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13.8: Improved Triangle Ratios

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  • 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,

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


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

    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 \(a_3\).

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