Question

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.

Short answer

1. The velocity of the planet is

2. The position of the planet is

   
   

3. 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.

Step-by-step solution

Identification of given data

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

Concept Introduction

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

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

Calculate the net gravitational force and acceleration of the planet

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,

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

Calculation of the velocity (a)

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

Thus the velocity of the planet is

Calculation of the position (b)

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

Thus the position of the planet is

Calculation of the position (c)

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.