Solution 3: Saturn Rings Width
1. Velocity Relation:
$v \propto \frac{1}{\sqrt{r}}$. Hence $v_{max}$ is at $r_{in}$ and $v_{min}$ is at $r_{out}$.
$v \propto \frac{1}{\sqrt{r}}$. Hence $v_{max}$ is at $r_{in}$ and $v_{min}$ is at $r_{out}$.
2. Given Condition:
\[ v_{max} = 1.05 v_{min} \implies \sqrt{\frac{1}{r_{in}}} = 1.05 \sqrt{\frac{1}{r_{out}}} \] \[ \frac{r_{out}}{r_{in}} = (1.05)^2 \approx 1.1025 \]
\[ v_{max} = 1.05 v_{min} \implies \sqrt{\frac{1}{r_{in}}} = 1.05 \sqrt{\frac{1}{r_{out}}} \] \[ \frac{r_{out}}{r_{in}} = (1.05)^2 \approx 1.1025 \]
3. Radial Width Ratio:
\[ \frac{w}{r_{in}} = \frac{r_{out} – r_{in}}{r_{in}} = \frac{r_{out}}{r_{in}} – 1 = 1.1025 – 1 \approx 0.10 \]
\[ \frac{w}{r_{in}} = \frac{r_{out} – r_{in}}{r_{in}} = \frac{r_{out}}{r_{in}} – 1 = 1.1025 – 1 \approx 0.10 \]
Answer: (c) 0.10