Question

What eccentricity value results in a circular orbit?

Answer 1

Answer:

Zero

Explanation:

Given the equation of an ellipse:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The eccentrity of an ellipse is given by:

$e=\sqrt{1-\frac{b^2}{a^2}}$

For a circle, we have

$a=b$

Therefore the eccentricity of a circle is

$e=\sqrt{1-\frac{1^2}{1^2}}=0$