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 $a_3$.