13.10: Higher-order Approximation
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The reason that we made the approximation to order τ3 was that, in evaluating the expressions for F1, G1, F3 and G3, we did not know the radial velocity ˙r2. Perhaps we can now evaluate it.
Exercise. Show that the radial velocity of a particle in orbit around the Sun, when it is at a distance r from the Sun, is
Ellipse:˙r=∓√GMa0(a2e2−(a−r)2ar2)1/2,
Parabola:˙r=∓√GM(r−q)a0,
Hyperbola:˙r=∓√GMa0((a+r)2−a2e2ar2)1/2.
Show that the radial velocity is greatest at the ends of a latus rectum.
Here a0 is the astronomical unit, a is the semi major axis of the elliptic orbit or the semi transverse axis of the hyperbolic orbit, q is the perihelion distance of the parabolic orbit, and e is the orbital eccentricity. The − sign is for pre-perihelion, and the + sign is for post-perihelion.
Unfortunately, while this is a nice exercise in orbit theory, we do not know the eccentricity, so these formulas at present are of no use to us.
However, we can calculate the heliocentric distances at the times of the first and third observations by exactly the same method as we used for the second observation. Here are the results for our numerical example, after one iteration. The units, of course, are au. Also indicated are the instants of the observations, taking t2=0 and expressing the other instants in units of 1/k (see section 13.8).
t1=−τ3=−0.086 010 494 75r1=3.419 52
t2=0r2=3.416 73
t3=+τ1=+0.172 020 989 5r3=3.410 82
We can fit a quadratic expression to this, of the form:
r=c0+c1t+c2t2
With our choice of time origin t2=0, c0 is obviously just equal to r2, so we have just two constants, c1 and c2 to solve for. We can then calculate the radial velocity at the time of the second observation from
˙r2=c1+2c2t2.
We can calculate A1, A2 and A3 in the same manner as before, up to τ4 rather than just τ3. The algebra is slightly long and tedious, but straightforward. Likewise, the results look long and unwieldy, but there is no difficulty in programming them for a computer, and the actual calculation is, with a modern computer, virtually instantaneous. The results of the algebra that I give below are taken from the book Determination of Orbits by A.D. Dubyago (which has been the basis of much of this chapter). I haven’t checked the algebra myself, but the conscientious reader will probably want to do so himself or herself.
A1=12√lτ1(1−τ216r32+τ314r42˙r2),
A2=12√lτ2(1−τ226r32+τ22(τ1−τ3)4r42˙r2),
A3=12√lτ3(1−τ236r32−τ334r42˙r2).
And from these,
a1=τ1τ2(1+τ3(τ2+τ1)6r32+τ3(τ3(τ1+τ3)−τ21)4r42˙r2).
and a3=τ3τ2(1+τ1(τ2+τ3)6r32−τ1(τ1(τ1+τ3)−τ234r42˙r2).
This might result in slightly better values for a1 and a3. I have not calculated this for our numerical example here, for reasons given in Section 13.9. We can move on to the next section, using our current vales of a1 and a3, namely
a1=0.666 770anda3=0.333 416.