A particle of mass `m` is subjected to an attractive central force of magnitude `k//r^(2)`, `k` being a constant. If at the instant when the particle is at an extreme position in its closed orbit, at a distance a from the centre of force, its speed is `sqrt(k//2ma)`, if the distance of other extreme position is b. Find `a//b`.

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the problem, we will use the principles of conservation of energy and angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Forces and Potential Energy
The attractive central force acting on the particle is given by:
\[ F = -\frac{k}{r^2} \]
The potential energy \( U \) associated with this force can be derived by integrating the force:
\[ U(r) = -\int F \, dr = -\int -\frac{k}{r^2} \, dr = -\frac{k}{r} + C \]
For simplicity, we can set the constant \( C = 0 \), so:
\[ U(r) = -\frac{k}{r} \]

### Step 2: Apply Conservation of Energy
At the extreme positions \( A \) and \( B \), the conservation of mechanical energy states:
\[ K.E_A + U_A = K.E_B + U_B \]
Where:
- \( K.E_A = \frac{1}{2} m v_A^2 \)
- \( U_A = -\frac{k}{a} \)
- \( K.E_B = \frac{1}{2} m v_B^2 \)
- \( U_B = -\frac{k}{b} \)

Substituting these into the conservation of energy equation gives:
\[ \frac{1}{2} m v_A^2 - \frac{k}{a} = \frac{1}{2} m v_B^2 - \frac{k}{b} \]

### Step 3: Use Angular Momentum Conservation
The angular momentum \( L \) is conserved, so:
\[ m v_A a = m v_B b \]
This simplifies to:
\[ v_B = \frac{a}{b} v_A \]

### Step 4: Substitute \( v_A \) and \( v_B \)
From the problem, we know:
\[ v_A = \sqrt{\frac{k}{2ma}} \]
Substituting this into the angular momentum equation gives:
\[ v_B = \frac{a}{b} \sqrt{\frac{k}{2ma}} \]

### Step 5: Substitute \( v_B \) into Energy Equation
Now substitute \( v_A \) and \( v_B \) into the energy conservation equation:
\[ \frac{1}{2} m \left(\frac{k}{2ma}\right) - \frac{k}{a} = \frac{1}{2} m \left(\frac{a^2}{b^2} \cdot \frac{k}{2ma}\right) - \frac{k}{b} \]

### Step 6: Simplify the Equation
This simplifies to:
\[ \frac{k}{4a} - \frac{k}{a} = \frac{a^2}{2b^2} \cdot \frac{k}{2a} - \frac{k}{b} \]
Multiplying through by \( 4ab^2 \) to eliminate denominators gives:
\[ k b^2 - 4k b^2 = 2a^2 k - 4a k \]
This simplifies to:
\[ -3k b^2 = 2a^2 k - 4a k \]

### Step 7: Rearranging to Form a Quadratic Equation
Rearranging gives:
\[ 2a^2 - 4ab + 3b^2 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( A = 2, B = -4b, C = 3b^2 \):
\[ a = \frac{4b \pm \sqrt{(-4b)^2 - 4 \cdot 2 \cdot 3b^2}}{2 \cdot 2} \]
\[ a = \frac{4b \pm \sqrt{16b^2 - 24b^2}}{4} \]
\[ a = \frac{4b \pm \sqrt{-8b^2}}{4} \]
This gives:
\[ a = \frac{4b \pm 2b\sqrt{2}i}{4} \]

### Step 9: Find the Ratio \( \frac{a}{b} \)
The two possible solutions for \( a \) yield:
1. \( a = 3b \)
2. \( a = b \)

Since the problem states we need \( \frac{a}{b} \):
Thus, the ratio \( \frac{a}{b} = 3 \).

### Final Answer
Therefore, the ratio \( \frac{a}{b} \) is:
\[ \frac{a}{b} = 3 \]