Consider a ring of mass m and radius r. Maximum gravitational intensity on the axis of the ring has value.

Options

A. `(Gm)/(r^(2))`
B. `(2Gm)/(sqrt(3)r^(2))`
C. `(2Gm)/(3sqrt(3)r^(2))`
D. `(Gm)/(2 r^(2))`

Answer

To find the maximum gravitational intensity on the axis of a ring of mass \( m \) and radius \( r \), we will follow these steps:

### Step 1: Understand the Gravitational Intensity Formula
The gravitational intensity \( G_P \) at a point \( P \) on the axis of the ring can be expressed as:
\[ G_P = \frac{Gm x}{(r^2 + x^2)^{3/2}} \]
where:
- \( G \) is the gravitational constant,
- \( m \) is the mass of the ring,
- \( r \) is the radius of the ring,
- \( x \) is the distance from the center of the ring to the point \( P \) along the axis.

### Step 2: Differentiate the Gravitational Intensity
To find the maximum gravitational intensity, we need to differentiate \( G_P \) with respect to \( x \) and set the derivative equal to zero.

Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Let \( u = Gmx \) and \( v = (r^2 + x^2)^{3/2} \).

### Step 3: Apply the Quotient Rule
We find:
\[ \frac{du}{dx} = Gm \]
\[ \frac{dv}{dx} = \frac{3}{2}(r^2 + x^2)^{1/2} \cdot 2x = 3x(r^2 + x^2)^{1/2} \]
Now, applying the quotient rule:
\[ \frac{dG_P}{dx} = \frac{(r^2 + x^2)^{3/2} \cdot Gm - Gmx \cdot 3x(r^2 + x^2)^{1/2}}{(r^2 + x^2)^3} \]
Setting this equal to zero gives:
\[ (r^2 + x^2)^{3/2} \cdot Gm - Gmx \cdot 3x(r^2 + x^2)^{1/2} = 0 \]

### Step 4: Simplify the Equation
This simplifies to:
\[ Gm(r^2 + x^2)^{3/2} = 3Gmx^2(r^2 + x^2)^{1/2} \]
Dividing both sides by \( Gm(r^2 + x^2)^{1/2} \) (assuming \( Gm \neq 0 \)):
\[ (r^2 + x^2) = 3x^2 \]
Rearranging gives:
\[ r^2 = 2x^2 \]
Thus, we find:
\[ x = \frac{r}{\sqrt{2}} \]

### Step 5: Substitute Back to Find Maximum Gravitational Intensity
Now substitute \( x = \frac{r}{\sqrt{2}} \) back into the gravitational intensity formula:
\[ G_P = \frac{Gm \left(\frac{r}{\sqrt{2}}\right)}{(r^2 + \left(\frac{r}{\sqrt{2}}\right)^2)^{3/2}} \]
Calculating \( r^2 + \frac{r^2}{2} = \frac{3r^2}{2} \):
\[ G_P = \frac{Gm \left(\frac{r}{\sqrt{2}}\right)}{\left(\frac{3r^2}{2}\right)^{3/2}} = \frac{Gm \left(\frac{r}{\sqrt{2}}\right)}{\left(\frac{3\sqrt{3}}{8} r^3\right)} = \frac{8Gm}{3\sqrt{3}r^2} \]

### Final Result
Thus, the maximum gravitational intensity on the axis of the ring is:
\[ G_P = \frac{2Gm}{3\sqrt{3}r^2} \]