Chapter 8.5, Problem 42E

Solution

To find: The perihelion and aphelion distances.

The perihelion and aphelion distances for this famous comet are $18.30\text{AU}$ and $20.06\text{AU}$ , respectively.

Given information:

The distance of semimajor axis is $a=19.18\text{AU}$ . The eccentricity is $e=0.0461$ .

Formula used:

The perihelion distance is the nearest distance from the Sun (focus) to the comet (one endpoint of the major axis), $a-c$ . The aphelion distance is the farthest distance from the Sun (focus) to the comet (one endpoint of the major axis), $a+c$. This is illustrated below:

  

Calculation:

Find the value of $c$ as follows.

  $\begin{array}{c}c=ae\\ =19.18\left(0.0461\right)\\ =0.884198\text{AU}\end{array}$

Find the perihelion distance as follows.

  $\begin{array}{c}\text{perihelion distance}=a-c\\ =19.18-0.884198\\ =18.30\text{AU}\end{array}$

Find the aphelion distance as follows.

  $\begin{array}{c}\text{aphelion distance}=a+c\\ =19.18+0.884198\\ =20.06\text{AU}\end{array}$

Therefore, the perihelion and aphelion distances for this famous comet are $18.30\text{AU}$ and $20.06\text{AU}$ , respectively.