Kepler's 2nd. Law
02-12-2018, 09:53 AM (This post was last modified: 02-12-2018 10:43 AM by Ángel Martin.)
Post: #1
 Ángel Martin Senior Member Posts: 895 Joined: Dec 2013
Kepler's 2nd. Law
From a recent conversation with a friend's on the Kepler's laws - his son's subject for a High school paper. The goal was to calculate the value of the swept area between two instants, as determined by the azimuth angles (a1, a2) of the segments linking the focus of the ellipse with the planet at those moments.

Initially I thought the formulas would involve the Elliptic functions, as elliptical sectors were involved - but it appears that's not the case when the coordinates are centered at the focal point, instead of at the center of the ellipse. I found that fact interesting, as it only involves trigonometric functions (not even hyperbolic).

Here's the reference I followed to program it, a good article that describes an ingenious approach - avoiding painful integration steps. It may not be the simplest way to get this done, chime in if you know a better one.

And here's the FOCAL listing for a plain HP-41 - no extensions whatsoever. The result is the area swept between the two positions defined by the angles a1 and a2; a2 > a1. The parameters a,b are the semi-axis of the ellipse, a>b.

Code:
1    LBL "K2+"     2    RAD     3    "a^b=?"     4    PROMPT     5    X>Y?       b > a ? 6    X<>Y       yes, swap 7    STO 02     b 8    X^2        b^2 9    X<>Y     10   STO 01     a 11   X^2        a^2 12   -          a^2 - b^2 13   ABS     14   SQRT     15   RCL 01     a 16   /     17   STO 00     e 18   LBL 00     19   "<)1^<)2=?"     20   PROMPT     21   X<Y?       a2 < a1 ? 22   X<>Y       yes, swap 23   STO 04     a2 24   RDN     25   STO 03     a1     26   PI        27   X<=Y?      a1 >= pi ? 28   GTO 01     yes, -> case 2 29   RCL 04     a2 30   X<=Y?      a2 <= pi ? 31   GTO 01     yes, -> case1 32   X<>Y       no, it's case 3 33   XEQ 02     first between [pi, a2] 34   STO O      partial result 35   RCL 03     a1 36   PI         now between [a1, pi] 37   XEQ 02     38   RCL O     39   +     40   RTN     41   GTO 00     42   LBL 01     43   RCL 04     a2 44   RCL 03     a1 45   LBL 02     46   XEQ 05     47   STO  M     f1 48   X<>Y       a2 49   XEQ 05     50   STO  N     f2 51   RCL  M     52   -     53   2     54   /          (f2 - f1) /2 55   ENTER^     56   SIN     57   RCL  M     f1 58   RCL  N     f2 59   +     60   2     61   /          (f2 + f1) /2 62   COS     63   *     64   RCL 00     e 65   *     66   -     67   RCL 01     a 68   X^2        a^2 69   *     70   1     71   RCL 00     72   X^2     73   -          1-e^2 74   SQRT     75   *         76   ABS     77   RTN     78   GTO 00     79   LBL 05     80   COS        cos a 81   RCL 00     82   X<>Y     83   +          e + cos a 84   LASTX      cos a 85   RCL 00     e 86   *          e.cos a 87   1     88   +     89   /          cos f =  (e + cos a)/(1 + e.cos a) 90   ACOS       f 91   END

(*) the symbol "<)" is for the angle character.

Example: calculate the area swept between a1 = pi/4 and a2 = 3.pi/4, if the ellipse parameters are a= 2, b= 3

Solution: A = 1.989554087

Once the area is obtained, and knowing the period of the orbiting (T), it's straightforward to determine the time taken by the planet to travel between the two positions, with the direct application of Kepler's 2nd. law:

t = T . Area / pi.a.b

Cheers,
ÁM
02-13-2018, 01:07 AM (This post was last modified: 02-13-2018 02:41 AM by toml_12953.)
Post: #2
 toml_12953 Senior Member Posts: 711 Joined: Dec 2013
RE: Kepler's 2nd. Law
(02-12-2018 09:53 AM)Ángel Martin Wrote:  From a recent conversation with a friend's on the Kepler's laws - his son's subject for a High school paper. The goal was to calculate the value of the swept area between two instants, as determined by the azimuth angles (a1, a2) of the segments linking the focus of the ellipse with the planet at those moments.

Initially I thought the formulas would involve the Elliptic functions, as elliptical sectors were involved - but it appears that's not the case when the coordinates are centered at the focal point, instead of at the center of the ellipse. I found that fact interesting, as it only involves trigonometric functions (not even hyperbolic).

