Problem 55 - Precalculus with Limits

Question

The planets travel in elliptical orbits with the sun at one focus. Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is \(2 a\) (see figure). Show that the polar equation of the orbit is \(r=a\left(1-e^{2}\right) /(1-e \cos \theta),\) where \(e\) is the eccentricity.

Step-by-Step Solution

Answer

The polar equation of the orbital ellipse is indeed \(r = a(1 - e^2)/(1 - e \cos \theta)\). This was concluded by defining the ellipse equation, converting into polar coordinates, defining the eccentricity and making the related substitution.

Step 1: Defining the Ellipse

An ellipse in Cartesian coordinates is defined by \((x/a)^2 + (y/b)^2 = 1\) where \(2a\) is the major axis and \(2b\) is the minor axis. The center of ellipse is shifted to origin. Also, note that this ellipse has a horizontal major axis.

Step 2: Convert to Polar Coordinates

In polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute these values into the equation of the ellipse to transform it into polar coordinates. This gives you: \((r \cos \theta / a)^2 + (r \sin \theta / b)^2 = 1\). Note that since the major axis is along axis, only \(b\) will affect the value of \(r\) based on \(\theta\). We will focus on this.

Step 3: Definition of Eccentricity

Recall the definition of eccentricity of an ellipse. The eccentricity \(e\) is related to major and minor axis as \(e = \sqrt{1 - (b/a)^2}\) or \(b = a \sqrt{1 - e^2}\). Substitute this into the equation in step 2, yields: \((r \cos \theta / a)^2 + (r \sin \theta / a \sqrt{1 - e^2})^2 = 1\). Simplifying gives: \(r = a(1 - e^2)/(1 - e \cos \theta)\)

Step 4: Conclusion

This is the polar equation of the ellipse that represents the orbit, showing that the polar equation of the orbit is \(r = a(1 - e^2)/(1 - e \cos \theta)\).