Kepler's 2nd. Law

[toml_12953]

(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