Kepler's Second Law of Planetary Motion

Diagram of Kepler's laws of planetary motion

 I wanted to play around with Celestia some more so I decided to see if I could practice understanding Kepler's Second Law of Planetary Motion that says

"A line joining a planet and the Sun sweeps out equal areas during equal intervals of time."

Diagram of Kepler's second law of planetary motion

To do this I decided to measure the area swept out of two sections of equal time of a planet's orbit around the Sun. To do this, I wanted to find the angle across the solar sky that the planet swept across in a certain amount of days and compare it to a different area of the orbit swept across in the same amount of days.


I also wanted to pick a planet with a high eccentricity and the planet with the most eccentric orbit is Mercury at .206.

According to Celestia, Mercury just passed aphelion on January 18 and will reach perihelion on March 2. So I arbitrarily chose two dates both 4.5 Earth days on either side of these dates and measured the distance from the Sun to Mercury and the angle on the Ecliptic grid Mercury was located at on these days. Because these dates are equally spaced from perihelion and aphelion, the distance from Mercury to the Sun will be equal in each pair.

ecliptic longitude = λ, distance = d

• t1 = January 13: λ1 = 245˚, d1 = 0.45954 AU
• t2 = January 22: λ2 = 269˚, d2 = 0.45954 AU
• t3 = February 26: λ3 = 49˚, d3 = 0.30962 AU
• t4 = March 2: λ4 = 105˚, d4 = 0.30962 

∆t12 = 9 days, ∆λ12 = 24˚ = 2π/15, ∆d12 = 0 AU
∆t34 = 9 days, ∆λ34 = 56˚ = 14π/45, ∆d34 = 0 AU


The equation to find these swept out areas is given by


A = (1/2)*(r^2)*(θ)


A12 = (1/2)*((0.45954 AU)^2)*(2π/15) = 0.0468483 (AU)^2
A34 = (1/2)*((0.30962 AU)^2)*(14π/45) = 0.0442288 (AU)^2


So indeed, it does seem that with equal time an equal elliptical area will be swept out by the orbiting planet in different sections of the orbit.


There was definitely some rounding and estimation with ecliptic longitude measurements done here to lose information which would account for my error, but I am satisfied with my results. Also I do realize it's pretty silly to get my "observational readings" from a computer simulation based on the actual laws I'm trying to prove, but this is pretty much the only feasible means I can make these measurements right now. I hope that someday I can use much more advanced equipment. :D

The Amazing Binary Systems of Castor

Looking at Gemini from Earth with the naked eye, one might conclude that Alpha Geminorum, aka Castor is one very bright star. However, astronomers have known since the 17th century that Castor is at least a binary system, meaning two stars gravitationally bound to each other. A little later, it was discovered that each of the two stars of Castor were themselves binary systems, making the whole thing at least a quadruple system. Today we now recognize a faint third binary system that is also gravitationally bound with the quadruple system. So, in all, we can call Castor a sextuple system.

Using my favorite 3D astronomy program Celestia with an add-on created by "Chuft-Captain" from the Celestia Motherlode, I've recorded how the system looks when sped up to 10 billion times the normal rate.