ody describes an orbit in virtue of a force varying inversely as the square of the distance, the square of the velocity at the distance \( r \) is given by the equation
\[ \frac{1}{r} - \frac{a}{d^2} = \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}, \]
in which \( a \) is the major axis of the orbit, and the sum of the masses is supposed equal to unit. In the case of the parabola, the major axis is infinite, and the equation becomes simply
\[ \frac{1}{r} = \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}. \]
Let \( D \) be the perihelion distance \( SP \), and \( \psi \) the true anomaly, or the angle \( PSC \). We have then \( x = r \cos \psi \), \( y = r \sin \psi \), whence \( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} = dr^2 + r^2 d\psi^2 \), and consequently
\[ \frac{2}{r} = \frac{dr^2}{dt^2} + r^2 d\psi^2. \]
But from the property of the parabola \( r = 2D - r \cos \psi \); whence \( r(1 + \cos \psi) = 2D \), or \( r \cos \psi = D \), therefore
\[ r = D(1 + \tan^2 \frac{1}{2}\psi). \]
Let \( \tan \frac{1}{2}\psi = u \); then \( dr = 2Du \cos^2 \frac{1}{2}\psi \)
\[ = \frac{1}{2}d\psi (1 + \tan^2 \frac{1}{2}\psi), \text{ whence } d\psi = \frac{2du}{1 + u^2}. \]
Substituting these expressions in the equation \( \frac{2}{r} = \frac{dr^2}{dt^2} + r^2 d\psi^2 \), it becomes \( dt^2 = 2D^2 (1 + u^2)^2 du^2 \), whence \( dt = \sqrt{2D^2 (1 + u^2)} du \); which, on being integrated, becomes
\[ t = \sqrt{2D^2 (u + \frac{1}{3}u^3)}. \]
Let \( u_0, u' \) be the values of \( u \) in this equation at the times \( t + \theta, t - \theta \) respectively; we have then
\[ t + \theta = \sqrt{2D^2 (u' + \frac{1}{3}u'^3)} \]
\[ t - \theta = \sqrt{2D^2 (u^2 + \frac{1}{3}u^3)}, \]
and, by subtracting,
\[ \theta = \sqrt{2D^2 (u^2 - u'^2 + \frac{1}{3}(u^3 - u'^3))}. \]
In order to transform this expression into another containing only \( r', r \) and \( e \), let \( \phi \) be the anomaly corresponding to \( C' \), \( \psi \) that corresponding to \( C \), \( \phi = \psi - \psi' = \angle C'SC' \), and \( \tan \frac{1}{2}\phi = r \). We have then \( \psi = \psi' + \phi \), and consequently \( \tan \frac{1}{2}\psi = \tan \frac{1}{2}\psi' + \tan \frac{1}{2}\phi = \frac{\tan \frac{1}{2}\psi' + \tan \frac{1}{2}\phi}{1 - \tan \frac{1}{2}\psi' \tan \frac{1}{2}\phi} \), that is, \( u = \frac{u' + v}{1 - u'u} \). Now, if we put
\[ Z = u' - u^2 + \frac{1}{3}(u^3 - u'^3), \]
we shall have evidently
\[ Z = (u' - u^2) \left(1 + \frac{1}{3}(u^2 + u'u^2 + u'^2)\right) \]
\[ = (u' - u^2) \left(1 - \frac{1}{3}(u^2 - u'^2)^2 + 3u'u^2\right), \]
therefore, substituting \( \frac{u^2 + v}{1 - u'u} \) for \( u' \),
\[ Z = \frac{v(1 + u'^2)}{1 - u'u} \left(1 + \frac{u'^2 + u'u^2}{1 - u'u^2} + \frac{v^2(1 + u'^2)}{(1 - u'u^2)^2}\right); \]
that is,
\[ Z = \frac{v(1 + u'^2)}{1 - u'u} \left(1 + \frac{u'^2(1 + u'^2)}{1 - u'u^2}\right); \]
hence the formula for the time becomes
\[ \theta + \theta' = \sqrt{2D^2} \cdot \frac{v(1 + u'^2)}{1 - u'u^2} \left(1 + \frac{u'^2(1 + u'^2)}{1 - u'u^2}\right). \]
Again, the triangle \( C'SC' \) gives the equation
\[ c^2 = r^2 + r'^2 - 2rr'\cos \phi, \]
but \( \tan \frac{1}{2}\phi = v \); therefore, from the trigonometrical formulae, \( \cos \phi = \frac{1 - v^2}{1 + v^2} \), whence, substituting and reducing,
