So I decided to have students draw a scale model of the solar system, but there’s a problem. I just realized that the far distance and near approach of a planet is usually at opposite sides of the orbit. If I have the perihelion and aphelion, it seems I have two points on the narrow ends of the ellipse. How can I find the distance between the “flat” sides of the ellipse?
Well, it’s double the Semi-minor axis of the orbit. Apparently this is hard to find directly, but can be calculated based on better known orbital stats (aphelion, perihelion, semi-major, eccentricity…)
The simplest method for your purposes is probably going to be to use stretched string. If you pin down the two ends of a piece of string at the two foci, and pull them tight with a pencil in the corner, the pencil will trace out an ellipse.
If you know the length of the major axis (you do) and eccentricity you can compute the length of the minor axis.
Prepare to be disappointed. I think all but Mercury will be visually indistinguishable from circles.
I think this is going to be more difficult than you may realize. If you make Mercury’s orbit a circle with about a half-inch diameter, the orbit of Neptune is going to be a circle that is somewhere in the neighborhood of about 3 feet wide. The eccentricity of many of the orbits will be less than the width of whatever you are using to draw the lines.
How would I know what the foci are?
One focus is at the Sun. The other focus is at a distance equal to the aphelion minus the perihelion. The other foci are also all in different directions from the Sun-- You can find this by looking up the orbital elements of the planets.
Yes. The Wikipedia page on apsishas diagrams similar to what you’re interested in. The orbits look virtually circular. What’s most striking is that the apsides for the different planets don’t line up, so an accurate diagram of the solar system looks like a collection of off-center circles.
Use a logarithmic scale, not a linear one, so you can get all the planets on one sheet of paper.
Perihelion and aphelicon will be exactly opposing (neglecting precession of the orbit, which is only signfiicant for Mercury) by definition. The easiest way to draw this physically is using the string attached to the foci mentioned by Chronos, but Lance Turbo and engineer_comp_geek are correct that this won’t be visible for any planet other than Mercury (and the Pluto/Charon doublet, if you want to go there).
In real world orbit determination, the orbital elements are determined by observing locations at various times in the epoch and the distance between observations, which can be used to create state vectors. These are used to first determine the inclination and longitude of the ascending node giving the reference for the orbital plane, and then the other arguments (mean anomaly and argument of pariapsis) are calculated, and then the elliptic parameters (eccentricity and semimajor axis) are determined. (The elliptic parameters usually can’t be measured directly because they are dependant upon measurements of distance from the observer, and it usually isn’t possible to get precise direct measurements of distance at interplanetary distances.) All of this is somewhat complicated by the fact that the observer is themselves in motion around the barycenter, and so those motions and perturbations have to be considered when calculating the state vectors.
It might be more interesting to have students draw the orbit of a long period comet, which will have a highly elliptical orbit which can be drawn by the string-at-foci methods, demonstrating Kepler’s motions in an obviously non-circular case.
Eccentric orbits in a log scale are going to look very peculiar and non-intuitive. Unless students have already been introduced to logrithms in extensive detail, this will just be confusing.
Depending on what you’re trying to show it may be interesting to show orbits of different eccentricities at different scales so the main difference is eccentricity.
Venus(0.0068), Earth(0.0167), Neptune(0.0097): Pretty much circular.
Mercury(0.2056), Pluto(0.2482): Noticeably elliptical.
Haley’s Comet(0.967): Whoa!
You can get them all on one sheet, regardless. (It’s just a question of how easily you want to be able to see the inner orbits.)
I suggest Eris at 0.44 eccentricity. A nice intermediate value.
If you don’t want to use a log scale, just make two charts: one for the inner planets and one for the outer. If you want an eccentric orbit to plot against the inner planets, add the asteroid 433 Eros, which has an eccentricity of 0.22, so its out of roundness would contrast with the circles of the planets’ orbits. Eros has actually been orbited by a spacecraft (NEAR-Shoemaker), although most people are unaware of that. It didn’t get the massive publicity of New Horizons at Pluto.
Eris is actually too far away to fit conveniently on the outer planets’ chart. Instead add 2060 Chiron, which orbits between Saturn and Uranus with an eccentricity of 0.38. Again it will be an interesting contrast with the planets’ circular orbits.
Damn. Just missed the perihelion–1/6, at 22:49 UT.
I think a lot of you are wrong about Mercury being noticeable at the scale the OP is talking about. A eccentricity of 0.2 translates into the minor axis being right about 98% of the major axis or in other words if the major axis were one foot the minor axis would be 11 3/4 inches.
For the OP, the ratio of the minor axis to the major axis is sqrt(1-e[sup]2[/sup])
Mercury and Mars both have noticeable eccentricity in that the Sun is visibly separated from the center of the orbit. They’re considerably less noticeable if you just look at the shape in isolation, without the Sun marked.
Well yes if you label center here and focus (Sun) there but I don’t think the OP was planning that. And the implication from many of the posters seem like “eccentricity of 0.2? Wow! That is a noticeable ellipse.” when it really isn’t.
This is a good website for the discussion. Scroll down to see this exact discussion with Pluto as the example.