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# 13.8: Improved Triangle Ratios

The equation of motion of the orbiting body is

$\ddot{\textbf{r}} = - \frac{GM}{r^3} \textbf{r} . \label{13.8.1} \tag{13.8.1}$

If we recall equation 13.4.2, this can be written

$\ddot{\textbf{r}} = -k^2 \left( \frac{a^3}{r^3} \right) \textbf{r} . \label{13.8.2} \tag{13.8.2}$

If we now agree to express $$r$$ in units of $$a$$ (i.e. in Astronomical Units of length (au)) and time in units of $$1/k$$ $$(1/k = 58.132 \ 440 \ 87$$ mean solar days), this becomes merely

$\ddot{\textbf{r}} = -\frac{1}{r^3} \textbf{r} . \label{13.8.3} \tag{13.8.3}$

In these units, $$GM$$ has the value $$1$$.

Now write the $$x$$- and $$y$$- components of this equation, where $$(x , \ y)$$ are heliocentric coordinates in the plane of the orbit (see sections 13.5 or 10.7).

$\ddot{x} = -\frac{x}{r^3} \label{13.8.4} \tag{13.8.4}$

and $\ddot{y} = - \frac{y}{r^3}, \label{13.8.5} \tag{13.8.5}$

where $x^2 + y^2 = r^2 . \label{13.8.6} \tag{13.8.6}$

The areal speed is $$\frac{1}{2}h = \frac{1}{2} \sqrt{GMl}$$ or, in these units, $$\frac{1}{2} \sqrt{l}$$ where l is the semi latus rectum of the orbit in $$\text{au}$$

Let the planet be at $$(x, \ y)$$ at time $$t$$. Then at time $$t + δt$$ it will be at $$(x + δx , y + δy)$$, where

$δx = \dot{x} δt + \frac{1}{2!} \ddot{x} (δt)^2 + \frac{1}{3!} \dddot{x} (δt)^3 + \frac{1}{4!}\ddddot{x} (δt)^4 + ... \label{13.8.7} \tag{13.8.7}$

and similarly for y.

Now, starting from equation \ref{13.8.4} we obtain

$\dddot{x} = \frac{3x \dot{r}}{r^4} - \frac{\dot{x}}{r^3} \label{13.8.8} \tag{13.8.8}$

and $\ddot{\ddot{x}} = 3 \left( \frac{\dot{x}\dot{r}}{r^4} + \frac{x \ddot{r}}{r^4} - \frac{4x \dot{r}^2}{r^5} \right) - \frac{r^3 \ddot{x} - 3r^2 \dot{x} \dot{r}}{r^6} . \label{13.8.9} \tag{13.8.9}$

(The comment in the paragraph preceding equation 3.4.16 may be of help here, in case this is heavy-going.)

Now $$\ddot{x}$$ and $$x$$ are related by equation \ref{13.8.4}, so that we can write equation \ref{13.8.9} with no time derivatives of $$x$$ higher than the first, and indeed it is not difficult, because equation \ref{13.8.4} is just $$r^3 \ddot{x} = -x$$. We obtain

and $\ddot{\ddot{x}} = x \left( \frac{1}{r^6} - \frac{12 \dot{r}^2}{r^5} + \frac{3 \ddot{r}}{r^4} \right) + \frac{6 \dot{x} \dot{r}}{r^4} . \label{13.8.10} \tag{13.8.10}$

In a similar fashion, because of the relation \ref{13.8.4}, all higher time derivatives of $$x$$ can be written with no derivatives of $$x$$ higher than the first, and a similar argument holds for $$y$$.

