Chapter 10.6, Problem 46AYU

Orbit of Mercury The planet Mercury travels around the Sun in an elliptical orbit given approximately by $r=\frac{\left(3.442\right){10}^7}{1-0.206\cos \theta }$ where $r$ is measured in miles and the Sun is at the pole. Find the distance from Mercury to the Sun at aphelion (greatest distance from the Sun) and at perihelion (shortest distance from the Sun). See the figure. Use the aphelion and perihelion to graph the orbit of Mercury using a graphing utility.

To determine

To find: The distance from Mercury to the Sun at aphelion (greatest distance from the Sun) and at perihelion (shortest distance from the Sun).

Use the aphelion and perihelion to graph the orbit of Mercury using a graphing utility.

Answer to Problem 46AYU

$\text{At Aphelion: Dist }\approx \left(4.335\right){10}^7$.

$\text{At perihelion}:\text{ Dist }\approx \left(2.854\right){10}^7$.

Explanation of Solution

Given:

The planet Mercury travels around the Sun in an elliptical orbit given approximately by $r=\frac{\left(3.442\right){10}^7}{1-0.206\text{cos}\theta }$.

Calculation:

$r =\frac{\left(3.442\right){10}^7}{1-0.206\text{cos}\theta }$

At Aphelion, the greatest distance from the sun is when $\theta =0$.

$r=\frac{\left(3.442\right){10}^7}{1-0.206\text{cos}\theta }=\frac{\left(3.442\right){10}^7}{1-0.206\left(1\right)}=\frac{\left(3.442\right){10}^7}{.794}\approx (4.335){10}^7$

$r\approx (4.335){10}^7\text{miles}$.

At perihelion, the shortest distance from the sun is when $\text{cos}\theta =-1$.

$r=\frac{\left(3.442\right){10}^7}{1-0.206\text{cos}\theta }=\frac{\left(3.442\right){10}^7}{1+0.206\left(1\right)}=\frac{\left(3.442\right){10}^7}{1.206}\approx \left(2.854\right){10}^7$

Graph of the orbit of Mercury.