91影视

The ring-particle system.

We can find the gravitational forceof attraction exerted by the ring on a particle of mass located on the ring鈥檚 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鈥檚 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

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