91Ó°ÊÓ

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

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

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

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

$\frac{\mathrm{μ}{\mathrm{ν}}^2}{r}=\frac{G\mathrm{μ}\left(m_1+m_2\right)}{r^2}----\left(1\right)$

Here, the reduced mass of the system $\mathrm{μ}=\frac{m_1{m}_2}{m_1+m_2}$and $v=\frac{2\mathrm{Ï€°ù}}{T}$.

Substitute the the formula for $\mathrm{μ}$and $v$into $\left(1\right)$and obtain the formula for $T$.

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

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

Substitute the values to obtain $T$.

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

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

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

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

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

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

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

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

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

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

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

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

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