\[ (r' + r^2)^2 - c^2 = \frac{4rr'\phi}{1 + v^2}. \]
For the sake of abridging, we shall put
\[ r' + r^2 + c = 2s, \quad r' + r^2 - c = 2d, \]
whence \( r' + r^2 = s + d, \quad c = s - d \), and \( (r' + r^2)^2 - c^2 = 4sd \). Comparing these with the above equation, we have
\[ \frac{r'^2}{1 + v^2} = sd, \quad \text{whence } v = \frac{(r'^2 - sd)^2}{(sd)^2}. \]
Now the polar equation to the parabola gives
\[ r^2 = D(1 + u'^2), \quad r' = D(1 + u^2) = D \left(\frac{1 + u'^2}{1 - u'^2}\right), \]
therefore \( sd = \frac{r'^2}{1 + v^2} = D^2 \left(\frac{1 + u'^2}{1 - u'^2}\right)^2 \), and consequently,
\[ \frac{1 + u'^2}{1 - u'^2} = \frac{(sd)^2}{D^2}. \]
From the same values of \( r^2 \) and \( r' \) we get
\[ r' + r^2 = D(1 + u'^2) \left\{1 + \frac{1 + v^2}{(1 - u'^2)^2}\right\} \]
\[ = \frac{2D(1 + u'^2)}{1 - u'^2} + \frac{De^2(1 + u'^2)^2}{1 - u'^2}, \]
but \( r' + r^2 = s + d \), and \( \frac{1 + u'^2}{1 - u'^2} = \frac{\sqrt{sd}}{D} \), therefore
\[ s + d = 2\sqrt{sd} + \frac{v^2\sqrt{sd}}{D}, \quad \text{whence } v = \sqrt{D} \left(\frac{\sqrt{s} - \sqrt{d}}{\sqrt{sd}}\right). \]
Substituting these values of \( v \) and \( \frac{1 + u'^2}{1 - u'^2} \) in the equation for the time, it becomes
\[ \theta + \theta' = \sqrt{2\sqrt{sd}} \left(\sqrt{s} - \sqrt{d}\right) \left\{1 + \frac{1}{3} \left(\frac{\sqrt{s} + \sqrt{d}}{\sqrt{sd}}\right)^2\right\}, \]
that is
\[ \theta + \theta' = \frac{\sqrt{2}}{3} (\sqrt{s} - \sqrt{d}) (\sqrt{s} + \sqrt{d} + d), \]
or
\[ \theta + \theta' = \frac{\sqrt{2}}{3} (\sqrt{s} - \sqrt{d}) (\sqrt{s} + \sqrt{d} + d), \]
and, finally, on restoring the values of \( s \) and \( d \),
\[ \theta + \theta' = \frac{1}{6}(r' + r^2 + c) - \frac{1}{6}(r' + r^2 - c) \quad (24). \]
This is the very elegant theorem found by Euler, but commonly attributed to Lambert, by means of which we are enabled to correct the hypothetical values successively given to \( \phi \), till a value of that quantity is found which will satisfy all the equations as nearly as the observations will admit.
13. We shall now collect into one view the different processes indicated in the preceding investigations; and in order to throw every necessary light on the subject, we shall give at the same time an example of the actual computation of an orbit from Legendre. The comet is that of 1805, and the data from which the computation is made are contained in the following table:
| Epochs | Longitudes | Latitudes | |--------|------------|-----------| | October 22-68488 | \( a = 163^\circ 20' 53'' \) | \( b = 22^\circ 59' 53'' \) | | 30-68670 | \( a = 183^\circ 48' 32'' \) | \( b = 15^\circ 37' 21'' \) | | 34-72088 | \( a = 191^\circ 46' 15'' \) | \( b = 12^\circ 2' 29'' \) | I. The first step is to compute the arc described by the mean motion of the earth in the time \( t + \delta \). Let this arc be represented by \( \Theta \); then since \( \delta \) and \( \theta \) are expressed in mean solar days, we have \( \Theta = \frac{2\pi(\delta + \theta)}{365.25638} \), \( 2\pi \) being the circumference of a circle whose radius is unit, and 365.25638 the mean solar days in a sidereal year. The logarithm of \( \Theta \) will therefore be found by adding the constant logarithm 8.2355821 (which is the logarithm of \( \frac{2\pi}{365.25638} \)) to the logarithm of \( \delta + \theta \).
