Chapter 9.7, Problem 59E

To determine

To calculate:

The polar equation, perihelion and aphelion distance of the following orbit:

Name of orbit of the planet: Mars

  $\begin{array}{l}a\approx 1.4164\times {10}^{8}miles\\ e\approx 0.0934\end{array}$

Answer to Problem 59E

The Mars planet has:

The polar equation: $r=\frac{1.4040\times {10}^8}{1-\left(0.0934\right)\cos \theta }$

The perihelion distance: $1.2841\times {10}^{8}mi$

The aphelion distance: $1.5487\times {10}^{8}mi$

Explanation of Solution

Given information:

Name of orbit of the planet: Mars

  $\begin{array}{l}a\approx 1.4164\times {10}^{8}miles\\ e\approx 0.0934\end{array}$

Formula used:

The equation of the orbit of the planet is given by the following expression:

  $r=\frac{\left(1-e^2\right)a}{1-e\cos \theta }$

The perihelion distance is given by the following expression:

  $r=a\left(1-e\right)$

The aphelion distance is given by the following expression:

  $r=a\left(1-e\right)$

Calculation:

The polar equation of the Mars is derived as:

  $r=\frac{\left(1-e^2\right)a}{1-e\cos \theta }$

Put the given data:

  $\begin{array}{l}r=\frac{\left(1-{\left(0.0934\right)}^2\right)1.4164\times {10}^8}{1-\left(0.0934\right)\cos \theta }\\ r=\frac{1.4040\times {10}^8}{1-\left(0.0934\right)\cos \theta }\end{array}$

The perihelion distance is calculated as:

  $r=a\left(1-e\right)$

Put the given data:

  $\begin{array}{l}r=a\left(1-e\right)\\ r=1.4164\times {10}^8\left(1-0.0934\right)\\ r=1.2841\times {10}^{8}mi\end{array}$

The aphelion distance is calculated as::

  $r=a\left(1+e\right)$

Put the given data:

  $\begin{array}{l}r=a\left(1+e\right)\\ r=1.4164\times {10}^8\left(1+0.0934\right)\\ r=1.5487\times {10}^{8}mi\end{array}$