Elliptical Orbits for Dummy Astrophysicists

Calculating perihelion

We can't be certain, of course, but we suspect that students majoring in humanities and liberal arts – the few still required to take a science elective, that is – usually head for one of a few so-called "gut" science courses: Geology 101¹, Meteorology 101, or Astronomy 101. Again, there's no proof, but we don't think eHowian Soren Bagley, already a six-time DotD, didn't choose astronomy. Our evidence? it's in his post "How to Calculate Perihelion,"² which is currently on display at Sciencing.com.

We'll note right up front that Bagley managed to find an authoritative reference and somehow transcribe its content correctly (for the most part) while still meeting the infamous Demand Media minimum word count. Since the answer is actually quite straightforward when expressed as a formula, Soren had to convert it to words – lots of 'em – to meet MWC. Even so, he never actually gets around to explaining why his answer works. If he had, he would have said something to the effect of

"Perihelion and aphelion occur at the foci of an elliptical orbit. Perihelion occurs at the focus nearer the sun; aphelion occurs at the farther focus. The two points' distances from the sun can be calculated using the geometry of the ellipse."

But Soren most likely didn't understand that. So instead, he just rattled off the process (as he understood it):

1. Determine the semi-major axis of the orbit.
2. Determine the eccentricity of the orbit.
3. Subtract the ellipse's eccentricity from the number 1.
4. Multiply the result of Step 3 by the result of Step 1.

That, in fact, is correct – though we're pretty sure it would have been clearer if Soren had said "the semi-major axis" instead of "the result of Step 1." But what's really wrong with Bagley's post is found further along in his second step. You see, his instructions for calculating eccentricity of an ellipse are to simply

"[divide] the length of the ellipse's major-axis [sic] by the distance between the ellipse's two foci..."

...which, unfortunately, is wrong. The eccentricity of an ellipse is the ratio of c, the distance from the center to a focus; and a, the distance between the focus and the end of the semi-minor axis. A, then, is the hypotenuse of a triangle with sides c and b, where b is the semi-minor axis (see image above). Using the Pythagorean Theorem, then, the value Bagley actually needed is the square root of  b² + c².

The old saw says you should never to send a boy to do a man's job. We say you should never send an English Literature graduate to do a scientist's job (and we don't give a rat's hiney about the scientist's gender). Bagley blew it, plain and simple, which means that it gives us great pleasure to award Soren his seventh Dumbass of the Day.

¹ aka "rocks for jocks"
² Leaf Group sent this one to one of their rewrite team. We might get to that one later...

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