From the above table we have \( \delta = 8.00182 \) days, \( \theta = 4.03418 \) days, \( \delta + \theta = 12.03600 \) days; hence taking the logarithms of these numbers, and adding 8.2355821, we have the following results expressed in logarithms:
\[ \begin{align*} \delta + \theta &= 9.3160643 \\ \delta + \theta &= 9.3160643 \\ \delta + \theta &= 9.827065 \\ \delta + \theta &= 9.5252730 \end{align*} \]
It may be here remarked, that although the computations are considerably shortened when \( \delta = \theta \), a condition which may in general be satisfied by interpolation, yet it is preferable to avoid any preliminary calculations, and to employ the direct results of observation. The intervals, however, should not be greatly different; and as the solution is founded on the supposition that quantities of the fourth order in respect of the time may be neglected, the longest interval cannot safely exceed ten days. In the present example, \( \delta \) is considerably greater than \( \theta \); it therefore affords no facilities for computation, and the approximations must be carried as far as the method admits.
II. The places of the sun at the time of the respective observations are next to be taken from the solar tables. These places, increased by 180°, give the three longitudes of the earth \( A^o, A, A' \). The same tables also give the corresponding distances \( R^o, R, R' \); and the quantities \( A_1 \) and \( R_1 \), which denote the longitude and radius vector of the earth at the time \( t + \frac{1}{2}(\delta + \theta) \), corresponding to the mean epoch of the three observations, will be obtained with sufficient accuracy by assuming \( A_1 = \frac{1}{2}(A^o + A + A') \), and \( R_1 = \frac{1}{2} \log(R^oRR') \).
At the epochs given in the above table, we find from the solar tables,
\[ \begin{align*} A^o &= 29^\circ 10' 50'' & \log. R^o &= 9.997421 \\ A &= 37^\circ 19' 48'' & \log. R &= 9.996496 \\ A' &= 41^\circ 22' 26'' & \log. R' &= 9.996053 \end{align*} \]
By taking a mean among these numbers we shall have
\[ A_1 = 36^\circ 0' 41'' & \log. R_1 = 9.9966567. \]
III. With the results now found, we may proceed to compute the quantities \( P, Q \) from the expressions given in § 5, namely,
\[ P = \sin(A_1 - a^o) \tan b^o - \sin(A_1 - a') \tan b'^o \\ Q = \sin(a^o - a') \tan b + \sin(a' - a) \tan b'^o + \sin(a - a^o) \tan b'. \]
For this purpose the following logarithms are required:
\[ \begin{align*} \sin(a^o - a) &= 9.1414978 & \tan b^o &= 9.6278110 \\ \sin(a' - a) &= 9.6775831 & \tan b &= 9.4465812 \\ \sin(a - a^o) &= 9.5453505 & \tan b' &= 9.3290147 \\ \sin(a^o - A_1) &= 9.6133857 \\ \sin(a' - A_1) &= 9.9004139 \end{align*} \]
On computing \( P \) and \( Q \) by means of these numbers, we find \( P = 0.0046583 \), and \( Q = 0.0002617 \). The signs of both are positive, and though \( Q \) is thus a very small number, yet its sign could not become negative without supposing errors in the observations amounting to four or five minutes. By reason of the smallness of \( P \) and \( Q \), no use can be made of the equation (16); we can only infer from it, since \( P \) and \( Q \) have the same sign, that \( r < R \); that is to say, the distance of the comet from the sun is less than the sun's distance from the earth.
