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Physics LibreTexts

13.7: Geocentric and Heliocentric Distances - First Attempt

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Let us write down the three heliocentric equatorial components of Equation 13.2.1:

ξ2=a1ξ1+a3ξ3,

η2=a1η1+a3η3,

ζ2=a1ζ1+a3ζ3.

Now write lxo for ξ, etc., from Equations 13.5.1,2,3 and rearrange to take the solar coordinates to the right hand side:

l1a11l22+l3a33=a1xo1xo2+a3xo3,

m1a11m22+m3a33=a1yo1yo2+a3yo3,

n1a11n22+n3a33=a1zo1zo2+a3zo3.

As a very first, crude, approximation, we can let a1=b1 and a3=b3, for we know b1 and b3 (in our numerical example, b1=23, b3=13), so we can solve Equations 13.7.4,5,6 for the three geocentric distances. However, we shall eventually need to find the correct values of a1 and a3.

When we have solved these Equations for the geocentric distances, we can then find the heliocentric distances from Equations 13.5.1,2 and 3. For example,

ξ1=l11xo1

and of course

r21=ξ21+η21+ζ21.

In our numerical example, we have

l1=+0.722 980 907
l2=+0.715 380 933
l3=+0.698 125 992

m1=0.631 808 343
m2=0.641 649 261
m3=0.664 816 398

n1=+0.279 493 876
n2=+0.276 615 882
n3=+0.265 780 465

As a check on the arithmetic, the reader can - and should - verify that

l21+m21+n21=l22+m22+n22=l23+m23+n23=1

This does not verify the signs of the direction cosines, for which care should be taken.

From The Astronomical Almanac for 2002, we find that

xo1=306 728 3yo1=+0.889 290 0zo1=+0.385 549 5AU
xo2=386 194 4yo2=+0.862 645 7zo2=+0.373 999 6
xo3=536 330 8yo3=+0.791 387 2zo2=+0.343 100 4

(For a fraction of a day, which will usually be the case, these coordinates can be obtained by nonlinear interpolation – see chapter 1, section 1.10.)

Equations 13.7.4,5,6 become

+0.481 987 27110. 715 380 9332+0. 232 708 6643=0.002 931 933

0.421 205 5621+0.641 649 26120.221 605 4663=0.005 989 967

+0.186 329 25110.276 615 88220.088 593 4883=0.002 599 800

I give below the solutions to these Equations, which are our first crude approximations to the geocentric distances in AU, together with the corresponding heliocentric distances. I also give, for comparison, the correct values, from the published MPC ephemeris

First crude estimatesMPC1=2.725 71r1=3.485 321=2.644r1=3.4062=2.681 60r2=3.481 332=2.603r2=3.4043=2.610 73r3=3.474 713=2.536r3=3.401

This must justifiably give cause for some satisfaction, because we now have some idea of the geocentric distances of the planet at the instants of the three observations, though it is a little early to open the champagne bottles. We still have a little way to go, for we must refine our values of a1 and a3. Our first guesses, a1=b1 and a3=b3, are not quite good enough.

The key to finding the geocentric and heliocentric distances is to be able to determine the triangle ratios a1=A1/A2, a3=A3/A2 and the triangle/sector ratios a/b. The sector ratios are found easily from Kepler’s second law. We have made our first very crude attempt to find the geocentric and heliocentric distances by assuming that the triangle ratios are equal to the sector ratios. It is now time to improve on that assumption, and to obtain better triangle ratios. After what may seem like a considerable amount of work, we shall obtain approximate formulas, Equations 13.8.35a,b, for improved triangle ratios. The reader who does not wish to burden himself with the details of the derivation of these Equations may proceed directly to them, near the end of Section 13.7


This page titled 13.7: Geocentric and Heliocentric Distances - First Attempt is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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