Question

Consider a double star system under the influence of the gravitational force between the stars. Star 1 has mass m1 = 1.42 × 1031 kg and Star 2 has mass m2 = 2.62 × 1031 kg. Assume that each star undergoes uniform circular motion about the center of mass of the system (cm). In the figure below r1 is the distance between Star 1 and cm, and r2 is the distance between Star 2 and cm.
If the stars are always a fixed distance s=r1+r2 = 3.48 × 1018 m apart, what is the period of the orbit (in s)?

Answer 1

Consider a double star system under the influence of the gravitational force between the stars. Star 1 has mass $m1 = 1.42 \times 1031 \, \mathrm{kg}$ and Star 2 has mass $m2 = 2.62 \times 1031 \, \mathrm{kg}$ . Assume that each star undergoes uniform circular motion about the center of mass of the system (cm). In the figure below r1 is the distance between Star 1 and cm, and r2 is the distance between Star 2 and cm. If the stars are always a fixed distance $s = r1 + r2 = 3.48 \times 1018 \, \mathrm{m}$ apart, what is the period of the orbit (in s)?

Solution

Choose radial coordinates for each star with origin at center of mass. Let $\hat{r}_1$ be a unit vector at Star 1 pointing radially away from the center of mass. Let $\hat{r}_2$ be a unit vector at Star 2 pointing radially away from the center of mass. The force diagrams on the two stars are shown in the figure below.

From Newton's Second Law, $\overline{F_1} = m_1\overline{a_1}$ , for Star 1 in the radial direction is

$$\hat {r} _ {1} \colon - G \frac {m _ {1} m _ {2}}{s ^ {2}} = - m _ {1} r _ {1} \omega^ {2}.$$

We can solve this for $r_1$

$$r _ {1} = G \frac {m _ {2}}{\omega^ {2} s ^ {2}}.$$

Newton's Second Law, $\overline{F_2} = m_2\overline{a_2}$ , for Star 2 in the radial direction is

$$\hat {r} _ {2} \colon - G \frac {m _ {1} m _ {2}}{s ^ {2}} = - m _ {1} r _ {1} \omega^ {2}.$$

We can solve this for $r_2$

$$r _ {2} = G \frac {m _ {1}}{\omega^ {2} s ^ {2}}.$$

Since $s$, the distance between the stars, is constant

$$s = r_1 + r_2 = G \frac{m_1}{\omega^2 s^2} + G \frac{m_1}{\omega^2 s^2} = G \frac{(m_2 + m_1)}{\omega^2 s^2}.$$

Thus the angular velocity is

$$\omega = \left(G \frac{(m_2 + m_1)}{s^3}\right)^{\frac{1}{2}}$$

and the period is then

$$T = \frac{2\pi}{\omega} = \left(\frac{4\pi^2 s^3}{G (m_2 + m_1)}\right)^{\frac{1}{2}}.$$ $$T = \sqrt{ \frac{4\pi^2 \left(3.48 \times 10^{18}\right)^3}{6.67 \times 10^{-11} \left(1.42 \times 10^{31} + 2.62 \times 10^{31}\right)} } = 7.85 * 10^{17} s.$$

Answer: $7.85 * 10^{17} s$.