IV. It is now necessary to make a supposition with regard to the value \( \phi \). In general a few trials will suffice to give an approximate value, by means of which \( \phi \) and \( \phi' \) may be computed from the formulae (19) or (20), according to the different cases. In the first trials we may suppose \( E = 0 \),
\[ e = 0, \lambda = \frac{\delta}{\delta + \theta}, \lambda' = \frac{\theta}{\delta + \theta}; \]
but when approximations to the values of \( \phi \) and \( \phi' \) have been obtained on this supposition, we must employ the more exact expressions given in the formulae (6). In the present case, on account of the great inequality between \( \delta \) and \( \theta \), the values of \( \phi \) and \( \phi' \) deduced from the supposition of \( E = 0, e = 0 \), &c., would not have that degree of approximation which would render them of any use in directing the subsequent hypotheses; we may therefore have recourse at once to the second degree of approximation. The formulae (6) give
\[ \begin{align*} e &= \frac{\delta}{2R^3}, E = \frac{\delta}{2R^3} \\ \lambda &= \frac{\delta}{\delta + \theta} \left(1 - \frac{\theta(\delta - \theta)}{6r^3}\right), \lambda' = \frac{\theta}{\delta + \theta} \left(1 + \frac{\delta(\delta - \theta)}{6r^3}\right), \end{align*} \]
from which we deduce
\[ \begin{align*} \frac{1-e}{\lambda} &= \frac{\delta + \theta}{\delta} \left(1 - \frac{\theta(2\delta + \theta)}{6r^3}\right), \\ \frac{1-e}{\lambda'} &= \frac{\delta + \theta}{\theta} \left(1 - \frac{\theta(2\delta + \theta)}{6r^3}\right), \\ \frac{e}{\lambda} &= \frac{\theta(\delta + \theta)}{2r^3}, \quad \frac{e}{\lambda'} = \frac{\delta(\delta + \theta)}{2r^3}, \\ \frac{E}{\lambda} &= \frac{\theta(\delta + \theta)}{2R^3}, \quad \frac{E}{\lambda'} = \frac{\delta(\delta + \theta)}{2R^3}. \end{align*} \]
In order to convert these formulae into numbers, we must recollect that if \( b \) is a small number in comparison of \( a \), we have \( \log.(a + b) = \log.a + \frac{b}{a}M \), where \( M \) denotes the modules of the tables, or the number 4342945, the logarithm of which is 6.6377843; and reciprocally, if we have \( \log.x = \log.a + b \), then \( x = a + \frac{ab}{M} \). According to these formulae
\[ \log.\frac{1-e}{\lambda} = \log.\frac{\delta + \theta}{\delta} - \frac{\theta(2\delta + \theta)}{6r^3}M. \]
Now \( \log.\frac{\delta + \theta}{\delta} = 0.1772935 \); and to find the number \( \frac{\theta(2\delta + \theta)}{6r^3}M \), we have \( 2\delta + \theta = 20.03782 \) days, therefore
\[ \begin{align*} \log.(2\delta + \theta) &= 1.3018505 \\ \log.\theta &= 8.2355821 \\ \log.\delta &= 8.8413373 \\ \log.6 &= 0.7781513 \\ \log.\frac{\theta(2\delta + \theta)}{6r^3} &= 7.6006186 \end{align*} \]
If therefore we denote by \( L \) the number whose logarithm is \( L \), we shall have
\[ \log.\frac{1-e}{\lambda} = 0.1772935 - (7.6006186)\frac{M}{r^3}; \] and passing to numbers
\[ \frac{1-e}{\lambda} = (0.1772935) - (7.7779121) \frac{1}{r^3} \]
By proceeding in the same manner, we find
\[ \log \frac{1-e}{\lambda} = 0.4747270 - (7.8022223) \frac{M}{r^3}, \]
whence
