Solution 4: Kepler’s Third Law
1. Period Ratio:
Using $T^2 \propto R^3$: \[ \frac{T_B}{T_A} = \left( \frac{R_B}{R_A} \right)^{3/2} = \left( \frac{4r}{r} \right)^{3/2} = 8 \]
Using $T^2 \propto R^3$: \[ \frac{T_B}{T_A} = \left( \frac{R_B}{R_A} \right)^{3/2} = \left( \frac{4r}{r} \right)^{3/2} = 8 \]
2. Given Time Interval:
$t = \frac{1}{4} T_B$.
$t = \frac{1}{4} T_B$.
3. Revolutions of A:
\[ n_A = \frac{t}{T_A} = \frac{T_B/4}{T_A} = \frac{1}{4} \times 8 = 2 \]
\[ n_A = \frac{t}{T_A} = \frac{T_B/4}{T_A} = \frac{1}{4} \times 8 = 2 \]
Answer: (b) 2