91影视

The given data can be listed below as,

• The mass of the particle is m.
• The mass of the thin ring is M.
• The particle is placed at a distance x.
• The radius of the ring is a.

Newton鈥檚 gravitational law states the force exerted is inversely proportional to the distances鈥� square and directly proportional to the products of the masses.

The equation of the force gives the gravitational potential energy, direction and the magnitude of the force, and the values of the potential energy.

a)

Considering a mass segment on the ring that is expressed as:

$$\begin{gathered} dm=\left(\frac{M}{2\mathrm{蟺补}}\right)dI \\ =\left(\frac{M}{2\mathrm{蟺补}}\right)dI \end{gathered}$$

Here,$蠒$is the integrated angle.

From the Newton鈥檚 gravitational law, the differential potential energy between the mass m and the segment of the ring is expressed as:

$$\begin{gathered} dU=-G\frac{mdM}{\sqrt{a^2+x^2}} \\ =-G\frac{m{\displaystyle \frac{M}{2\mathrm{蟺}}}d蠒}{\sqrt{a^2+x^2}} \end{gathered}$$

Hence, the potential energy of this system is-

localid="1668309765806" $$\begin{gathered} U=鈭�dU \\ ={鈭�}_0^{2\mathrm{蟺}}G-\frac{m{\displaystyle \frac{M}{2\mathrm{蟺}}}d蠒}{\sqrt{a^2+x^2}} \\ =-G\frac{mM}{\sqrt{a^2+x^2}} \end{gathered}$$

Thus, the potential energy of the system is localid="1668309752201" $-G\frac{mM}{\sqrt{a^2+x^2}}$.

b)

As x>>a , thenlocalid="1657952274488" $\sqrt{a^2+x^2}$ becomes x.

Thus, the potential energy becomes$U鈮�-G\frac{mM}{x}$.

c)

We know that,$F=-\frac{dU}{dx}$

Hence, using the value of the potential energy, the above equation becomes,

$$\begin{gathered} F_x=-\frac{d}{dx}\left[G-\frac{mM}{\sqrt{a^2+x^2}}\right] \\ =-GmM\frac{x}{{\left(x^2+a^2\right)}^{3/2}} \end{gathered}$$

Thus, the magnitude of the force on the particle is $GmM\frac{x}{{\left(x^2+a^2\right)}^{3/2}}$and the direction is negative as it is an attractive force character.

d)

As x>>a , then ${\left(x^2+a^2\right)}^{3/2}$becomes $x^3$,

Hence, the equation of the force becomes localid="1657952524059" $F_x鈮�-G\frac{mM}{x^2}$.

Thus, the magnitude of the force becomes $G\frac{mM}{x^2}$and it is in the negative direction.

e)

At x=0, the potential energy becomes-

$U=-G\frac{mM}{a}$

At x=0 , the force becomes-

$F_{net}\left(x=0\right)=0$

Thus, the values of localid="1657952683335" $U\mathrm{and}{F}_x$when x=0 are $-G\frac{mM}{a}$and 0, respectively.