\[ \frac{1-e}{\lambda} = (0.4747270) - (8.2769493) \frac{1}{r^3}. \]
The following logarithms are found without difficulty:
\[ \log_e \frac{\theta(\phi + \delta)}{2r^3} = 7.8563716 \]
\[ \log_e \frac{\theta(\phi + \delta)}{2r^3} = 8.1538051 \]
\[ \log_E \frac{\theta(\phi + \delta)}{2R^3} = 7.8668836 \]
\[ \log_E \frac{\theta(\phi + \delta)}{2R^3} = 8.1643171. \]
In the present example the motion in longitude is more considerable than in latitude; it is therefore preferable to employ the equations (20), namely,
\[ \begin{align*} \epsilon &= \frac{(1-e)\sin(\alpha' - \alpha) + (E-e)R_1\sin(\alpha' - A_1)}{\lambda\sin(\alpha' - \alpha')} \\ \epsilon' &= \frac{(1-e)\sin(\alpha' - \alpha') + (e-E)R_1\sin(\alpha' - A_1)}{\lambda\sin(\alpha' - \alpha')} \end{align*} \]
to compute which, the following logarithms are required in addition to those already found, namely,
\[ \begin{align*} \log_e \frac{\sin(\alpha' - \alpha)}{\sin(\alpha' - \alpha')} &= 9.4639147 \\ \log_e \frac{R_1\sin(\alpha' - A_1)}{\sin(\alpha' - \alpha')} &= 9.9324593 \\ \log_e \frac{\sin(\alpha' - \alpha')}{\sin(\alpha' - \alpha')} &= 9.8659474 \\ \log_e \frac{R_1\sin(\alpha' - A_1)}{\sin(\alpha' - \alpha')} &= 9.2194875, \end{align*} \]
by adding which to the values of \(\frac{1-e}{\lambda}\), &c., the above two equations become
\[ \begin{align*} \epsilon &= (9.9386417) - \frac{1}{r^3}(7.7408640) \\ &+ (8.0967764) - \frac{1}{r^3}(8.0862644) \\ \epsilon' &= (9.0432409) - \frac{1}{r^3}(7.6438595) \\ &- (8.0863711) + \frac{1}{r^3}(8.0758591) \end{align*} \]
V. Having assumed a hypothetical value of \(e\), the radius vector \(r\) will be obtained from the equation (17), namely,
\[ r^2 = R^2 + 2eR\cos(A-a) + e^2\sec^2b, \]
which, on taking the logarithms of the known quantities, becomes
\[ r^2 = 0.9839930 - (0.2185267) + (0.0326962); \]
and on substituting this value of \(r\) in the above two equations, we shall obtain the values of \(\epsilon\) and \(\epsilon'\).
It is now necessary to calculate \(r^2\), \(r'\), and \(c\). With regard to \(r^2\) and \(r'\), we have the two equations, corresponding to equation (17),
\[ \begin{align*} r'^2 &= R'^2 + 2e'R'\cos(A'-a') + e'^2\sec^2b', \\ r'^2 &= R'^2 + 2e'R'\cos(A'-a') + e'^2\sec^2b'; \end{align*} \]
which being reduced to numbers, become
\[ \begin{align*} r'^2 &= 0.9881934 - (0.1403596) + (0.0719354), \\ r'^2 &= 0.9819876 - (0.2363369) + (0.0193249). \end{align*} \]
The chord \(c\) is determined by the equation
\[ c^2 = r^2 + r'^2 - 2F, \]
in which \(F =\)
\[ R^2R'\cos(A-A') + e^2\cos(a-a') + \tan b\tan b' \]
\[ + e'^2R'\cos(A'-a') + e'^2R'\cos(A'-a'). \]
On taking the logarithms of the constant quantities, this becomes
\[ F = 0.9634042 - (9.7199492) + (9.9766975) + (9.9867701). \]
Lastly, the error of the hypothesis will be found from the equation of the time
\[ 6(\phi + \delta) = (r^2 + r'^2 + c)^2 + (r^2 + r'^2 - c)^2 = 1.2422687. \]
As the present object is merely to illustrate the method of computation, it is unnecessary to begin with an assumption very far from the truth. Suppose, in the first place, \(e = 0.6\) (9.7781512). This will give the following results:
