Chapter 13.4, Problem 63STP

To determine

To find:Equation for the orbit of Earth.

Answer to Problem 63STP

Required equation of ellipse is

  $\frac{x^2}{8.714\times {10}^{15}}+\frac{y^2}{8.712\times {10}^{15}}=1$

Explanation of Solution

Given information:

At its closest point, Earth is $91.8$ million miles from the center of the sun.

At its farthest point, Earth is $94.9$ million miles from the center of the sun.

The value of $a$ is one half the length of the major axis.

  $a=\frac{1}{2}\left(91.8+94.9\right)$

  $\Rightarrow a=\frac{1}{2}\left(186.7\right)$

  $\Rightarrow a=93.35$ million miles.

Squaring on both sides

We get

  $\Rightarrow a^2={\left(93.35\right)}^2$

  $\Rightarrow a^2=8714.22$ million miles.

  $\Rightarrow a^2=8.714\times {10}^{15}$

The value of $c$ is the distance from the center of the ellipse to the focus.

This distance is equal to $a$ minus the perihelion.

  $c=93.35-91.8$

  $\Rightarrow c=1.55$ million miles.

We know

  $c^2=a^2-b^2$

On substituting values

We get

  $\begin{array}{l}{c}^2=a^2-b^2\\ \Rightarrow {\left(1.55\right)}^2={\left(93.35\right)}^2-b^2\\ \Rightarrow 2.4025=8714.22-b^2\\ \Rightarrow b^2=8714.22-2.4025\\ \Rightarrow b^2=8711.8175\\ \Rightarrow b=\sqrt{8711.8175}\\ \Rightarrow b=93.34\end{array}$

  $\begin{array}{l}{c}^2=a^2-b^2\\ \Rightarrow {\left(1.55\right)}^2={\left(93.35\right)}^2-b^2\\ \Rightarrow 2.4025=8714.22-b^2\\ \Rightarrow b^2=8714.22-2.4025\\ \Rightarrow b^2=8711.8175\\ \Rightarrow b^2=8712\end{array}$

  $\Rightarrow b^2=8.712\times {10}^{15}$

We know

Equation of ellipse is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

On substituting values

We get

  $\frac{x^2}{8.714\times {10}^{15}}+\frac{y^2}{8.712\times {10}^{15}}=1$

Therefore,

Required equation of ellipse is

  $\frac{x^2}{8.714\times {10}^{15}}+\frac{y^2}{8.712\times {10}^{15}}=1$