Thus we can write equation \ref{13.8.7} as

$x + δx = Fx + G \dot{x} \label{13.8.11} \tag{13.8.11}$

and similarly for $$y$$:

$y + δy = Fy + G \dot{y} , \label{13.8.12} \tag{13.8.12}$

where $F = 1 - \frac{1}{2r^3} (δt)^2 + \frac{\dot{r}}{2r^4} (δt)^3 + \frac{1}{24} \left( \frac{1}{r^6} - \frac{12 \dot{r}^2}{r^5} + \frac{3 \ddot{r}}{r^4} \right) (δt)^4 + ... \label{13.8.13} \tag{13.8.13}$

and $G = δt - \frac{1}{6r^3} (δt)^3 + \frac{\dot{r}}{4r^4} (δt)^4 + ... \label{13.8.14} \tag{13.8.14}$

Now we are going to look at the triangle and sector areas. From figure $$\text{XIII.1}$$ we can see that

$\textbf{A}_1 = \frac{1}{2} \textbf{r}_2 \times \textbf{r}_3 , \ \textbf{A}_2 = \frac{1}{2} \textbf{r}_1 \times \textbf{r}_3 , \ \textbf{A}_3 = \frac{1}{2} \textbf{r}_1 \times \textbf{r}_2 . \label{13.8.15a,b,c} \tag{13.8.15a,b,c}$

Also, angular momentum per unit mass is $$\textbf{r} \times \textbf{v}$$ and Kepler’s second law tells us that areal speed is half the angular momentum per unit mass and that it is constant and equal to $$\frac{1}{2} \sqrt{l}$$ (in the units that we are using), so that

$\dot{\textbf{B}}_1 = \frac{1}{2} \textbf{r}_1 \times \dot{\textbf{r}}_1 = \frac{1}{2} \textbf{r}_2 \times \dot{\textbf{r}}_2 = \frac{1}{2} \textbf{r}_3 \times \dot{\textbf{r}}_3 .\label{13.8.16a,b,c} \tag{13.8.16a,b,c}$

All four of these vectors are parallel and perpendicular to the plane of the orbit, to that their magnitudes are just equal to their $$z$$-components. From the usual formulas for the components of a vector product we have, then,

$A_1 = \frac{1}{2}(x_2 y_3 - y_2 x_3 ) , \quad A_2 = \frac{1}{2} (x_1 y_3 - y_1 x_3 ) , \quad A_3 = \frac{1}{2} (x_1 y_2 - y_1 x_2 ) \label{13.8.17a,b,c} \tag{13.8.17a,b,c}$

and

$\frac{1}{2} \sqrt{l} = \frac{1}{2} (x_1 \dot{y}_1 - y_1 \dot{x}_1 ) = \frac{1}{2}(x_2 \dot{y}_2 - y_2 \dot{x}_2 ) = \frac{1}{2} (x_3 \dot{y}_3 - y_3 \dot{x}_3 ) . \label{13.8.18a,b,c} \tag{13.8.18a,b,c}$

Now, start from the second observation $$(x_2 , y_2)$$ at instant $$t_2$$. We shall try to predict the third observation, using equations 13.8.11-14, in which $$x + δx$$ is $$x_3$$ and $$δt$$ is $$t_3 − t_2$$, which we are calling (see section 13.3) $$τ_1$$. I shall make the subscripts for $$F$$ and $$G$$ the same as the subscripts for $$τ$$. Thus the $$F$$ and $$G$$ that connect observations 2 and 3 will have subscript 1, just as we are using the notation $$τ_1$$ for $$t_3 − t_2$$.

Thus we have

$x_2 = F_1 x_2 + G_1 \dot{x}_2 \label{13.8.19} \tag{13.8.19}$

and $y_3 = F_1 y_2 + G_1 \dot{y}_2 , \label{13.8.20} \tag{13.8.20}$

where $F_1 = 1 - \frac{1}{2r_2^3}τ_1^2 + \frac{\dot{r}_2}{2r_2^4} τ_1^3 + \frac{1}{24} \left( \frac{1}{r_2^6} - \frac{12 \dot{r}_2^2}{r_2^5} + \frac{3 \ddot{r}_2}{r_2^4} \right) τ_1^4 + ... \label{13.8.21} \tag{13.8.21}$

and $G_1 = τ_1 - \frac{1}{6r_2^3} τ_1^3 + \frac{\dot{r}_2}{4r_2^3} τ_1^4 + ... \quad . \label{13.8.22} \tag{13.8.22}$