\[ \begin{align*} r^2 &= 0.3797824 = (9.5795120), r = (9.7897560) \\ r^2 &= 0.6695087, r = (9.8389800) \\ r^2 &= 0.6008290 = (9.7788893) \\ r^2 &= 0.2906613 = (9.4634172) \\ r^2 + r'^2 &= 0.8909903 \\ 2F &= 0.7537670 \\ c^2 &= 0.1372233 = (9.1374278) \\ r^2 &= (9.8891946) = 0.7748089 \\ r' &= (9.7317086) = 0.5391475 \\ r^2 + r'^2 &= 1.8139564 \\ c &= (9.5687139) = 0.3794366 \\ r^2 + r'^2 + c &= a = 1.6843930 \\ r^2 + r'^2 - c &= b = 0.9435208 \\ a^2 &= 2.1860775 \\ b^2 &= 0.9164888 \\ a^2 - b^2 &= 1.2695887 \\ \text{but } 6(\phi + \delta) &= 1.2422687 \\ \text{Error} &= -0.034176 \end{align*} \]
For a second hypothesis let us make \(e = 0.58\) (9.7634280). On going through the same calculations, there will result
\[ \begin{align*} r^2 + r'^2 + c &= a = 1.6833303, a^2 = 2.1840703 \\ r^2 + r'^2 - c &= b = 0.9631377, b^2 = 0.9452192 \\ a^2 - b^2 &= 1.2388511 \\ 6(\phi + \delta) &= 1.2422687 \\ \text{Error} &= -0.034176 \end{align*} \]
Let \(n = \frac{-0.034176}{0.02732 + 0.0034176} = 0.11119\), then the true value of any quantity which is denoted by \(A\) in the first hypothesis, and by \(B\) in the second, will be \(B - n(B-A)\).
Computing from this formula, there results,
\[ \begin{align*} \epsilon &= (9.6569463), r = (9.8906184) \\ \epsilon' &= (9.7650651), r' = (9.7935971) \\ \epsilon' &= (9.8259598), r' = (9.7632737) \\ c &= (9.5577858). \end{align*} \]
VI. In most cases the solution which has now been obtained would be sufficiently approximate; but in the present case, on account of the great inequality between the intervals \( \delta \) and \( \theta \), it is necessary to recommence the calculations, and to determine \( \lambda \), \( \lambda' \), and \( e \) from the more accurate expressions which form the third degree of approximation. Let
\[ \delta = \frac{Mg}{6(r^2)^3} - \frac{Mg}{6(r')^2}; \quad \text{the equations (7) give} \]
\[ \log \lambda = \log \left( \frac{\delta + \theta}{\delta + \theta} \right) - \frac{\delta + \theta}{\delta + \theta}, \quad \delta, e = \frac{\delta + \theta}{2r^2r'} \]
\[ \log \lambda' = \log \left( \frac{\delta + \theta}{\delta + \theta} \right) + \frac{\delta + \theta}{\delta + \theta}; \quad E = \frac{N}{2R'R'^2} \]
from which, employing the approximate values of \( r^2 \), \( r' \), and \( r'' \) already obtained, there results
\[ \delta = (9-7643107), \quad (e - E) = (9-1358552), \]
\[ \lambda = (9-8219310), \quad (1 - e) = (9-9919399), \]
\[ \lambda' = (9-5268112). \]
If we now proceed to substitute these results in the equations (17), we shall find the following expressions for \( r^2 \) and \( r' \):
\[ r^2 = (9-9290428) - (8-5415038), \]
\[ r' = (9-0359557) - (8-5334117), \]
in which the only indeterminate quantity is \( \varepsilon \). We must now make another hypothesis with respect to \( \varepsilon \). If we assume the value given by the last approximation, namely \( \varepsilon = (9-7650651) \), we shall find that it is considerably too small, and that the first approximation is consequently insufficient, owing to the imperfect values it gave of the quantities \( \lambda \), \( \lambda' \), and \( e \).
