Kepler's Data, Discrete Method by Hand

We may follow Kepler to derive the pitch sequence for a planet without the use of ordinary differential equations. This approach uses a discrete form of Kepler's Second Law.

Discrete Form of Kepler's Second Law.

To take a concrete example, consider Mercury. The period of Mercury, in mean Earth days, is approximately 88 days. As we have seen above regarding Kepler's constant, we expect -- on average -- a daily increment in the observed solar angle nu of Mercury of about 246 minutes of arc. On the other hand, this increment (Dnu) should be smaller when
  • planet near aphelion,
  • nu is near zero (or 2*PI, 360 degrees),
  • the planet is farther from the sun (rho > 1), and
  • the planet moves more slowly.
And Dnu should be larger when
  • planet near perihelion,
  • nu is near PI or 180 degrees,
  • the planet is closer to the sun (rho < 1), and
  • the planet moves more quickly.
The precise form of Kepler's Second Law in this context may be obtained from equation [4] by replacing the instanteneous rate of change, nu-dot, by a discrete approximation, Dnu/Dt, where Dt is one mean earth day. Thus we have,
[4d] Dnu ~= K/rho2
or in this case,
[4d] Dnu ~= 245.5410/rho2 for Mercury
This suggests the following method to create a table of daily values for Dnu, which -- for Kepler -- determines the relative pitch of the music of the spheres.

Step 1 Determine the current value of nu by adding together the solar angle nu of the preceding day, and the corresponding increment, Dnu.

nu(today) = nu(yesterday) + Dnu(yesterday)

Step 2 Determine the current value of the solar distance, rho(today), from nu(today) using equation,

[6] rho = 1-e2/(1-e*cos(nu))

Step 3 Determine the current Dnu from equation [4d] above.

A table for Mercury

We will make a table in which each row corresponds to a single day,
so there will be 88 rows in all. In each row there will be four columns:
Day Dnu Nu Rho

The first row will be the initial data. It is a special case, following which
we may continue with Steps 1,2,3. Let us
begin with the planet at the top, or nu=0 on day 0. Then by equation [6] or by consulting the figures, rho = 1+e. For Mercury, this will be about 1.2. We now need Dnu from [4d],
Dnu = 245.5410/(1.2)2 ~= 170.5 minutes
Now we have the first (initial) row complete:

Day Nu Rho Dnu
0 0 1.2 170.5

Now we apply Step 1 to find ,

nu(1) = nu(0)+Dnu(0) = 0 + 170.5 = 170.5
and Step 2 to find (using the computer or hand calculator)
rho(1) = 1-e2/(1-e*cos(nu))= 0.96/(1-0.2*0.9)~=1.171
and finally Step 3 to find,
Dnu(1) ~= 245.5410/(1.171)2 = 179.1
and now our table, with complete row for day 1, is:
Day Nu Rho Dnu
0 0 1.2 170.5
1 170.5 1.171 179.1

It is easiest to continue in a spreadsheet program, in which these equations are set into the headings
of the columns. And remember, K may have to be adjusted, and the whole spreadsheet repeated, so
that we get a full revolution of 21600 minutes for nu on the 88-th day.

Revised 10 jan 2002 by ralph abraham