Question

# A very long (length $L$ ) cylindrical galaxy is made of uniformly distributed mass and has radius $R\left( R<< L \right)$ . A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its center. If the time period of star is ”t” and its distance from the galaxy’s axis is r, then(A) $T\propto {{r}^{2}}$ (B) $T\propto r$ (C) $T\propto \sqrt{r}$ (D) ${{T}^{2}}\propto {{r}^{3}}$

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Hint: We can find the gravitational potential due to cylindrical mass distribution by comparing it with the electric field due to cylindrical charge distribution.
The gravitational field provides the necessary centripetal force. So, on comparing the equations of gravitational field and centripetal force, we can find the desired relation.

Formula used:
Gravitational force $=\dfrac{2G\lambda }{r}$ where
$G=\dfrac{1}{4\pi E}$
Centripetal force $=m{{w}^{2}}r$ .

Complete step by step solution
According to Gauss law for cylindrical charge distribution,
\begin{align} & E\times 2\pi rl=\dfrac{\lambda L}{E} \\ & E=\dfrac{\lambda }{2\pi rE} \\ \end{align}
The gravitational potential due to cylindrical mass distribution can be found by comparing it with electric field due to cylindrical charge distribution.
So gravitational field $=\dfrac{\lambda }{2\pi rE}$
$=\dfrac{2\lambda }{4\pi Er}$
$E=\dfrac{2G\lambda }{r}$
Here $G=\dfrac{1}{4\pi E}$
Now centripetal force is given by:
$F=m{{w}^{2}}r.........(1)$
Also,
As gravitational field is inversely proportional to distance from center, the centripetal force is
\begin{align} & F=mE \\ & F=m\dfrac{2G\lambda }{r}.........(2) \\ \end{align}
On comparing (1) and (2), we get
$m{{w}^{2}}r=m\dfrac{2G\lambda }{r}$
or $w=\dfrac{1}{2}\sqrt{2G\lambda }$
Now as we know that Time period T is given by
\begin{align} & T=\dfrac{2\pi }{w} \\ & T=\dfrac{r}{\sqrt{2G\lambda }} \\ \end{align}
or $T\text{ }\propto \text{ }r$ .

Note
Gauss law for cylindrical charge distribution is equivalent to electric field due to line charge.
To find electric charge at a point, a cylindrical surface is drawn about the line charge as its axis. This cylindrical surface may be treated as the Gaussian surface for the line charge.
Electric flux through the Gaussian surface=magnitude of electric field $\times$ area of curved surface of cylindrical of radius r and length l.
Flux= $E\times 2\pi rl$ ……… (A)
According to gauss theorem, we have
Electric flux
$=\dfrac{q}{E}$
Now, charge enclosed by Gaussian surface $q=\lambda l$
So, Flux $=\dfrac{\lambda l}{E}$ ……. (B)
From equations (A) and (B) we get
\begin{align} & E\times 2\pi rl=\dfrac{\lambda l}{E} \\ & E=\left( \dfrac{1}{2\pi E} \right)\left( \dfrac{\lambda }{r} \right)=\left( \dfrac{1}{4\pi E} \right)\left( \dfrac{2\lambda }{r} \right) \\ \end{align} .