Question

Two planets, A and B orbit around a star such that the time period of A is 8 times the time period of B. The ratio of orbital velocities of the planets A and B is:

• A. \(4:1\)
• B. \(1:4\)
• C. \(2:1\)
• D. \(1:2\)

Hint:

For planets revolving around the same star: \[ T^2\propto r^3 \] and \[ v\propto \frac{1}{\sqrt{r}}. \] First find the orbital radius ratio using Kepler’s law, then calculate the velocity ratio.

Answer 1

The Correct Option is D

Step 1: Recall Kepler's third law and derive radius from period.
Kepler's third law states $T^2 \propto r^3$, so $r \propto T^{2/3}$.

Step 2: Write orbital velocity in terms of period.
For a circular orbit, $v = \frac{2\pi r}{T}$.
Substituting $r \propto T^{2/3}$:
\[ v \propto \frac{T^{2/3}}{T} = T^{-1/3} \]

Step 3: Express the ratio of orbital velocities.
\[ \frac{v_A}{v_B} = \left(\frac{T_A}{T_B}\right)^{-1/3} \]

Step 4: Substitute the given ratio of time periods.
$T_A = 8T_B$, so $T_A/T_B = 8$:
\[ \frac{v_A}{v_B} = 8^{-1/3} = \frac{1}{8^{1/3}} = \frac{1}{2} \]

Step 5: Write the ratio.
\[ v_A : v_B = 1 : 2 \]

Step 6: Interpret the result.
Planet A has a larger orbit (longer period) and moves more slowly. Planet B is closer to the star and moves faster. This matches the known pattern: inner planets orbit faster.
\[ \boxed{1 : 2} \]