1. Given force law

The central force is inversely proportional to $R^3$:

$F \propto \frac{1}{R^3}$

Let

$F = \frac{k}{R^3}$

for some constant $k$.

2. Use centripetal force condition

For uniform circular motion, the required centripetal force is:

$F = \frac{mv^2}{R}$

So,

$\frac{mv^2}{R} = \frac{k}{R^3}$

Multiply both sides by $R$:

$mv^2 = \frac{k}{R^2}$

Hence,

$v^2 \propto \frac{1}{R^2}$

So,

$v \propto \frac{1}{R}$

3. Relate speed to time period

For circular motion,

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

Thus,

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

Since $v \propto \frac{1}{R}$,

$T \propto R \cdot R = R^2$

Therefore,

$\boxed{T \propto R^2}$

4. Check options

• A: $T \propto R^{4/3}$ — incorrect
• B: $T \propto R^{5/2}$ — incorrect
• C: $T \propto R^{3/2}$ — incorrect
• D: $T \propto R^2$ — correct

So the correct option is:

$\boxed{\text{D}}$