Chapter 11.4, Problem 62SR

a.

To determine

Write an equation for the orbit of each planet.

Answer to Problem 62SR

  $\boxed{\frac{x^2}{{\left(67.685\right)}^2}+\frac{y^2}{{\left(67.684\right)}^2}=1}$

  $\boxed{\frac{x^2}{{\left(483.685\right)}^2}+\frac{y^2}{{\left(483.13\right)}^2}=1}$

Explanation of Solution

Given information:

The table at the right shows the closest and farthest distances of Venus and Jupiter from the Sun in millions of miles.

  $\begin{array}{ccc}Planet& Closest& Farthest\\ Venus& 66.8& 67.7\\ Jupiter& 460.1& 507.4\end{array}$

Write an equation for the orbit of each planet. Assume that the centre of the orbit is the origin and the centre of the Sun is a focus that lies on the $x-axis$ .

Calculation:

Consider the standard equation of the horizontal ellipse which is centred to the origin,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Length of major axis, $2a$ units, length of minor axis $2b$ unit and foci are at $\left(c,0\right)$ and $\left(-c,0\right)$

The valid equation is,

  $c^2=a^2-b^2\mathrm{......}\left(1\right)$

Consider the table for distances,

Assume that planet Venus revolves around the Sun in an elliptical orbit with the Sun at its one of its focus on $x-axis$ .

The length of major axis will be the sum of closest and farthest distances and diameter of the Sun which is $0.87million$ miles. So now,

  $\begin{array}{l}2a=66.8+0.87+67.7\\ 2a=135.37\\ a=\frac{135.37}{2}\\ a=67.685\end{array}$

Now find $c$ which is the distance from centre of ellipse to the focus, equal to minus the close sets distance and radius of the Sun $\left(0.87/2=0.435\right)$ .

  $\begin{array}{l}c=67.685-66.8-0.435\\ c=0.45\end{array}$

Now put values of $a=67.685,c=0.45$ in equation $\left(1\right)$ .

  $\begin{array}{l}{\left(0.45\right)}^2={\left(67.685\right)}^2-b^2\\ b^2=4581.056725\\ b=67.684\end{array}$

Now put $a=67.685,b=67.684$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get,

  $\frac{x^2}{{\left(67.685\right)}^2}+\frac{y^2}{{\left(67.684\right)}^2}=1$

Hence, the equation of the orbit of Venus is , $\boxed{\frac{x^2}{{\left(67.685\right)}^2}+\frac{y^2}{{\left(67.684\right)}^2}=1}$ .

Now consider the closest and farthest distance of Jupiter from the Sun,

  $\begin{array}{l}2a=460.1+\mathrm{0.870506.4}\\ 2a=967.37\\ a=483.685\end{array}$

  $\begin{array}{l}c=483.685-460.1-0.435\\ c=23.15\end{array}$

Now substitute $a=483.685$ , $c=23.15$ in equation $\left(1\right)$ .

  $\begin{array}{l}{\left(23.15\right)}^2={\left(483.685\right)}^2-b^2\\ b^2=233,415.2567\\ b=483.13\end{array}$

Now put $a=483.685,b=483.13$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ we get,

  $\frac{x^2}{{\left(483.685\right)}^2}+\frac{y^2}{{\left(483.13\right)}^2}=1$

Hence equation for orbit of Jupiter is $\boxed{\frac{x^2}{{\left(483.685\right)}^2}+\frac{y^2}{{\left(483.13\right)}^2}=1}$

b.

To determine

Which planet has an orbit that is closer to the circle?

Answer to Problem 62SR

  $\boxed{Venus}$

Explanation of Solution

Given information:

The table at the right shows the closest and farthest distances of Venus and Jupiter from the Sun in millions of miles.

  $\begin{array}{ccc}Planet& Closest& Farthest\\ Venus& 66.8& 67.7\\ Jupiter& 460.1& 507.4\end{array}$

Which planet has an orbit that is closer to the circle?

Calculation:

Consider the standard equation of eccentricity of ellipse is,

  $e=\frac{c}{a}$

For Venus planet put $a=67.685,c=0.45$ in above equation,

  $\begin{array}{l}{e}_v=\frac{0.45}{67.685}\\ e_v=6.6484\times {10}^{-3}\end{array}$

For Jupiter planet, $a=483.685$ , $c=23.15$ ,

  $\begin{array}{l}{e}_J=\frac{23.15}{483.685}\\ e_J=47.8617\times {10}^{-3}\end{array}$

Thus lesser the value of $e$ more circular the orbit of ellips,

Hence, the orbit of $\boxed{Venus}$ is more circular.