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 the star is $$T$$ and its distance from the galaxy's axis is $$r$$, then

Consider a very long cylindrical galaxy of length $$L$$ and radius $$R$$ (with $$R \ll L$$), with mass uniformly distributed. A star orbits the galaxy in a plane perpendicular to the galaxy's axis and passing through its center, at a distance $$r$$ from the axis (where $$r > R$$ since the star is outside the galaxy). We need to find the relationship between the orbital time period $$T$$ and $$r$$.

Due to the cylindrical symmetry and the condition $$R \ll L$$, we can approximate the galaxy as an infinitely long cylinder for gravitational effects at distances $$r$$ (where $$r \ll L$$). The mass per unit length $$\lambda$$ is constant and given by the total mass divided by $$L$$. For an infinitely long cylinder, the gravitational field outside the cylinder depends only on $$\lambda$$.

Using Gauss's law for gravity, consider a cylindrical Gaussian surface of radius $$r$$ and length $$l$$, coaxial with the galaxy. The mass enclosed by this surface is $$\lambda l$$. The gravitational flux through the curved surface is $$g \cdot 2\pi r l$$ (since the field is radial and uniform), and through the ends is zero (as the field is parallel to the ends). Gauss's law states:

$$\int \mathbf{g} \cdot d\mathbf{A} = -4\pi G M_{\text{enc}}$$

Substituting the values:

$$g \cdot 2\pi r l = -4\pi G (\lambda l)$$

Solving for the magnitude of the gravitational field $$g$$ (ignoring the negative sign for direction):

$$g = \frac{2G\lambda}{r}$$

This gravitational force provides the centripetal force for the star's circular orbit. Let $$m$$ be the mass of the star and $$v$$ its orbital speed. The centripetal force is:

$$F_c = \frac{m v^2}{r}$$

The gravitational force is:

$$F_g = m g = m \cdot \frac{2G\lambda}{r}$$

Setting $$F_g = F_c$$:

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

Canceling $$m$$ and $$r$$ (assuming $$m \neq 0$$ and $$r \neq 0$$):

$$\frac{2G\lambda}{r} = \frac{v^2}{r}$$

Multiplying both sides by $$r$$:

$$2G\lambda = v^2$$

Thus:

$$v^2 = 2G\lambda$$

Since $$\lambda$$ is constant for the galaxy, $$v$$ is constant and independent of $$r$$.

The orbital period $$T$$ is the time for one complete orbit, which is the circumference divided by the speed:

$$T = \frac{2\pi r}{v}$$

As $$v$$ is constant, $$T$$ is proportional to $$r$$:

$$T \propto r$$

Comparing with the options:

A. $$T \propto \sqrt{r}$$

B. $$T \propto r$$

C. $$T \propto r^2$$

D. $$T^2 \propto r^3$$

Option B matches the derived relationship.

Hence, the correct answer is Option B.