Question

Consider two planets, A and B, of masses $M_A = 1.0 \times 10^{24}$ kg and $M_B = 2.0 \times 10^{24}$ kg, in circular orbits of radius $R_A = 1.0 \times 10^{11}$ m and $R_B = 4 \times 10^{11}$ m around a common star of mass $M_S = 24 \times 10^{31}$ kg. The two planets are orbiting in the same direction. You may assume that the planets are point masses and have no rotation about their own individual axes. (Take: $G = \frac{20}{3} \times 10^{-11} N-m^2/kg^2$)

3. The planets have angular momentum $L_A$ and $L_B$ about the star. The value of $\frac{L_A}{L_B}$ is:

4. The inhabitants of the planets have constructed a light indestructible link bridge, which they connect when planets A and B are closest to each other. This joins the two planets instantaneously. Determine the angular velocity $\omega$ (in rad/year) of the two planets about their centre of mass at the instant just after the connection is made.

Answer 1

Problem 3: 1/4 Problem 4: 21.0 rad/year

Answer 2

Problem 3: Ratio of angular momenta $L_A/L_B$

1. Formula for Angular Momentum: For a planet in a circular orbit, angular momentum $L = mvr$.
2. Orbital Speed: The gravitational force provides the centripetal force, so $GM_S m/r^2 = mv^2/r$, which simplifies to $v = \sqrt{GM_S/r}$.
3. Substitute $v$ into $L$: $L = m \sqrt{GM_S r}$.
4. Calculate Ratio: $\frac{L_A}{L_B} = \frac{M_A \sqrt{G M_S R_A}}{M_B \sqrt{G M_S R_B}} = \frac{M_A}{M_B} \sqrt{\frac{R_A}{R_B}}$.
5. Substitute values: $\frac{L_A}{L_B} = \frac{1.0 \times 10^{24}}{2.0 \times 10^{24}} \sqrt{\frac{1.0 \times 10^{11}}{4.0 \times 10^{11}}} = \frac{1}{2} \sqrt{\frac{1}{4}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Problem 4: Angular velocity $\omega$ (in rad/year) of the two planets about their centre of mass

1. Orbital Speeds: Calculate $v_A$ and $v_B$ using $v = \sqrt{GM_S/r}$.
   $v_A = 4 \times 10^5$ m/s, $v_B = 2 \times 10^5$ m/s.
2. Center of Mass Position: Calculate the position of the center of mass of the two planets, $R_{CM} = \frac{M_A R_A + M_B R_B}{M_A + M_B} = 3.0 \times 10^{11}$ m from the star.
3. Distances from CM: Calculate the distances of planets A and B from their common center of mass: $r_A' = |R_A - R_{CM}| = 2.0 \times 10^{11}$ m and $r_B' = |R_B - R_{CM}| = 1.0 \times 10^{11}$ m.
4. Moment of Inertia about CM: Calculate the moment of inertia of the two-planet system about its center of mass: $I_{CM} = M_A (r_A')^2 + M_B (r_B')^2 = 6 \times 10^{46}$ kg m$^2$.
5. Angular Momentum about CM (Spin Angular Momentum): The angular momentum of the system about its center of mass just before connection is conserved (assuming no external torque on the system of planets about their CM). It's the sum of individual angular momenta relative to the CM: $L_{spin} = M_A v_A r_A' - M_B v_B r_B'$ (the negative sign indicates they are on opposite sides of the CM, and their contributions to spin are in opposite senses relative to the CM, but based on the velocities, they contribute to rotation in the same direction).
   More accurately, $L_{spin} = M_A v_A' r_A' + M_B v_B' r_B'$, where $v_A'$ and $v_B'$ are velocities relative to CM.
   $V_{CM} = \frac{M_A v_A + M_B v_B}{M_A + M_B} = \frac{8}{3} \times 10^5$ m/s.
   $V_A' = v_A - V_{CM} = \frac{4}{3} \times 10^5$ m/s.
   $V_B' = v_B - V_{CM} = -\frac{2}{3} \times 10^5$ m/s.
   The angular velocity $\omega_{spin} = V_A'/r_A' = (\frac{4}{3} \times 10^5) / (2 \times 10^{11}) = \frac{2}{3} \times 10^{-6}$ rad/s.
   Alternatively, using angular momentum conservation for the spin:
   $L_{spin} = |M_A (R_A - R_{CM}) v_A + M_B (R_B - R_{CM}) v_B| = |(1 \times 10^{24})(-2 \times 10^{11})(4 \times 10^5) + (2 \times 10^{24})(1 \times 10^{11})(2 \times 10^5)| = |-8 \times 10^{40} + 4 \times 10^{40}| = 4 \times 10^{40}$ kg m$^2$/s.
   Then $\omega_{spin} = L_{spin} / I_{CM} = (4 \times 10^{40}) / (6 \times 10^{46}) = \frac{2}{3} \times 10^{-6}$ rad/s.
6. Convert to rad/year: Multiply by the number of seconds in a year ($3.15576 \times 10^7$ s/year).
   $\omega = \frac{2}{3} \times 10^{-6} \times 3.15576 \times 10^7 \approx 21.0384$ rad/year.

The final answer is $\boxed{\text{21.0 rad/year}}$