13.15: Calculating the Elementss
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We can now immediately calculate the semi latus rectum from Equation 13.12.6a (recalling that 2f3=v2−v1, so that everything except l in the Equation is already known.) In fact we have three opportunities for calculating the semi latus rectum by using each of Equations 13.12.6a,b,c, and this serves as a check on the arithmetic. For our numerical example, I obtain
l=2.61779
identically (at least to eleven significant figures) for each of the three permutations.
Now, on referring to Equation 2.3.37, we recall that the polar Equation to an ellipse is
r=l1+ecosv.
We therefore have, for the first and third observations,
ecosv1=l/r1−1
and, admitting that v3=v1+2f2,
ecos(v1+2f2)=l/r3−1.
We observe that, in Equations 13.15.2 and 13.15.3, the only quantities we do not already know are v1 and e – so we are just about to find our first orbital element, the eccentricity!
A hint for solving Equations 13.15.2 and 3: Expand cos(v1+2f2). Take esinv to the left hand side, and Equation 13.15.3 will become
esinv1=(l/r1−1).cos2f2−(l/r3−1)sin2f2.
After this, it is easy to solve Equations 13.15.2 and 13.15.4 for e and for v1. The other true anomalies are given by v2=v1+2f3 and v3=v1+2f2. A check on the arithmetic may (and should) be performed by carrying out the same calculation for the first and second observations and for the second and third observations. For all three, I obtained
e=0.23875_
We have our first orbital element!
(The MPC value for the eccentricity for this epoch is 0.22994 – but this is based on all available observations, and we cannot expect to get the MPC value from just three hypothetical “observations”.)
The true anomalies at the times of the three observations are
v1=191∘.99814v2=192∘.68221v3=194∘.05377
After that, the semi major axis is easy from Equation 2.3.10, l=a(1−e2), for the semi latus rectum of an ellipse. We find
a=2.77602 au_
The period in sidereal years is given by P2=a3, and is therefore 4.62524 sidereal years. This is not one of the six independent elements, since it is always related to the semi major axis by Kepler’s third law, so it doesn’t merit the extra dignity of being underlined. However, it is certainly worth converting it to mean solar days by multiplying by 365.25636. We find that P=1689.39944 days.
The next element to yield will be the time of perihelion passage. We find the eccentric anomalies for each of the three observations from any of Equations 2.3.16, 17a, 17b or 17c. For example:
cosE=e+cosv1+ecosv.
Then the time of perihelion passage will come from Equations 9.6.4 and 9.6.5:
T=t−P2π(E−esinE)+nP.
With n=1 I make this T=t1+756d.1319_
The next step is to calculate the Ps and Qs. These are defined in Equation 10.9.40. They are the direction cosines relating the heliocentric plane-of-orbit basis set to the heliocentric equatorial basis set.
Exercise. Apply Equation 10.9.50 to the first and third observations to show that
Px=ξ1r3sinv3−ξ3r1sinv1r1r3sin2f2
and Qx=ξ3r1cosv−ξ1r3cosv3r1r2sin2f2
From Equations 10.9.51 and 52, find similar Equations for Py, Qy, Pz, Qz.
The numerical work can and should be checked by calculating these direction cosines also from the first and second, and from the second and third, observations. Check also that P2x+P2y+P2z=Q2x+Q2y+Q2z=1. I get
Px=−0.48044Py=+0.86568Pz=−0.14059
Qx=−0.87392Qy=−0.45907Qz=+0.15978
(Remember that my computer is carrying all significant figures to double precision, though I print out here only a limited number of significant figures. You will not get exactly my numbers unless you, too, carry all significant figures and do not prematurely round off.)
The direction cosines are related to the Eulerian angles, of course, by Equations 10.9.41- 46 (how could you possibly forget?!). All (!) you have to do, then, is to solve these six Equations for the Eulerian angles. (You need six Equations to remove quadrant ambiguity from the angles. Remember the ATAN2 function on your computer – it’s an enormous help with quadrants.)
Exercise. Show that (or verify at any rate) that:
sinωsini=Pzcosε−Pysinε
and cosωsini=Qzcosε−Qysinε.
You can now solve this for the argument of perihelion ω. Don’t yet try to solve it for the inclination. (Why not?!) Using ε=23∘.438 960 for the obliquity of the ecliptic of date (calculated from page B18 of the 2002 Astronomical Almanac), I get
ω=304∘.81849_
Exercise. Show that (or verify at any rate) that:
sinΩ=(Pycosω−Qysinω)secε
and cosΩ=Pxcosω−Qxsinω.
From these, I find:
Ω=172∘.64776_
One more to go!
Exercise. Show that (or verify at any rate) that:
cosi=−(Pxsinω+Qxcosω)cscΩ.
You can now solve this with Equation 13.15.9 or 13.15.10 (or both, as a check on the arithmetic) for the inclination. I get
i=35∘.20872_
Here they are, all together:
a=2.77602 AUi=35∘.20872e=0.23875Ω=172∘.64776T=t1+756d.1319ω=304∘.81849
Have we made any mistakes? Well, presumably after you read chapter 10 you wrote a program to generate an ephemeris. So now, use these elements to see whether they will reproduce the original observations! Incidentally, to construct an ephemeris, there is no need actually to use the elements – you can use the Ps and Qs instead.