Similarly, the first observation is given by

$x_1 = F_3 x_2 + G_3 \dot{x}_2 \label{13.8.23} \tag{13.8.23}$

and $y_1 = F_3 y_2 + G_3 \dot{y}_2 , \label{13.8.24} \tag{13.8.24}$

where, by substitution of $$−τ_3$$ for $$δt$$,

$F_3 = 1 - \frac{1}{2r_2^3} τ_3^2 - \frac{\dot{r}_2}{2r_2^4} τ_3^3 + \frac{1}{24} \left( \frac{1}{r_2^6} - \frac{12 \dot{r}_2^2}{r_2^5} + \frac{3\ddot{r}_2}{r_2^4} \right) τ_3^4 + ... \label{13.8.25} \tag{13.8.25}$

and $G_3 = -τ_3 + \frac{1}{6r_2^3} τ_3^3 + \frac{\dot{r}_2}{4r_2^4}τ_3^4 + ... \quad . \label{13.8.26} \tag{13.8.26}$

From equations 13.8.17,18,19,20,23,24, we soon find that

$A_1 = \frac{1}{2} G_1 \sqrt{l} , \quad A_2 = \frac{1}{2} ( F_3 G_1 - F_1G_3 ) \sqrt{l} , \quad A_3 = -\frac{1}{2} G_3 \sqrt{l} . \label{13.8.27a,b,c} \tag{13.8.27a,b,c}$

Now we do not yet know $$\dot{r}$$ or $$\ddot{r}$$, but we can take the expansions of $$F$$ and $$G$$ as far as $$τ^2$$ . We then obtain, correct to $$τ^3$$ :

$A_1 = \frac{1}{2} \sqrt{l} τ_1 \left( 1 - \frac{τ_1^2}{6_2^3} \right) , \label{13.8.28} \tag{13.8.28}$

$A_2 = \frac{1}{2} \sqrt{l} τ_2 \left( 1 - \frac{τ_1^2}{6r_2^3} \right) , \label{13.8.29} \tag{13,8.29}$

and $A_3 = \frac{1}{2} \sqrt{l} τ_3 \left( 1 - \frac{τ_3^2}{6r_2^3} \right) . \label{13.8.30} \tag{13.8.30}$

Thus the triangle ratio $$a_1 = A_1/A_2$$ is

$a_1 = \frac{τ_1}{τ_2} \left( 1 - \frac{τ_1^2}{6r_2^3} \right) \left( 1 - \frac{τ_2^2}{6r_2^3} \right)^{-1} , \label{13.8.31} \tag{13.8.31}$

which, to order $$τ^3$$ , is $a_1 = \frac{τ_1}{τ_2} \left( 1 + \frac{(τ_2^2 - τ_1^2)}{6r_2^3} \right) , \label{13.8.32} \tag{13.8.32}$

or, with $$τ_2 - τ_1 = τ_3$$, $a_1 = \frac{τ_1}{τ_2} \left( 1 + \frac{τ_3 (τ_2 + τ_1)}{6r_2^3} \right) . \label{13.8.33} \tag{13.8.33}$

Similarly, $a_3 = \frac{τ_3}{τ_2} \left( 1 + \frac{τ_1 (τ_2 + τ_3)}{6r_2^3} \right) . \label{13.8.34} \tag{13.8.34}$

Further, with $$τ_1 /τ_2 = b_1$$ and $$τ_3 /τ_2 = b_3$$,

$a_1 = b_1 + \frac{τ_1 τ_3}{6r_2^3} (1 + b_1) \quad \text{and} \quad a_3 = b_3 + \frac{τ_1 τ_3}{6r_2^3} (1 + b_3) . \label{13.8.35a,b} \tag{13.8.35a,b}$

These will serve as better approximations for the triangle ratios. Be aware, however, that equations 13.8.35a,b are only approximations, and do not give the exact values for $$a_1$$ and $$a_3$$.