Question: Several planets (Jupiter, Saturn, and Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass Mand outer radius R(Fig. 13-52). (a) What gravitational attraction does it exert on a particle of mass mlocated on the ring’s central axis a distance xfrom the ring center? (b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring?

The ring-particle system.

We can find the gravitational forceof attraction exerted by the ring on a particle of mass located on the ring’s central axis at a distance x from the ring centerby taking the derivative ofgravitational P.E of the particle. Then, using the law of conservation of energy, we can find the K.E of the particle. From this, we will get the speed of the particle.

Formula:

$\text{U}=-\frac{\text{GMm}}{\text{r}}$

$F=\frac{dU}{dx}$

The particle is at a distance x from the center of the ring of radius R. So, its distance from the particles on the ring is

$\sqrt{{\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}}\text{.}$

The relation between gravitational P.E and the gravitational force of attraction is

$F=\frac{dU}{dx}$

But, the gravitational P.E of the particle is

$\text{U}=\frac{-\text{GMm}}{\sqrt{{\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}}}$

So, the gravitational force of attraction exerted by the ring on a particle is

$$\begin{gathered} d\text{F}=\frac{\text{d}\left(\frac{-\text{GMm}}{\sqrt{{\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}}}\right)}{\text{dx}} \\ =\frac{\text{GMmx}}{\left({\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}\right)} \end{gathered}$$

Therefore, the gravitational force of attraction exerted by the ring on a particle of mass located on the ring’s central axis at a distance x from the ring center is

$\text{F =}\frac{\text{GMmx}}{{\left({\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}\right)}^{3/2}}$

The gravitational P.E of the particle as it falls to the center is$\text{U}=-\frac{\text{GMm}}{\text{r}}$

So, the change in the gravitational P.E of the particle as it falls to the center is

$\text{U}=-\frac{\text{GMm}}{\text{r}}+\frac{\text{GMm}}{\sqrt{{\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}}}$

According to the conservation of energy, this should be converted into the K.E of the particle.

$$\begin{gathered} \frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}=\frac{-\text{GMm}}{\text{R}}+\frac{\text{GMm}}{\sqrt{{\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}}} \\ v=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}\right)} \end{gathered}$$

Therefore, the speed of the particle with which it passes through the center of the ring is

$v=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}\right)}$

The speed of the particle with which it passes through the center of the ring is

$v=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{\sqrt{x^2+R^2}}\right)}$