A star of mass $\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{30}}\mathbf{}\mathbf{kg}$ is located at $<5\times {10}^{12},2\times {10}^{12},0>\mathbf{m}$.A planet of mass $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathbf{kg}$ and is located at $<3\times {10}^{12},4\times {10}^{12},0>\mathbf{m}$ and is moving with a velocity of $<0.3\times {10}^4,1.5\times {10}^4,0>\mathbf{m}\mathbf{/}\mathbf{s}$ (a) At a time $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{s}$ later what is the new velocity of the planet? (b) Where is the planet at this later time? (c) Explain briefly why the procedures you followed in parts (a) and (b) were able to produce usable results but wouldn’t work if the later time had been $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{s}$ instead of $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{s}$ after the initial time. Explain briefly how you could use a computer to get around this difficulty.

The mass of the star is ${\mathrm{m}}_{\mathrm{star}}=7\times {10}^{30}\mathrm{kg}$

The location of the star is

The mass of the planet is ${\mathrm{m}}_{\mathrm{planet}}=3\times {10}^{24}\mathrm{kg}$

The location of the planet is

The initial velocity of the planet is

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

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

Distance between the star and the planet can be given as,

Thus the magnitude of the distance is,

$$\begin{gathered} \left|\overrightarrow{\mathrm{r}}\right|=\sqrt{{\left(-2\times {10}^{12}\mathrm{m}\right)}^2+{\left(2\times {10}^{12}\mathrm{m}\right)}^2+0} \\ =2.83\times {10}^{12}\mathrm{m} \end{gathered}$$

The unit vector of the distance is,

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

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{Planet}}=\mathrm{G}\frac{{\mathrm{m}}_{\mathrm{star}}{\mathrm{m}}_{\mathrm{Planet}}}{{\left|\overrightarrow{\mathrm{r}}\right|}^2} \\ =\frac{6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\times 7\times {10}^{30}\mathrm{kg}\times 3\times {10}^{24}\mathrm{kg}}{{\left(2.83\times {10}^{12}\mathrm{m}\right)}^2} \\ =1.75\times {10}^{20}\mathrm{N} \end{gathered}$$

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

$$\begin{gathered} {\overrightarrow{\mathrm{a}}}_{\mathrm{planet}}=\frac{{\overrightarrow{\mathrm{F}}}_{\mathrm{Planet}}}{{\mathrm{m}}_{\mathrm{planet}}}\hat{\mathrm{r}} \\ =\frac{1.75\times {10}^{20}\mathrm{N}}{3\times {10}^{24}\mathrm{kg}}\hat{\mathrm{r}} \\ =\left(5.84\times {10}^{-5}\mathrm{m}/{\mathrm{s}}^2\right)\hat{\mathrm{r}} \end{gathered}$$

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

Thus the velocity of the planet is

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

Thus the position of the planet is

If the later time is $1\times {10}^9$ instead of $1\times {10}^6$ after the initial time then we can see from the calculations done in parts (a) and (b) that the values obtained for the velocity and the location of the planet will change very rapidly (due to the larger values). Therefore, these results will not be useful.