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Step-by-Step Solution
Step 1: Express the Mass Element
Given that the mass density of the spherical galaxy varies with distance $r$ from its center as
$ \rho = \frac{k}{r}.$
For a spherical shell of thickness $dr$ at radius $r$, the volume element $dV$ is
$4 \pi r^2 \, dr.$
Thus, the mass element $dm$ is:
$dm = \rho \, dV = \left(\frac{k}{r}\right)\,(4 \pi r^2\,dr) = 4\pi k \, r \, dr.$
Step 2: Integrate to Find Total Mass up to Radius R
To find the total mass $M$ enclosed within radius $R$, integrate $dm$ from $r = 0$ to $r = R$:
$M = \int_{0}^{R} dm \;=\; \int_{0}^{R} 4\pi k \, r \, dr
= 4\pi k \int_{0}^{R} r \, dr
= 4\pi k \left[\frac{r^2}{2}\right]_{0}^{R}
= 2 \pi k R^2.$
Step 3: Apply the Condition for Circular Motion
A star of mass $m$ moving in a circular orbit of radius $R$ experiences a gravitational force that provides the necessary centripetal force:
$ \frac{G M m}{R^2} \;=\; \frac{m v^2}{R}. $
Substitute $M = 2 \pi k R^2\,$ into the equation:
$ \frac{G \left(2 \pi k R^2\right)m}{R^2} \;=\; \frac{m v^2}{R}. $
Canceling out common factors gives:
$ 2 \pi G k \,=\, \frac{v^2}{R}
\;\;\; \Longrightarrow
\;\;\; v \;=\; \sqrt{2\pi G k R}.
$
Step 4: Determine the Orbital Period
The orbital period $T$ for a circular orbit of radius $R$ is given by:
$ T \;=\; \frac{\text{circumference of orbit}}{\text{speed}}
\;=\; \frac{2 \pi R}{v}
\;=\; \frac{2 \pi R}{\sqrt{2\pi G k R}}. $
Simplifying, we get:
$ T \;\propto\; \sqrt{R}
\;\;\;\Longrightarrow\;\;\;
T^2 \;\propto\; R. $
Step 5: Conclusion
Since $T^2$ is directly proportional to $R$, the correct relationship is
$ T^2 \;\propto\; R.$
Therefore, the correct answer is: $T^2 \;\propto\; R.$
