[math-fun] orbit trivia
Responding to a highschooler's Shuttle questions, some rough calculations turned up a) at roughly 49.2 degrees north and south lattitude, there are countours where g = 79000 mph/hr, exactly. b) although the equation of motion of an elliptical orbit requires higher transcendental functions, the shape, in Earth centered polar coordinates, is simply 2 r = ------------------------------, (k0 - k1) cos(theta) + k1 + k0 where k0 = 1/perigee and k1 = 1/apogee. (The hard part is theta(t), which apparently even transcends Lambert_W.) One question was what happens if you shoot a 1000mph bullet straight down from ISS orbit (225 mi). The line extrapolating the velocity vector sum comes nowhere near the Earth, so I was a little surprised to find the orbit grazing the surface about 1/4 the way around, recrossing the ISS orbit about halfway around, reaching an apogee of 484 mi at about the 3/4 point. So there ought to be a continuum of slightly lower muzzle velocities and slightly rearward aimpoints wherein you shoot yourself an orbit later. These unfortunate orbits also cross yours about halfway around, but "safely" ahead. Each provides an additional self-shooting solution--the vector difference between the shooter's orbital velocity and that of the first outbound crossing of the bullet, which will turn out to be slightly backward from straight "up" (away from the Earth.) This will cross once behind you and then hit you on the upswing after one full orbit. A couple of other shuttle trivia: c) the deorbit dv is only about 158 mph. d) reentry is a lot more like the old capsule/heat-shield scheme than I realized. The Shuttle comes in almost belly-first (rolling alternately almost + and - 90 degrees to avoid regaining altitude with the very high angle of attack), with the black undersurface acting as heat shield. (I always wondered why the windows didn't melt.) Similarly, the capsules had a deliberately eccentric center of mass so that the heat shield maintained a significant angle of attack. An alternative to the careening, serpentine descent would be multiple skips (with roll ~0), which might increase the number of orbit counts in reach of KSC, and reduce maximal tile heating. I'm guessing this is dangerously over-reliant on the rather feeble attitude thrusters, which in the current scheme are used once to get the attitude exactly right well in advance of atmospheric effects, after which most of the hypergolic fuel is dumped. The Earth turns about 22 degrees per Shuttle orbit, so they get a shot at KSC when the ground track passes just east of Florida, and another when it passes just west on the succeeding orbit. After the second waveoff Monday morning, NASA told the public that the next opportunity would be about 24 hr after the first one. However, the groundtracks passed Florida again after only 10 hrs or so, heading SE instead of NE. I asked some NASA people why nobody was considering this and got a rather defensive "I can't answer that." Could it be that they calculated that another reentry accident would only kill the program if the debris landed in the Midwest instead of the Caribbean? Maybe it was just a circadian or sonic boom decision. Lastly, the NASA PR types who showed the reentry live in the Ames visitor center theater also mentioned that the Mars rovers recently got life-extended when a friendly dust devil cleaned off their solar panels! --rwg PS, about 200 people (including children) stayed up both nights to watch. There were also scads of TV newscrews to tape the horror-stricken faces in case of an accident. PPS: The center has a moon rock, some meteoritic martian basalt, and what they claim is a piece of Vesta. How the he** do they know which asteroid it came from? Nobody taped the horror-stricken faces of the PR types I asked.
I said
b) although the equation of motion of an elliptical orbit requires higher transcendental functions,[...] (The hard part is theta(t), which apparently even transcends Lambert_W.)
Solving Kepler's equation, E - epsilon sin(E) = M, for E also solves the old math teacher embarrasser of finding the thickness x of the circular segment of given area a, e.g., 2 acos(1 - x) - sin(2 acos(1 - x)) = 2 a. The analytic solution is inf ==== sin(2 a n) Bessel_J (n) \ n x = 1 - cos( > ----------------------- + a). / n ==== n = 1 E.g., to slice a unit disk into four equal areas using parallel cuts requires "heel" thickness inf n ==== (- 1) Bessel_J (2 n - 1) %pi \ 2 n - 1 x = 1 - cos(--- - > -------------------------------) ~ 0.5960 . 4 / 2 n - 1 ==== n = 1 Convergence is lousy and the big Bessels have bad cancellation--you're way better off with Newton's method. Alternatively, revert the a(x) series (c192) sphere_segment_volume(n,1,x) n - 1 n - 1 n + 1 ----- ----- ----- 2 2 n + 1 1 - n n + 3 x 2 2 %pi hyper_2f1(-----, -----, -----, -) x 2 2 2 2 (d192) -------------------------------------------------------- n + 1 (-----)! 2 with (dimension) n = 2, yielding 2 3 4 5 6 b 11 b 823 b 150653 b 3362377 b (d294)/T/ x = b + -- + ----- + ------ + --------- + ---------- + . . . 10 350 63000 24255000 1051050000 2 9 a 1/3 with b = (----) . 32 (x ~ area^(2/3) ?!) Or you could even use the "continued sin" iteration 2 a + sin(2 a + sin(2 a + sin(2 a + . . .))) x = 1 - cos --------------------------------------------. 2 For the thickness of a spherical segment, you need only solve the cubic x 2 v = pi (1 - -) x . 3 E.g., a depth of pi x = 1 - 2 sin -- ~ 0.6527 18 quarter-fills a spherical tank of unit radius. Apropos the cyclotomic discussion, I assume everybody knows about the 2s in factor(x^105-1). What is the lowest degree polynomial with coeffs in {-1,0,1} with a coeff of 2 in its irreducible factorization? --rwg PS (reminder): a really obscure typo on p 573 of Watson's Bessel treatise: oo ==== \ 2 1 1 (2) > Bessel_J (n z) = -------------- - - / n 2 2 ==== 2 sqrt(1 - z ) n = 1 is OK, (although the summand looks more like Bessel_J (n z) 2 n ) but the rhs of oo ==== 2 2 \ 2 2 z (z + 4) (3) > n Bessel_J (n z) = -------------- X X / n 2 1/2 <<< X ==== 16 (1 - z ) X X n = 1 should have been the much less plausible 2 2 z (z + 4) --------------. 2 7/2 16 (1 - z ) Luckily TeX, though written at an AI lab, contains no AI. (Actually, really good AI would flag the inconsistent sqrt notations in the denominators, or even test the formulas!)
rwg wrote:
Apropos the cyclotomic discussion, I assume everybody knows about the 2s in factor(x^105-1). What is the lowest degree polynomial with coeffs in {-1,0,1} with a coeff of 2 in its irreducible factorization?
(1-x)(1+2x+x^2+x^3+x^4+...+x^n) will, of course, only have {-1,0,1} coeffs. Irreducibility fails for n=2, but the n=3 example x^4+x^2-x-1 works. (I'll let you dispatch degree 3 by hand yourself :-). --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
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Michael Kleber -
R. William Gosper