object with mass \(2x\) in order to maximize the gravitational force between them, we can follow these steps:

### Step 1: Define the masses after transfer
Let \(y\) be the mass transferred from \(3x\) to \(2x\). After the transfer:
- The mass of the first object (originally \(3x\)) becomes \(3x - y\).
- The mass of the second object (originally \(2x\)) becomes \(2x + y\).

### Step 2: Write the formula for gravitational force
The gravitational force \(F\) between two masses is given by the formula:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Substituting the new masses into this equation, we have:
\[ F = \frac{G \cdot (3x - y) \cdot (2x + y)}{r^2} \]

### Step 3: Simplify the expression
We can simplify this expression by factoring out the constant terms:
\[ F = \frac{G}{r^2} \cdot (3x - y)(2x + y) \]
Let \(k = \frac{G}{r^2}\), then:
\[ F = k \cdot (3x - y)(2x + y) \]

### Step 4: Expand the expression
Now, we expand the product:
\[ F = k \cdot (6x^2 + 3xy - 2xy - y^2) = k \cdot (6x^2 + xy - y^2) \]

### Step 5: Differentiate with respect to \(y\)
To find the maximum force, we need to differentiate \(F\) with respect to \(y\) and set the derivative equal to zero:
\[ \frac{dF}{dy} = k \cdot (x - 2y) = 0 \]

### Step 6: Solve for \(y\)
Setting the derivative to zero gives:
\[ x - 2y = 0 \implies y = \frac{x}{2} \]

### Step 7: Verify it's a maximum
To confirm that this value of \(y\) gives a maximum, we can take the second derivative:
\[ \frac{d^2F}{dy^2} = -2k \]
Since \(k > 0\), \(\frac{d^2F}{dy^2} < 0\), indicating that we have a maximum.

### Conclusion
Thus, the amount of mass that should be transferred from \(3x\) to \(2x\) to maximize the gravitational force between them is:
\[ \boxed{\frac{x}{2}} \]