Chapter 9, Problem 103CR

To determine

To find:

The polar equation for the orbit and also find the closeness of the comet to the sun.

Answer to Problem 103CR

The polar equation for the orbit is $r=\frac{0.424}{1+0.9\sin \theta }$ and the closeness of the comet to sun is approximately $20,000,000\text{ miles}$ .

Explanation of Solution

Given information:

A comet has an elliptical orbit around the sun with an eccentricity of $e\approx 0.9$ .

The length of the major axis of the orbit is approximately 20 astronomical units.

Calculation:

By the use of vertical major axis, choose an equation of the form,

$r=\frac{ep}{1+e\sin \theta }$

Because the vertices of the ellipse occurs at $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$ , it can be determined that the length of the major axis is the sum of the $r$ values of the vertices. That is,

$\begin{array}{l}2a=\frac{0.9p}{1+0.9}+\frac{0.9p}{1-0.9}\\ 2a=0.473p+9p\\ 2a=9.473p\end{array}$

As the length of major axis is 20 astronomical units. So, $20\approx 9.437p$

Hence,

$\begin{array}{l}20\approx 9.437p\\ p\approx 0.471\end{array}$

Now, the value of $ep$ will be,

$\begin{array}{l}ep\approx \left(0.9\right)\cdot \left(0.471\right)\\ ep\approx 0.424\end{array}$

Use this value of $ep$ to form the required equation,

$r=\frac{0.424}{1+0.9\sin \theta }$

Here, $r$ is measured in astronomical units.

To find the closest point to the sun, substitute $\theta =\frac{\pi }{2}$ into this equation to obtain,

$\begin{array}{l}r=\frac{0.424}{1+0.9\sin \left(\frac{\pi }{2}\right)}\\ r=\frac{0.424}{1+0.9\times 1}\\ r=\frac{0.424}{0.9}\\ r\approx 0.22\text{ astronomical units}\\ r\approx 20,000,000\text{ miles}\end{array}$

Therefore, the polar equation for the orbit is $r=\frac{0.424}{1+0.9\sin \theta }$ and the closeness of the comet to sun is approximately $20,000,000\text{ miles}$ .