Sometimes it's near, sometimes it's far

Geometry Level 3

A planet revolves around its star in an elliptical orbit of eccentricity $0.6$ , the star being at one of the ellipse's foci.

The planet is said to be at aphelion when the distance between the planet and the star its maximum; at perihelion, when the distance between the star and the planet is minimum.

If $r_\text{ap}$ and $r_\text{per}$ are the distances between the planet and the star at aphelion and perihelion respectively, then find $\dfrac{r_\text{ap}}{r_\text{per}}$ .

The answer is 4.

1 solution

Julian Yu
Apr 5, 2019

Let $c$ be the distance between the center of the ellipse and the foci, and let $a$ be half the length of the major axis. The eccentricity of an ellipse is given by $\dfrac{c}{a}$ .

By the conditions of the problem, $c=0.6a$ .

The quantity that we are being asked for is $\dfrac{a+c}{a-c}$ , which is equal to $\dfrac{a+0.6a}{a-0.6a}=\dfrac{1.6a}{0.4a}=\boxed{4}.$

Sometimes it's near, sometimes it's far

The answer is 4.

1 solution

0 pending reports