Understanding Gravitational Field on the Ring Axis

The question asks for the maximum gravitational field intensity on the axis of a uniform circular ring. A uniform circular ring of mass M and radius R creates a gravitational field at points along its axis. The field varies depending on the distance from the center of the ring.

Gravitational Field Formula on the Axis

For a uniform circular ring of mass M and radius R, the gravitational field intensity \(E(x)\) at a point on its axis at a distance \(x\) from the center is given by the formula:

$$ E(x) = \frac{GMx}{(R^2 + x^2)^{3/2}} $$

Here:

• \(G\) is the gravitational constant.
• \(M\) is the total mass of the ring.
• \(R\) is the radius of the ring.
• \(x\) is the distance of the point from the center of the ring along the axis.

We need to find the value of \(x\) for which \(E(x)\) is maximum and then calculate this maximum value.

Finding the Maximum Gravitational Field Intensity

To find the maximum value of \(E(x)\), we need to differentiate the function \(E(x)\) with respect to \(x\) and set the derivative equal to zero. This process helps us find the critical points where the function reaches its maximum or minimum value.

The formula is \(E(x) = GMx (R^2 + x^2)^{-3/2}\).

Let's differentiate \(E(x)\) with respect to \(x\) using the product rule \((uv)' = u'v + uv'\):

Let \(u = GMx\) and \(v = (R^2 + x^2)^{-3/2}\).

$$ \frac{dE}{dx} = \frac{d}{dx} \left( GMx (R^2 + x^2)^{-3/2} \right) $$

$$ \frac{dE}{dx} = GM \left[ \frac{d}{dx}(x) \cdot (R^2 + x^2)^{-3/2} + x \cdot \frac{d}{dx}(R^2 + x^2)^{-3/2} \right] $$

$$ \frac{dE}{dx} = GM \left[ 1 \cdot (R^2 + x^2)^{-3/2} + x \cdot \left(-\frac{3}{2}\right) (R^2 + x^2)^{-3/2 - 1} \cdot \frac{d}{dx}(R^2 + x^2) \right] $$

$$ \frac{dE}{dx} = GM \left[ (R^2 + x^2)^{-3/2} - \frac{3}{2}x (R^2 + x^2)^{-5/2} \cdot (2x) \right] $$

$$ \frac{dE}{dx} = GM \left[ (R^2 + x^2)^{-3/2} - 3x^2 (R^2 + x^2)^{-5/2} \right] $$

Factor out \((R^2 + x^2)^{-5/2}\):

$$ \frac{dE}{dx} = GM (R^2 + x^2)^{-5/2} \left[ (R^2 + x^2)^{(-3/2) - (-5/2)} - 3x^2 \right] $$

$$ \frac{dE}{dx} = GM (R^2 + x^2)^{-5/2} \left[ (R^2 + x^2)^{1} - 3x^2 \right] $$

$$ \frac{dE}{dx} = GM (R^2 + x^2)^{-5/2} (R^2 + x^2 - 3x^2) $$

$$ \frac{dE}{dx} = GM (R^2 + x^2)^{-5/2} (R^2 - 2x^2) $$

Set the derivative to zero to find the value of \(x\) for maximum intensity:

$$ GM (R^2 + x^2)^{-5/2} (R^2 - 2x^2) = 0 $$

Since \(GM\) and \((R^2 + x^2)^{-5/2}\) are non-zero for any \(x\) along the axis (except potentially at infinity, which isn't where the max occurs), the term causing the derivative to be zero must be \((R^2 - 2x^2)\).

$$ R^2 - 2x^2 = 0 $$

$$ 2x^2 = R^2 $$

$$ x^2 = \frac{R^2}{2} $$

$$ x = \pm \frac{R}{\sqrt{2}} $$

Since \(x\) represents a distance along the axis from the center, we consider the magnitude \(x = \frac{R}{\sqrt{2}}\). The maximum gravitational field intensity occurs at a distance of \(R/\sqrt{2}\) on either side of the center along the axis.

