Chapter 9, Problem 151RE

To determine

Calculate the polar equation for the orbit and its perihelion and aphelion distances.

Answer to Problem 151RE

The two polar equation are $\frac{3.05}{2}(1+e)=1.667$ and $\frac{3.05}{2}(1-e)=1.383$ .

Explanation of Solution

Given: It is given in the question that the mars planet has an elliptical orbit with an eccentricity of $e\approx 0.093$ .The length of the major axis of the orbit is approximately $3.05$ astronomical units.

Concept Used:

In this, use the concept of polar equation of the orbit of a planet and the formula used is $r=\frac{(1-e^2)a}{1-e\cos \theta }$ .

Calculation:

Here, use the formula $r=\frac{(1-e^2)a}{1-e\cos \theta }$ .

  $\begin{array}{l}r=\frac{(1-e^2)a}{1-e\cos \theta }\\ r=\frac{(1-{0.093}^2)\frac{3.05}{2}}{1-0.093\cos \theta }\\ r=\frac{1.512}{1-0.093\cos \theta }\end{array}$

Now,

For Perihelion: $a(1-e)$ : $a(1-e)=\frac{3.05}{2}(1-e)=1.383$

For Aphelion: $a(1+e)=\frac{3.05}{2}(1+e)=1.667$

Conclusion:

The two equation is $\frac{3.05}{2}(1+e)=1.667$ and $\frac{3.05}{2}(1-e)=1.383$ .