Hyperbolic Kepler’s Equation

Description

by Dr. D.G. Simpson, http://www.pgccphy.net/1030/software.html

Given the mean anomaly M (in degrees) and the orbit eccentricity e, this program solves the hyperbolic Kepler equation
M = e·sinh F - F

to find the variable F . This is a very simple implementation — it includes no convergence test, and simply solves the hyperbolic Kepler equation by performing 15 iterations of Newton’s method.
To run the program, enter:

M ENTER e f A

where M is in degrees. The program returns the variable F .

Example:
Let M = 60°, e = 1.15. Enter the above program, then type:

60 ENTER 1.15 f A

The program returns F = 1.5555.

Program Resources

Labels

Name Description
 A Main program, initialise registers
 0 Kepler’s Equation

Storage Registers

Name Description
.0 Anomaly M (in degrees)
.1 Initial orbit eccentricity e
.2 Initial orbit eccentricity e
.3 Control variable

Program

Line Display Key Sequence Line Display Key Sequence
000 018 45 .2 RCL . 2
001 42,21,11 f LBL A 019 42,22,23 f HYP SIN
002 44 .0 STO . 0 020 45 .0 RCL . 0
003 34 x↔y 021 20 ×
004 42 3 f → RAD 022 30
005 44 .1 STO . 1 023 45 .2 RCL . 2
006 44 .2 STO . 2 024 42,22,24 f HYP COS
007 1 1 025 45 .0 RCL . 0
008 48 . 026 20 ×
009 0 0 027 34 x↔y
010 1 1 028 30
011 4 4 029 10 ÷
012 44 .3 STO . 3 030 30
013 45 .2 RCL . 2 031 44 .2 STO . 2
014 42,21, 0 f LBL 0 032 42, 6, .3 f ISG . 3
015 45 .1 RCL . 1 033 22 0 GTO 0
016 45 .2 RCL . 2 034 43 32 g RTN
017 40 +