The eccentricity is just a little more than 1, why can't everything be regarded as a parabola?

Let the track go from slightly closed to slightly open, and compare the positions at the same time.

One orbit report writes e=0.999, and the other writes 1.001. Both look long, but one is still an ellipse and the other is a hyperbola. We retain this difference and truly find the position before and after perihelion.

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Guess first

Can both e=0.999 and e=1.001 be directly regarded as e=1 without losing information?

Show answer and explanation

Can't — Closing and opening are different in nature, and the position difference will accumulate over time.

Then connect experience and science

Don't write "very close to 1" directly as 1

e corresponds to different orbits on both sides of 1: less than 1 is still a very oblate ellipse, and greater than 1 is a hyperbola. Small differences may add up to different positions.

e<1, e=1, e>1 must be distinguished

Use parabola solution as starting point

Chapter 35 of the original book starts from a close parabolic solution and then uses Landgraf/Stumpff series to correct it.

g=(1−e)/(1+e)

Iterate until stable

In each round, the series is accumulated first, and then the cubic equation is solved once; if the number of iterations is exceeded or the term is too large, a non-convergence report must be made.

Outer level iteration + inner Barker type root finding

Compare it with the independent two-body solution

Independently find roots of elliptic or hyperbolic equations, checked against near-parabolic algorithms. All small differences cannot be regarded as display errors.

r=q(1+e)/(1+e cosν)

First find the initial values of g and the parabola, then accumulate Stumpff corrections in the outer layer and solve the Barker type equation in the inner layer until convergence. The final formula only replaces ν with r, which cannot replace the previous iteration; see below for the scope of this lesson and failure handling.

give it a try

Keep q and the number of days unchanged, and choose 0.98, 1, and 1.02 in order. The results are calculated iteratively from the original book and compared with independent two-body root finding.

Take a look at it →

No need to memorize symbols

g: (1−e)/(1+e)
s: tan(ν/2)
Iteration residuals: The remaining error in the candidate answer substitution equation

Algorithm, accuracy and scope of application

Implement Equation 35.1 of Printed Pages 245–247 and the two-level iteration of the Landgraf program. Maximum of 50 terms/50 rounds. If the terms exceed 10,000 or fail to converge, it will be a clear failure and no secret substitution of methods will be allowed. The independent conicReference uses stable small series and bisection methods to solve elliptic/hyperbolic equations for verification, and the UI display position is poor. The six sets of exercises in the original book cover q=.1, e=.987, t=254.9, etc., and the test range can be larger than the UI. AA+ CalculateTrueAnomalyAndRadius implements another expansion, so the two are not required to be equal bit by bit.

my understanding check

e slightly greater than 1 belongs to?

hyperbola — 理想两体模型中e>1为hyperbola。

What should I do when the series does not converge?

Report failure explicitly — Numerical failure cannot be disguised as legal ephemeris.

Is there another approximate branch of AA+ considered as exactly the same algorithm in this lesson?

No — This lesson implements the main iteration of Landgraf in the original book, using independent two-body solutions for verification; AA+ also uses near-parabolic expansions.

Getting close to the critical value does not mean that the difference can be erased. The algorithm should preserve the orbit type and also be able to identify non-convergence.

Jean Meeus · Astronomical Algorithms, 2nd ed. · No. 35 chapter

Planets and Small Solar System Objects · All interactive courses