Let us suppose \( \varepsilon = 0-6045 = (9-7813963) \). By proceeding as before with the corrected values now found, we obtain
\[ r^2 = (9-6799602), \quad r' = (9-8393705), \]
\[ r'' = (9-5973194) = (9-7762067), \]
\[ r^2 = (9-290501) = (9-4631582) \]
\[ c^2 = (9-1318662) = (9-1184836) \]
whence
\[ r^2 + r' + c = a = 1-6742972, \quad a^2 = 2-1664512 \]
\[ r^2 + r' - c = b = 0-9494078, \quad b^2 = 0-9250798 \]
\[ a^2 - b^2 = 1-2421374 \]
\[ 6(\delta + \theta) = 1-2422687 \]
Error — 0-0008973
Suppose, therefore, \( \varepsilon = 0-6055 = (9-7821141) \). Computing as above, we shall find
\[ r^2 + r' + c = a = 1-6742754, \quad a^2 = 2-1664090 \]
\[ r^2 + r' - c = b = 0-9485998, \quad b^2 = 0-9238902 \]
\[ a^2 - b^2 = 1-2425188 \]
\[ 6(\delta + \theta) = 1-2422687 \]
Error + 0-002501
Let \( n = \frac{2501}{8973 + 2501} = 0-218 \); the mean between the two values of the same quantity in the two hypotheses is \( B + (A - B)(0-218) \), by taking which, there results
\[ r^2 = (9-8880427), \quad r' = (9-7886828), \quad r'' = (9-7314028). \]
These values being considerably different from those found by the first approximation, it becomes necessary to have recourse to a third operation.
The values of \( \lambda \), \( \lambda' \), and \( 1 - e \), deduced from the values of \( r^2 \), \( r' \), now found, are
\[ \lambda = (9-8219155), \quad \lambda' = (9-5268419), \quad 1 - e = (9-9918188). \]
In order to find a more exact value of \( A_1 \) than results from taking the mean of the three longitudes of the earth, we may make use of the equations (18), namely,
\[ 0 = \lambda m' + \lambda' m'' - (1 - e) mg + M, \]
\[ 0 = \lambda m' + \lambda' m'' - (1 - e) mg + N, \]
in which \( M = (e - E) R_1 \cos A_1 = \pi \cos A_1 \),
\( N = \pi \sin A_1 \), and we shall have in the present example
\[ \pi \cos A_1 = \lambda(9-8713530) + \lambda'(9-9378419) - (1 - e) \]
\[ (9-77909484) \]
\[ \pi \sin A_1 = \lambda(9-8162347) + \lambda'(9-6874819) - (1 - e) \]
\[ (9-7792387) \]
whence, substituting for \( \lambda \), \( \lambda' \), and \( (1 - e) \), the above values,
\[ \pi \cos A_1 = (8-0399020), \quad \tan A_1 = (9-8969484) \]
\[ \pi \sin A_1 = (7-9115731), \quad A_1 = 36^\circ 39' 20" \]
\[ \pi = (8-1355982). \]
The value of \( A_1 \), thus found, differs considerably from that which was employed in the previous approximation, namely, \( A_1 = 36^\circ 0' 41" \).
On substituting \( \pi \) for \( (e - E) R_1 \), the equations (20) become
\[ \begin{align*} \frac{(1 - e)}{\lambda} \sin (\alpha - a) &= \sigma \sin (\alpha - A_1), \\ \frac{(1 - e)}{\lambda'} \sin (\alpha' - a) &= \sigma \sin (\alpha' - A_1), \end{align*} \]
whence, making \( A_1 = 36^\circ 39' 20" \),
\[ \begin{align*} \varepsilon &= (9-9288916) - (8-5552427), \\ \varepsilon' &= (9-0358507) + (8-9401948), \end{align*} \]
from which two equations we shall obtain the nearest approximation of which this method of solving the problem is susceptible.
