91Ó°ÊÓ

The mass of the star is ${\mathrm{m}}_{\mathrm{star}}=4×{10}^{30}\mathrm{kg}$

The velocity of the star is

Location of the star is

The mass of the second star is $\mathrm{m}{\text{'}}_{\mathrm{star}}=3×{10}^{30}\mathrm{kg}$

The velocity of the second star is

Location of the second star is

The gravitational attraction between two massive bodies of masses m₁and m₂ having a separation of r can be expressed as,

$\stackrel{\mathbf{→}}{\mathbf{F}}\mathbf{=}\mathbf{G}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{m}}_{\mathbf{2}}}{{\left|\stackrel{→}{r}\right|}^{\mathbf{2}}}\hat{\mathbf{r}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Distance between the first star and second star can be given as,

Thus the magnitude of the distance is,

$$\begin{gathered} \left|\stackrel{→}{r}\right|=\sqrt{{\left(3.00×{10}^{10}\mathrm{m}\right)}^2+{\left(6.00×{10}^{10}\mathrm{m}\right)}^2+{\left(-5.00×{10}^{10}\mathrm{m}\right)}^2} \\ =8.37×{10}^{10}\mathrm{m} \end{gathered}$$

The unit vector of the distance is,

$\hat{\mathrm{r}}=\frac{\left|\stackrel{→}{\mathrm{r}}\right|}{\stackrel{→}{\mathrm{r}}}$

role="math" localid="1668432412555"

Thus the net gravitational force on the first star due to the second star can be given using equation (1) such that,

$$\begin{gathered} {\stackrel{→}{\mathrm{F}}}_{\mathrm{star}}=\mathrm{G}\frac{{\mathrm{m}}_{\mathrm{star}}\mathrm{m}{\text{'}}_{\mathrm{star}}}{{\left|\stackrel{→}{\mathrm{r}}\right|}^2} \\ =\frac{6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2×4×{10}^{30}\mathrm{kg}×3×{10}^{30}\mathrm{kg}}{{\left(8.37×{10}^{10}\mathrm{m}\right)}^2} \\ =1.14×{10}^{29}\mathrm{N} \end{gathered}$$

Thus the acceleration of the first star due to this attractive gravitational force will be,

$$\begin{gathered} {\stackrel{→}{\mathrm{a}}}_{\mathrm{star}}=\frac{{\stackrel{→}{\mathrm{F}}}_{\mathrm{star}}}{{\mathrm{m}}_{\mathrm{star}}}\hat{\mathrm{r}} \\ =\frac{1.14×{10}^{29}\mathrm{N}}{4×{10}^{30}\mathrm{kg}}\hat{\mathrm{r}} \\ =\left(0.0285\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{r}} \end{gathered}$$

Thus the acceleration of the second star due to this attractive gravitational force will be,

$$\begin{gathered} {\stackrel{→}{\mathrm{a}}}_{\mathrm{star}}=\frac{{\stackrel{→}{\mathrm{F}}}_{\mathrm{star}}}{{\mathrm{m}}_{\mathrm{star}}}\hat{\mathrm{r}} \\ =\frac{1.14×{10}^{29}\mathrm{N}}{3×{10}^{30}\mathrm{kg}}\hat{\mathrm{r}} \\ =\left(0.038\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{r}} \end{gathered}$$

According to Newton’s first law of motion, the velocity can be calculated as,

Thus the velocity of the planet is

Hence, the momentum of the star will be,

Since the both stars are in motion therefore the calculations performed in part (a) are performed considering the initial values of the positions and velocities of the first and second stars. Therefore, the obtained values can be approximate.

According to Newton’s first law of motion, the velocity can be calculated as,

Thus the velocity of the planet is

Hence, the momentum of the star will be,

Since the both stars are in motion therefore the calculations performed in part (a) are performed considering the initial values of the positions and velocities of the first and second stars. Therefore, the obtained values can be approximate.