Question

The two stars in a binary star system have masses $2.0\times {10}^{30}kg$and $6.0\times {10}^{30}kg$. They are separated by $2.0\times {10}^{12}m$. What are

a. The system’s rotation period, in years?

b. The speed of each star?

Short answer

a. Rotation period of the system is $7.7\times {10}^{8}years$.

b. The speed of the first and second stars are $1.4\times {10}^{4}m/s$and $8.2\times {10}^{3}m/s$respectively.

Step-by-step solution

Part (a) Step 1 : Given Information

The first star has mass, $m_1=2.0\times {10}^{30}kg$.

The second star has mass, $m_2=6.0\times {10}^{30}kg$.

The two stars are separated by$r=2.0\times {10}^{10}m$distance.

Part (a) Step 2 : Explanation 

For the revolution, we know the centripetal force between the two stars = Gravitational force between the two stars.

$\frac{\mu {\nu }^2}{r}=\frac{G\mu \left(m_1+m_2\right)}{r^2}----\left(1\right)$

Here, the reduced mass of the system $\mu =\frac{m_1{m}_2}{m_1+m_2}$and $v=\frac{2\mathrm{\pi r}}{T}$.

Substitute the the formula for $\mu$and $v$into $\left(1\right)$and obtain the formula for $T$.

localid="1648182136695" $T={\left[\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{G\left(m_1+m_2\right)}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Here, the gravitational constant $G=6.67\times {10}^{-11}N·m^2/k{g}^2$

Part (a) Step 3 : Calculation 

Substitute the values to obtain $T$.

localid="1648182491757" $$\begin{gathered} T={\left[\frac{4\times 3.{14}^2\times {\left(2.0\times {10}^{12}\right)}^3}{6.67\times {10}^{-11}\times \left(2.0\times {10}^{30}+6.0\times {10}^{30}\right)}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =7.7\times {10}^{8}years \end{gathered}$$

Part (a) Step 4 : Final Answer

Rotation period of the system is $7.7\times {10}^{8}years$.

Part (b) Step 5 : Given Information

The first star has mass,$m_1=2.0\times {10}^{30}kg$.

The second star has mass,$m_2=6.0\times {10}^{30}kg$.

The two stars are separated by$r=2.0\times {10}^{10}m$distance.

Part (b) Step 6 : Explanation

For the revolution, we know the centripetal force of the first star = Gravitational force between the two stars.

$$\begin{gathered} \frac{m_1{{\nu }_1}^2}{r}=\frac{G{m}_1{m}_2}{r^2} \\ {v}_1=\sqrt{\frac{G{m}_2}{r}} \end{gathered}$$

the centripetal force of the second star = Gravitational force between the two stars.

$$\begin{gathered} \frac{m_2{{\nu }_2}^2}{r}=\frac{G{m}_1{m}_2}{r^2} \\ {v}_2=\sqrt{\frac{G{m}_1}{r}} \end{gathered}$$

Here, the gravitational constant$G=6.67\times {10}^{-11}N·m^2/k{g}^2$

Part (b) Step 7 : Calculation

Substitute the values to obtain $v_1$and $v_2$.

role="math" localid="1648183977860" $$\begin{gathered} v_1=\sqrt{\frac{G=6.67\times {10}^{-11}N·m^2/k{g}^2\times 6\times {10}^{30}kg}{2.0\times {10}^{12}m}} \\ =1.4\times {10}^{4}m/s \\ {v}_2=\sqrt{\frac{G=6.67\times {10}^{-11}N·m^2/k{g}^2\times 2\times {10}^{30}kg}{2.0\times {10}^{12}m}} \\ =8.2\times {10}^{3}m/s \end{gathered}$$

Part (b) Step 8 : Final Answer

The speed of the first and second stars are$1.4\times {10}^{4}m/s$androle="math" localid="1648184239664" $8.2\times {10}^{3}m/s$respectively.