Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1031: 45

Answer

See graph and explanations.
1585443791

Work Step by Step

Step 1. Using the given equation with parameter values, for Mercury, we have $r=\frac{(1-e^2)a}{1-e\ cos\theta}=\frac{(1-0.2056^2)(36.0\times10^6)}{1-0.2056\ cos\theta}$ Step 2. For Earth, we have $r=\frac{(1-0.0167^2)(92.96\times10^6)}{1-0.0167\ cos\theta}$ Step 3. Graph both equations (red for Mercury) as shown in the figure. Step 4. We can see that both orbits are close to circles and they should have a common focus (the Sun).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.