Find the gravitational force of attraction between the ring and sphere as shown in the figure, where

Q: Why can we treat the sphere as a point mass?

Newton's Shell Theorem proves that any spherically symmetric mass distribution exerts a gravitational force on external objects exactly as if all its mass were concentrated at its geometric center.

Q: Can we treat the ring as a point mass too?

No! The ring is not spherically symmetric. Its mass is distributed in a circle, so the distance from each point on the ring to the sphere's center varies, requiring the specific ring field formula.

Q: What if the distance was much larger than $R$?

If the distance was extremely large, the $R^2$ term in the denominator would become negligible, and the field would approximate $Gm/d^2$, making the ring behave like a point mass.

Q: Why does the field formula have a $3/2$ power in the denominator?

The $3/2$ power comes from resolving the gravitational force vectors. The distance from any mass element on the ring to the point is $\sqrt{d^2 + R^2}$, giving a $1/(d^2+R^2)$ force dependence. The cosine component along the axis introduces an additional $d/\sqrt{d^2+R^2}$ factor, combining to the $3/2$ power.

Q: Is the force attractive or repulsive?

Gravitational force is always universally attractive. The ring pulls the sphere towards its center, and the sphere pulls the ring equally in the opposite direction.

Question

Find the gravitational force of attraction between the ring and sphere as shown in the figure, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8} R$ is the distance between the centres of a ring (of mass $m$) and a sphere (of mass $M$), where both have equal radius $R$.

Options:

(A) $\frac{\sqrt{8}}{9} \cdot \frac{GmM}{R^2}$

(B) $\frac{2\sqrt{2}}{3} \cdot \frac{GMm}{R^2}$

(C) $\frac{1}{3\sqrt{8}} \cdot \frac{GMm}{R^2}$

(D) $\frac{\sqrt{8}}{27} \cdot \frac{GmM}{R^2}$

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the System

Imagine a cosmic dance between two massive objects: a uniform ring and a solid sphere. They are separated by a specific distance, $d = \sqrt{8}R$. Our goal is to find the exact gravitational pull they exert on each other.

### The Master Equation: Field of a Ring

To solve this, we break the problem into two parts. First, we find the gravitational field created by the ring at the exact location of the sphere.

The gravitational field $E$ on the axis of a uniform ring of mass $m$ and radius $R$ at a distance $d$ is given by:

$$E = \frac{Gmd}{(d^2 + R^2)^{3/2}}$$

This formula is derived by integrating the field contributions from infinitesimally small mass elements around the ring.

### Substituting the Cosmic Distance

We are given that the distance $d$ is $\sqrt{8}R$. Let's plug this into our master equation.

$$E = \frac{Gm(\sqrt{8}R)}{((\sqrt{8}R)^2 + R^2)^{3/2}}$$

Now, we simplify the denominator. The square of $\sqrt{8}R$ is simply $8R^2$. Adding the $R^2$ term, the expression inside the parentheses becomes $9R^2$.

$$E = \frac{Gm(\sqrt{8}R)}{(9R^2)^{3/2}}$$

### The Power of Three Halves

Next, we need to evaluate $(9R^2)^{3/2}$. This might look intimidating, but it's just a square root followed by a cube.

The square root of $9R^2$ is $3R$. Cubing this gives $(3R)^3 = 27R^3$. Substituting this back into our field equation:

$$E = \frac{\sqrt{8}GmR}{27R^3}$$

Canceling one $R$ from the numerator and denominator, we get the simplified field:

$$E = \frac{\sqrt{8}Gm}{27R^2}$$

### The Shell Theorem: Simplifying the Sphere

Now, how do we calculate the force on the solid sphere? Here is where **Newton's Shell Theorem** comes to our rescue.

It states that a spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. Therefore, we can treat the entire sphere of mass $M$ as a single point mass located exactly at distance $d$.

### The Final Calculation

The gravitational force $F$ is simply the mass of the object multiplied by the gravitational field it sits in.

$$F = M \cdot E$$

Substituting our calculated field $E$:

$$F = M \cdot \left( \frac{\sqrt{8}Gm}{27R^2} \right)$$

Rearranging the terms, we arrive at our final, elegant result:

$$F = \frac{\sqrt{8}GmM}{27R^2}$$

This perfectly matches option (d). The beauty of this problem lies in combining a standard formula with a powerful theorem to turn a complex 3D setup into a straightforward calculation.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$E = \frac{G m d}{( d ^{2} + R ^{2} ) ^{3/2}}$

$Substitute d = 8 R$

$Simplify Denominator$

$Evaluate Power$

$Simplify Field$

$F = M \cdot E$

$Final Force$

$Conclusion$

The Sigma Insight: Gravitational Force

The Setup

The Master Equation

Substituting the Cosmic Distance

The Power of Three Halves

The Shell Theorem

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=(d2+R2)3/2Gmd​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute d=8​R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify Denominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=M⋅E

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion