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 solves Kepler’s equation
M = E - e·sin E

to find the eccentric anomaly E. This is a very simple implementation—it includes no convergence test, and simply solves Kepler’s 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 eccentric anomaly E in degrees.
After running the program, the calculator will be set to degrees mode.

Example:
Let M = 60°, e = 0.15. Enter the above program, then type:

60 ENTER .15 f A

The program returns E = 67.9667°.

Program Resources

Labels

Name Description
 A Main program, initialise registers
 0 Newton’s algorithm

Storage Registers

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

Program

Line Display Key Sequence Line Display Key Sequence
000 019 45 .2 RCL . 2
001 42,21,11 f LBL A 020 23 SIN
002 44 .0 STO . 0 021 45 .0 RCL . 0
003 34 x↔y 022 20 ×
004 42 3 f → RAD 023 40 +
005 44 .1 STO . 1 024 45 .2 RCL . 2
006 44 .2 STO . 2 025 24 COS
007 43 8 g RAD 026 45 .0 RCL . 0
008 1 1 027 20 ×
009 48 . 028 1 1
010 0 0 029 30
011 1 1 030 10 ÷
012 4 4 031 30
013 44 .3 STO . 3 032 44 .2 STO . 2
014 45 .2 RCL . 2 033 42, 6, .3 f ISG . 3
015 42,21, 0 f LBL 0 034 22 0 GTO 0
016 45 .1 RCL . 1 035 43 3 g →DEG
017 45 .2 RCL . 2 036 43 7 g DEG
018 30 037 43 32 g RTN