A particle of mass `m` is kept on the axis of a fixed circular ring of mass M and radius `R` at a distance `x` from the centre of the ring. Find the maximum gravitational force between the ring and the particle.

Answer

To find the maximum gravitational force between a particle of mass `m` placed on the axis of a fixed circular ring of mass `M` and radius `R` at a distance `x` from the center of the ring, we can follow these steps:

### Step 1: Gravitational Field Expression
The gravitational field \( E \) at a distance \( x \) from the center of the ring due to the ring can be expressed as:
\[ E = \frac{G M x}{(R^2 + x^2)^{3/2}} \]
where \( G \) is the gravitational constant.

### Step 2: Gravitational Force
The gravitational force \( F \) acting on the particle of mass \( m \) is given by:
\[ F = m \cdot E = \frac{G M m x}{(R^2 + x^2)^{3/2}} \]

### Step 3: Maximizing the Force
To find the maximum gravitational force, we need to maximize the expression for \( F \) with respect to \( x \). This involves taking the derivative of \( F \) with respect to \( x \) and setting it to zero:
\[ \frac{dF}{dx} = 0 \]

### Step 4: Differentiate the Force
Using the product and quotient rule, we differentiate \( F \):
\[ \frac{dF}{dx} = \frac{G M m}{(R^2 + x^2)^{3/2}} + \frac{G M m x \cdot \frac{d}{dx}(x)}{(R^2 + x^2)^{3/2}} - \frac{G M m x \cdot \frac{d}{dx}((R^2 + x^2)^{3/2})}{(R^2 + x^2)^3} \]

### Step 5: Setting the Derivative to Zero
Setting the derivative to zero to find critical points:
\[ \frac{G M m}{(R^2 + x^2)^{3/2}} - \frac{3 G M m x^2}{(R^2 + x^2)^{5/2}} = 0 \]

### Step 6: Solve for x
This simplifies to:
\[ \frac{1}{(R^2 + x^2)^{3/2}} = \frac{3 x^2}{(R^2 + x^2)^{5/2}} \]
Cross-multiplying gives:
\[ R^2 + x^2 = 3 x^2 \]
which simplifies to:
\[ R^2 = 2 x^2 \implies x = \frac{R}{\sqrt{2}} \]

### Step 7: Substitute x back into Force Equation
Now substitute \( x = \frac{R}{\sqrt{2}} \) back into the force equation:
\[ F_{\text{max}} = \frac{G M m \left(\frac{R}{\sqrt{2}}\right)}{\left(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\right)^{3/2}} \]
Calculating the denominator:
\[ R^2 + \frac{R^2}{2} = \frac{3R^2}{2} \]
Thus:
\[ F_{\text{max}} = \frac{G M m \left(\frac{R}{\sqrt{2}}\right)}{\left(\frac{3R^2}{2}\right)^{3/2}} = \frac{G M m \left(\frac{R}{\sqrt{2}}\right)}{\left(\frac{3\sqrt{3}R^3}{4}\right)} = \frac{4 G M m}{3\sqrt{3} R^2} \]

### Final Answer
The maximum gravitational force between the ring and the particle is:
\[ F_{\text{max}} = \frac{2 G M m}{3\sqrt{3} R^2} \]