1. Given force law

   The attractive force between the star and the planet is proportional to $F \propto R^{-3/2}.$

2. Use centripetal force for circular motion

   For a planet of mass $m$ moving in a circular orbit of radius $R$ with angular speed $\omega$, $F = m\omega^2 R.$

   Since $\omega = \frac{2\pi}{T},$ we get $F = m\left(\frac{2\pi}{T}\right)^2 R = \frac{4\pi^2 m R}{T^2}.$

3. Compare proportionalities

   Given $F \propto R^{-3/2},$ and also $F \propto \frac{R}{T^2}.$

   Therefore, $\frac{R}{T^2} \propto R^{-3/2}.$

4. Solve for $T^2$

   Rearranging, $\frac{1}{T^2} \propto R^{-3/2-1} = R^{-5/2}.$

   Hence, $T^2 \propto R^{5/2}.$

5. Check options

   • A: $T^2 \propto R^{7/2}$ ❌
   • B: $T^2 \propto R^3$ ❌
   • C: $T^2 \propto R^{5/2}$ ✅
   • D: $T^2 \propto R^{3/2}$ ❌

Therefore, the correct option is C.