So what’s the trajectory in velocity space?

@ Scythian: Oooh . . that's not the v-space(*) result I remember getting for r-space(*) ellipses, but it's qualitatively the same. And yes, we are on the same page. (BTW, how did you obtain that curve?)

I trust you realize I mean to include the gamut -- circle, ellipse, parabola, hyperbola -- of (non-degenerate) conics, for r-space orbits. But the ellipse is a good place to start. After the circle, which is trivial.

(*) Some term assignments:

All orbits:

M = mass of central body

G = univ. grav constant

k² = GM, a single parameter to describe the strength of the central body's gravity

ֿr = position vector

ֿv = velocity vector

e = eccentricity ≥ 0

q = periapsis distance

E = total energy (per mass) = T+V

• <0 for ellipse; =0 for parabola; >0 for hyperbola

T = kinetic energy/m = ½v²

V = potential energy/m = -k²/r

Ellipse (incl circle):

0 ≤ e < 1

a = semi-major axis

b = semi-minor axis

c = center-focus distance = ae

Distance, focus to end of minor axis = a, so

a² =

a² = b² + c²

q = a(1-e)

s = apoapsis distance = a(1+e)

⇒ b² = sq

E = -k²/2a

Perhaps a good convention to adopt for the general case, is to hold q constant.

Feel free to adopt other term assignments, of course.

[It was promising me 5000 chars total, 3814 more available; but it truncated me at 980.]

Anyway, note that, in your notation, √(a² - b²) = f = a(1-e)

Also, try working in polar coordinates.

@ Scythian: I'm coming to the same conclusion that there'll be too little space here for all this. But the result I recall is (at this point, I might as well give away this much), it's always a circle (closed orbits), or an arc thereof (open orbits).

Your response has prompted me to double down on trying to work the derivation again myself. For the ellipse, name points:

Q, periapsis

R, one end of the minor axis

S, apoapsis

Speeds at these, in multiples of k/√a are:

Q: v[Q] = (1+e)/(1-e), dir. +y

R: v[R] = 1, . . . . . . . .dir. +x

S: v[S] = (1-e)/(1+e), dir. -y

So plotting those 4 points (incl the pt symmetric to R, =1 in dir. -x), you'll see that since v[R] is the mean proportional to v[Q] & v[S], they lie on a circle.

This doesn't prove, of course, that the whole curve is a circle, but that's what I'm looking for.

Oh, and I'm still keenly interested to know what method you used to produce the "lemon drop" curve???

@ Al P: Tx for the response. I appreciate your well-wishes & points. Yours is one of the minds I was hoping to ensnare with this Q. ;-)

But I wasn't intending so complex a Q, so let me clarify a few things I maybe took for granted.

• The orbits are planar, so we can stay in xy

• No GR; strictly Newtonian analysis wanted

• "Infinitesimal mass" in these sorts of problems is straightforward; just take the limit as mass→0; i.e., neglect any effect of m on M. Are you seeing some ambiguity there?

As to the result sought, it's just like in r-space; you follow the track in space of the body, and describe that curve, with time-dependence removed; and that's the orbit -- the (r-space) trajectory. And in v-space, you can also trace the endpoint of the velocity vector with time-dependence removed, and describe the curve that results -- the v-space trajectory.

Other than that, I'm having trouble understanding your 2nd para.

@ Scyth: Tx for that -- Math'ica was an outstanding choice (I'm jealous!).

I think I've found the problem.

In your eqn for slope, s, the "/√(a²-b²)" should be "/√(a²-x²)".

With fingers crossed, I anxiously await a re-run. If you're of a mind to.

Meanwhile, I'm trying to see whether this revision allows any easier pencil-paper analysis.

I think the whole thing is easier with the origin at M, the focus, but it's certainly not necessary.

@ Scythian: OK, the typo didn't get into the orig. plot.

But that plot disagrees with my result that the x-intercept (v@minor axis) must be the mean proportional of the 2 y-intercepts (v@peri & apo). Something must still be amiss.

Wait, what about this -- conservation of ang. mtm., L

L(peri) = VR

But L(r) = vr sinφ ≠ vr

where φ = angle between ֿr and ֿv, which is 90º only at peri and apo.

@ Scy. Edit5: No.

K2Law says that the ang. mtm. vector, which is the cross product

ֿL = ֿv x ֿr

is constant. The magnitude of which is

L = vr sinφ

@ Scy. Edit5:

Sorry, make that

ֿL = ֿr x ֿv

OK, I've got the proof for the elliptical case, but it's too long to fit in this margin. Taking the central body as origin, periapsis at (+q,0), and counterclockwise orbit, the r-space trajectory (long known) is

r = a(1-e²)/(1 + e cosθ)

v-space trajectory, in Cart. coords. is

v_x = -v₀sinθ /√(1-e²)

v_y = -v₀(e + cosθ) /√(1-e²)

. . a circle, center = (0, v₀e/√(1-e²))

radius = v₀/√(1-e²)

where v₀ = k/√a

Oops again -- wrong sign on v_y !

Make that:

v_x = -v₀sinθ /√(1-e²)

v_y = v₀(e + cosθ) /√(1-e²)

@ Scy, Major Edit:

Great! I'm glad we ultimately agree on the result! You've helped confirm my dimly-recalled answer from the past, and you get BA. I'll hold it open a little longer, up to another 24 hr, for any final thoughts/events.

Your reaction was nearly identical to mine when I 1st encountered this -- why always a circle? Followed by, if this is really true, why have I never heard it?

Holding q (periapsis dist=your R) constant, and varying e (=your f/a) from 0 to ∞, makes for an interesting development of the v-curve. I agree with your hunch that the circular form is maintained throughout that process, but with ever-shorter arcs of it, once e gets > 1. The q=const results for ellipse:

r = q(1+e)/(1 + e cosθ)

v_x = -k sinθ /√q(1+e)

v_y = k(e + cosθ) /√q(1+e)

center = (0, ke/√q(1+e))

radius = k/√q(1+e)