Question

At $\mathbf{t}\mathbf{=}\mathbf{0}$ a star of mass $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{}\mathbf{kg}$ has the velocity $<7\times {10}^4,6\times {10}^4,-8\times {10}^4>\mathbf{m}\mathbf{/}\mathbf{s}$ and is located at $<2.00\times {10}^{12},-5.00\times {10}^{12},4.00\times {10}^{12}>\mathbf{m}$ relative to the center of a cluster of stars. There is only one nearby star that exerts a significant force on the first star. The mass of the second star is $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{}\mathbf{kg}$, its velocity is $<2\times {10}^4,-1\times {10}^4,9\times {10}^4>\mathbf{m}\mathbf{/}\mathbf{s}$, and this second star is located at $<2.03\times {10}^{12},-4.94\times {10}^{12},3.95\times {10}^{12}>$ relative to the center of the cluster of stars (a) At $\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{s}$, what is the approximate momentum of the first star? (b) Discuss briefly some ways in which your result for (a) is approximate, not exact. (c) At $\mathbf{t}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{s}$, what is the approximate position of the first star? (d) Discuss briefly some ways in which your result for (b) is approximate, not exact.

Short answer

1. The momentum of the first star is
2. Possibilities of the approximate answer are discussed.
3. The momentum of the first star is
4. Possibilities of the approximate answer are discussed.

Step-by-step solution

Identification of given data

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

The velocity of the star is

Location of the star is

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

The velocity of the second star is

Location of the second star is

Concept Introduction

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

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

Calculate the net gravitational force and acceleration of the planet

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

Thus the magnitude of the distance is,

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

The unit vector of the distance is,

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

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,

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

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

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

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

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

Calculation of the velocity and momentum of the first star (a)

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,

Discussion for the approximate results (b)

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.

Calculation of the velocity and momentum of the second star (c)

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,

Discussion for the approximate results (d)

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.