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Physics LibreTexts

13.12: Sector-Triangle Ratio

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We recall that it is easy to determine the ratio of adjacent sectors swept out by the radius vector. By Kepler’s second law, it is just the ratio of the two time intervals. What we really need, however, are the triangle ratios, which are related to the heliocentric distance by Equation 13.2.1. Oh, wouldn’t it just be so nice if someone were to tell us the ratio of a sector area to the corresponding triangle area! We shall try in this section to do just that.

Notation: Triangle ratios:a1=A1/A2,a3=A3/A2.

Sector ratios:b1=B1/B2,b3=B3/B2.

Sector-triangle ratios:R1=B1A1,R2B2A2,R3=B3A3,

from which it follows that

a1=R2R1b1,a3=R2R3b3.

We also recall that subscript 1 for areas refers to observations 2 and 3; subscript 2 to observations 3 and 1; and subscript 3 to observations 1 and 2. Let us see, then, whether we can determine R3 from the first and second observations.

Readers who wish to avoid the heavy algebra may proceed direct to Equations 13.12.25 and 13.12.26, which will enable the calculation of the sector-triangle ratios.

Let (r1,v1) and (r2,v2) be the polar coordinates (i.e. heliocentric distance and true anomaly) in the plane of the orbit of the planet at the instant of the first two observations. In concert with our convention for subscripts involving two observations, let

2f3=v2v1.

We have R3=B3/A3. From Equation 13.4.1, which is Kepler’s second law, we have, in the units that we are using, in which GM=1, ˙B=12l and therefore B3=12lτ3. Also, from the z-component of Equation 13.8.15c, we have A3=12r1r2sin(v2v1).

Therefore R3=lτ3r1r2sin(v2v1)=lτ3r1r2sin2f3.

In a similar manner, we have

R1=lτ1r2r3sin(v3v2)=lτ1r2r3sin2f1

R2=lτ2r3r1sin(v3v1)=lτ2r3r1sin2f2.

I would like to eliminate l from here.

I now want to recall some geometrical properties of an ellipse and a property of an elliptic orbit. By glancing at figure II.11, or by multiplying Equations 2.3.15 and 2.3.16, we immediately see that rcosv=a(cosEe), and hence by making use of a trigonometric identity we find

rcos212v=a(1e)cos212E,

and in a similar manner it is easy to show that

rsin212v=a(1+e)sin212E.

Here E is the eccentric anomaly.

Also, the mean anomaly at time t is defined as 2πP(tT) and is also equal (via Kepler’s Equation) to EesinE. The period of the orbit is related to the semi major axis of its orbit by Kepler’s third law: P2=4π2GMa3. (This material is covered on Chapter 10.) Hence we have (in the units that we are using, in which GM=1):

EesinE=tTa3/2,

where T is the instant of perihelion passage.

Now introduce 2f3=v2v1,

2F3=v2+v1,

2g3=E2E1,

2G3=E2+E1.

From Equation 13.12.7 I can write

r1r2cos12v1cos12v2=a(1e)cos12E1cos12E2

and from Equation 13.12.8 I can write

r1r2sin12v1sin12v2=a(1+e)sin12E1sin12E2.

I now make use of the sum of the sum-and-difference formulas from page 38 of chapter 3, namely cosAcosB=12(cosS+cosD) and sinAsinB=12(cosDcosS), to obtain

r1r2(cosF3+cosf3)=a(1e)(cosG3+cosg3)

and r1r2(cosf3cosF3)=a(1+e)(cosg3cosG3).

On adding these, we obtain

r1r2cosf3=a(cosg3ecosG3).

I leave it to the reader to derive in a similar manner (also making use of the formula for the semi latus rectum l=a(1e2))

r1r2sinf3=alsing

and r1+r2=2a(1ecosg3cosG3).

We can eliminate ecosG from Equations 13.12.18 and 13.12.20:

r1+r22r1r2cosf3cosg3=2asin2g3

Also, if we write Equation 13.12.9 for the first and second observations and take the difference, and then use the formula on page 35 of chapter 3 for the difference between two sines, we obtain

2(g3esing3cosG3)=τ3a3/2.

Eliminate ecosG3 from Equations 13.12.18 and 13.12.22:

2g3sin2g3+2r1r2asing3cosf3=τ3a3/2.

Also, eliminate l from Equations 13.12.6 and 13.12.19:

R3=τ32ar1r2cosf3sing3.

We have now eliminated F3, G3 and e, and we are left with Equations 13.12.21, 23 and 24, the first two of which I now repeat for easy reference:

r1+r22r1r2cosf3cosg3=2asin2g3

2g3sin2g3+2r1r2asing3cosf3=τ3a3/2.

In these Equations we already know an approximate value for f3 (we’ll see how when we resume our numerical example); the unknowns in these Equations are R3, a and g3, and it is R3 that we are trying to find. Therefore we need to eliminate a and g3. We can easily obtain a from Equation 13.12.24, and, on substitution in Equations 13.12.21 and 23 we obtain, after some algebra:

R23=M23N3cosg3

and R33R23=M23(g3sing3cosg3)sin3g3,

where M3=τ32(r1r2cosf3)3/2

and N3=r1+r22r1r2cosf3.

Similar Equations for R1 and R2 can be obtained by cyclic permutation of the subscripts. Equations 13.12.25 and 26 are two simultaneous Equations in R3 and g3. Their solution is given as an example in section 1.9 of chapter 1, so we can now assume that we can calculate the sector-triangle ratios.

We can then calculate better triangle ratios from Equations 13.12.4 and return to Equations 13.7.4, 5 and 6 to get better geocentric distances. From Equations 13.7.8 and 9 calculate the heliocentric distances. Make the light-time corrections. (I am not doing this in our numerical example because our original positions were not actual observations, but rather were ephemeris positions.) Then go straight to this section (13.12) again, until you get to here again. Repeat until the geocentric distances do not change.


This page titled 13.12: Sector-Triangle Ratio 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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