Here's the reference I followed to program it, a good article that describes an ingenious approach - avoiding painful integration steps. It may not be the simplest way to get this done, chime in if you know a better one.

Example: calculate the area swept between a1 = pi/4 and a2 = 3.pi/4, if the ellipse parameters are a= 2, b= 3

Solution: A = 1.989554087

For the parameters a=2, b=3 and a1=.785398..., a2=2.35619...

(Did I get the parameters right?)

I get 5.89676233948396 which agrees with the calculator at

http://keisan.casio.com/exec/system/1343722259

Code:
DECLARE EXTERNAL FUNCTION F INPUT PROMPT "Enter semimajor axis a: ":a INPUT PROMPT "Enter semiminor axis b: ":b INPUT PROMPT "Enter Theta0: ":theta0 INPUT PROMPT "Enter Theta1: ":theta1 LET S = F(a,b,Theta1) - F(a,b,Theta0) PRINT "Area =";S END EXTERNAL FUNCTION F(a,b,t) LET F = a*b/2*(t-ATN((b-a)*SIN(2*t)/(b+a+(b-a)*COS(2*t)))) END FUNCTION

Tom L

DM42 SN: 00025 (Beta)
SN: 00221 (Production)
02-13-2018, 02:01 AM
Post: #3
 SlideRule Senior Member Posts: 336 Joined: Dec 2013
RE: Kepler's 2nd. Law
Angel

The 143 page publication Introduction to Orbital Flight Planning (I) (NASA-CE-165052) may be of interest as it includes a well-documented {equations, diagrams, etc.}discussion with respect to orbital mechanics & Kepler as well as an HP-67 based program for same. The discussion starts with elliptic orbits on page 51 then transitions to Kepler on page 54 with an HP-67 program on page 57.

BEST!
SlideRule
02-13-2018, 08:06 AM (This post was last modified: 02-13-2018 08:07 AM by Ángel Martin.)
Post: #4
 Ángel Martin Senior Member Posts: 895 Joined: Dec 2013
RE: Kepler's 2nd. Law
(02-13-2018 01:07 AM)toml_12953 Wrote:
(02-12-2018 09:53 AM)Ángel Martin Wrote:  Example: calculate the area swept between a1 = pi/4 and a2 = 3.pi/4, if the ellipse parameters are a= 2, b= 3

Solution: A = 1.989554087

For the parameters a=2, b=3 and a1=.785398..., a2=2.35619...

(Did I get the parameters right?)

I get 5.89676233948396 which agrees with the calculator at

http://keisan.casio.com/exec/system/1343722259

Mmm... something's fishy. I consistently get 1.989554087 using a1 = pi/4 (45 degrees) and a2 = 3.pi/4 (135 degrees).
Besides, it's not possible to get the same as the web applet - which uses centered sectors, not focus-centered ones.

In fact, using the angles above the applet shows an Area result of 3.5280156212854

Could you double check?
02-13-2018, 08:11 AM
Post: #5
 Ángel Martin Senior Member Posts: 895 Joined: Dec 2013
RE: Kepler's 2nd. Law
(02-13-2018 02:01 AM)SlideRule Wrote:  The 143 page publication Introduction to Orbital Flight Planning (I) (NASA-CE-165052) may be of interest as it includes a well-documented {equations, diagrams, etc.}discussion with respect to orbital mechanics & Kepler as well as an HP-67 based program for same. The discussion starts with elliptic orbits on page 51 then transitions to Kepler on page 54 with an HP-67 program on page 57.

Great reference, thanks for the link. I'll check it out ASAP.
02-13-2018, 07:32 PM
Post: #6
 Luigi Vampa Member Posts: 192 Joined: Dec 2015
RE: Kepler's 2nd. Law
(02-13-2018 02:01 AM)SlideRule Wrote:  Angel
The 143 page publication Introduction to Orbital Flight Planning (I) (NASA-CE-165052) may be of interest as it includes a well-documented {equations, diagrams, etc.}discussion with respect to orbital mechanics & Kepler as well as an HP-67 based program for same. The discussion starts with elliptic orbits on page 51 then transitions to Kepler on page 54 with an HP-67 program on page 57.
BEST!
SlideRule

Big thanks for the information!

Saludos Saluti Cordialement Cumprimentos MfG BR + + + + +
Luigi Vampa +
Free42 BlackviewA7 '<3' I + + +
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