Distance between two foci - astronomy

Question

The maximum distance from Earth to the sun is $9.3\times10^7$ miles. The minimum distance is $9.1\times10^7$ miles. The sun is at one focus of the elliptical orbit. Find the distance from the sun to the other focus.

I tried to solve this using logic; I don't know if this works or if there is a mathematical solution.

My interpretation:

If this interpretation is correct then the distance from the sun to the other focus will be $0.2\times10^7$ or $2\times10^6$.

Is this correct? If not, what is the correct interpretation?

Answer 1

Using the geometry of ellipses, each focus point is displaced by $\epsilon a$ from the center where $\epsilon$ is the eccentricity of the ellipse and $a$ is the semimajor axis. The perihelion of an ellipse is given by $a(1-\epsilon)$ and the aphelion is given by $a(1+\epsilon)$. So you know that $$a(1-\epsilon)=9.1\times10^7\text{ miles}$$ $$a(1+\epsilon)=9.3\times10^7\text{ miles}$$ and you're looking for the distance between foci which will be $2\epsilon a$. Solving this system of equations I get that $2\epsilon a= 2.0 \times 10^6\text{ miles}$ so your answer looks good.

Another way to do this without all the ellipse properties it to notice that the total width of the ellipse is $18.4 \times10^7\text{ miles}$ so the center is located a distance of $9.2 \times 10^7\text{ miles}$ away from the left hand side and therefore the distance from the center of the ellipse to one foci is $1.0\times10^6\text{ miles}$ which you can multiply by $2$ to get the result.

Answer 2

In simple case, you calculation is true. For the sun and earth, which sun is on a focus of an ellipse and earth orbit around it, this model for simple calculation not bad. But in real world not so, the sun and the earth aren't two mathematical points and the orbit not a complete ellipse. Even a big asteroid affects on earth path, the axises change in time, more perturbations effect on earth path and their masses. Then such simple aren't correct values.