Question

A very long (length $L$) cylindrical galaxy is made of uniformly distributed mass and has radius $R(R<<L)$. A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. 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. $T^2\propto r^3$

Answer 1

The correct option is B $T\propto r$

The gravitational potential due to cylindrical mass distribution can be found by comparison with electric field due to cylindrical charge distribution.

Applying gauss law for cylindrical charge distribution, consider a cylindrical Gaussian surface of length L:

$\oint \overrightarrow{E}.d\overrightarrow{A}=\frac{Q_{en}}{ϵ}$

$E2\pi rL=\frac{\lambda L}{ϵ}$

$E=\frac{\lambda }{2\pi ϵr}$

Similarly gravitational field =

$E=\frac{2G\lambda }{r}$ where $\frac{1}{4\pi ϵ}=G$ from similarity

As the Gravitational field is inversely proportional to distance from center, The centripetal force is

$F_{centripetal}=mE=m\frac{2G\lambda }{r}$ ...(i)

$F_{centripetal}=m{\omega }^{2}r$...(ii)

substituting value of centripetal force in equation (i) from equation (ii)

$m{\omega }^{2}r=m\frac{2G\lambda }{r}$

$\omega =\frac{1}{r}\sqrt{2G\lambda }$

$T=\frac{2\pi }{\omega }=\frac{r}{\sqrt{2G\lambda }}\Rightarrow T\alpha r$ hence correct answer is option B