Calculating the Maximum Intensity Value

Now, substitute \(x = \frac{R}{\sqrt{2}}\) back into the formula for \(E(x)\):

$$ E_{max} = E\left(x = \frac{R}{\sqrt{2}}\right) = \frac{GM \left(\frac{R}{\sqrt{2}}\right)}{\left(R^2 + \left(\frac{R}{\sqrt{2}}\right)^2\right)^{3/2}} $$

$$ E_{max} = \frac{\frac{GMR}{\sqrt{2}}}{\left(R^2 + \frac{R^2}{2}\right)^{3/2}} $$

$$ E_{max} = \frac{\frac{GMR}{\sqrt{2}}}{\left(\frac{3R^2}{2}\right)^{3/2}} $$

$$ E_{max} = \frac{\frac{GMR}{\sqrt{2}}}{\left(\frac{3}{2}\right)^{3/2} (R^2)^{3/2}} $$

$$ E_{max} = \frac{\frac{GMR}{\sqrt{2}}}{\left(\frac{3\sqrt{3}}{2\sqrt{2}}\right) R^3} $$

$$ E_{max} = \frac{GMR}{\sqrt{2}} \cdot \frac{2\sqrt{2}}{3\sqrt{3} R^3} $$

Cancel out terms:

$$ E_{max} = \frac{2GM}{3\sqrt{3} R^2} $$

This is the maximum gravitational field intensity on the axis of the uniform circular ring.

Comparing with Options

Let's compare our calculated maximum intensity with the given options:

Our calculated value matches Option 1.

Revision Table: Gravitational Field Concepts

Additional Information: Gravitational Potential

Besides the gravitational field, another important concept is the gravitational potential. The gravitational potential \(V(x)\) on the axis of a uniform circular ring at a distance \(x\) from the center is given by:

$$ V(x) = -\frac{GM}{\sqrt{R^2 + x^2}} $$

The gravitational field is related to the negative gradient of the potential:

$$ E(x) = -\frac{dV}{dx} $$

Let's verify this for the ring:

$$ V(x) = -GM (R^2 + x^2)^{-1/2} $$

$$ -\frac{dV}{dx} = - \frac{d}{dx} \left( -GM (R^2 + x^2)^{-1/2} \right) $$

$$ -\frac{dV}{dx} = GM \frac{d}{dx} (R^2 + x^2)^{-1/2} $$

$$ -\frac{dV}{dx} = GM \left( -\frac{1}{2} (R^2 + x^2)^{-3/2} \cdot 2x \right) $$

$$ -\frac{dV}{dx} = GM \left( -x (R^2 + x^2)^{-3/2} \right) $$

$$ -\frac{dV}{dx} = -\frac{GMx}{(R^2 + x^2)^{3/2}} $$

Wait, there's a sign difference. Gravitational field is generally defined as pointing towards the mass. On the positive x-axis, the field is in the positive x direction for positive x, so the field is \(\vec{E}(x) = E(x) \hat{i}\) where \(E(x)\) is the magnitude. The relation is \(\vec{E} = -\nabla V\). In 1D along the axis, this is \(E_x = -\frac{dV}{dx}\). Let's recheck the derivative:

$$ \frac{dV}{dx} = -GM \left( -\frac{1}{2} \right) (R^2 + x^2)^{-3/2} (2x) = GMx (R^2 + x^2)^{-3/2} $$

So, \(E_x = -\frac{dV}{dx} = -GMx (R^2 + x^2)^{-3/2}\). The magnitude is \(|E_x| = \frac{GM|x|}{(R^2 + x^2)^{3/2}}\). For \(x > 0\), the magnitude is \(\frac{GMx}{(R^2 + x^2)^{3/2}}\), which matches our field formula used for the maximum intensity calculation.

The gravitational potential is minimum where the field is maximum or minimum magnitude (excluding points at infinity). The potential is related to the work done by gravity.