Suppose \( \varepsilon = (9-7790000) \). On going through the same computation as before, we shall find
\[ \begin{align*} r^2 + r' + c &= a = 1-6758479, \\ a^2 &= 2-1694623 \\ r^2 + r' - c &= b = 0-9509163, \\ b^2 &= 0-9272854 \\ a^2 - b^2 &= 1-2421769 \\ 6(\delta + \theta) &= 1-2422687 \end{align*} \]
Error — 918
Suppose, secondly, \( \varepsilon = (9-7791000) \). This supposition will give
\[ \begin{align*} r^2 + r' + c &= a = 1-6758426, \\ a^2 &= 2-1694511 \\ r^2 + r' - c &= b = 0-9508021, \\ b^2 &= 0-9271186 \\ a^2 - b^2 &= 1-2423325 \\ 6(\delta + \theta) &= 1-2422687 \end{align*} \]
Error + 638
From these errors, the mean to take between A and B, the two values of the same quantity in the two hypotheses, is \( B + (A - B)(0-41) \), whence the following solution:
\[ \begin{align*} \varepsilon &= (9-6763680), \\ \varepsilon' &= (9-8884811) = 0-77353695, \\ \varepsilon' &= (9-83738895), \\ \varepsilon' &= (9-7322409) = 0-53981000, \\ \varepsilon &= (9-5593056) = 0-3624980. \end{align*} \]
VI. It now remains only to indicate the method of computing, from the foregoing solution, the elements of the parabolic orbit.
It was shown in § 12 that \( v = \tan \frac{1}{2} \varphi = \frac{\sqrt{r^2 - sd}}{\sqrt{sd}} \),
and also that \( v = \sqrt{D} \cdot \frac{\sqrt{s} - \sqrt{d}}{\sqrt{sd}} \). Comparing these values, \( \sqrt{r^2 - sd} = \sqrt{D} (\sqrt{s} - \sqrt{d}) \), and consequently,
\[ D = \frac{r^2 - sd}{(\sqrt{s} - \sqrt{d})^2} = \frac{(s - r^2)(s - r)}{(\sqrt{s} - \sqrt{d})^2}. \]
Having obtained the perihelion distance \( D \), the true anomaly \( \varphi \) of the point \( C \) will be given by the equation of the parabola, namely, \( r^2 \cos^2 \frac{1}{2} \varphi = D \), from which
\[ \cos^2 \frac{1}{2} \varphi = \frac{(s - r^2)(s - r)}{r^2 (\sqrt{s} - \sqrt{d})}. \]
The calculation of \( D \) and \( \varphi \) is as follows:
\[ \begin{align*} \frac{1}{2}(r^2 + r' + c) &= s = (9-9232039) \\ \frac{1}{2}(r^2 + r' - c) &= d = (9-6770815) \\ s - r^2 &= (8-60878805) \\ s - r' &= (9-47438015) \\ (s - r^2)(s - r') &= (8-2831682) \\ \sqrt{d} &= (9-8385407) \\ \sqrt{s} &= (9-9616019) \\ \frac{\sqrt{d}}{\sqrt{s}} &= (9-8769388) = 7532495 \\ 1 - \frac{\sqrt{d}}{\sqrt{s}} &= 2467505 = (9-39225808) \\ \sqrt{s} &= (9-9616019) \\ \sqrt{s} - \sqrt{d} &= \sqrt{s}(1 - \frac{\sqrt{d}}{\sqrt{s}}) = (9-35385998) \\ (s - r^2)(s - r') &= (8-2831682) \\ D &= (9-5754482) \\ r^2 &= (9-8884811) \\ \cos^2 \frac{1}{2} \varphi &= (9-8869671) \\ \tan \frac{1}{2} \varphi &= (0-0118412). \end{align*} \]
Having thus obtained the perihelion distance and the true anomaly, we must now find the time required for describing this anomaly. For this purpose it is usual to employ a table constructed from the following property of bodies moving in a parabolic orbit, namely, if \( T \) and \( T' \) be the times of describing the same anomaly in two parabolas whose perihelion distances are respectively \( D \) and \( D' \), then \( T : T' : D : D' \). Hence, if a table is formed showing the time \( T' \) in which every degree, &c., is described in a parabola whose perihelion distance is \( D' = 1 \), it will give the time \( T' \) in which the same anomaly is described in the parabola whose perihelion distance is \( D \), by simply multiplying the tabular time by \( D' \); for when \( D' = 1 \), we have \( T = D'T' \).
The time \( T \) must be added to the epoch of the first observation, if the comet is approaching the perihelion of its orbit, but subtracted from that epoch if it has already passed the perihelion.
But instead of employing a table of parabolic motion, the time may be computed directly from the formula
\[ T = \sqrt{2D} (\tan \frac{1}{2} \varphi + \frac{1}{2} \tan^2 \frac{1}{2} \varphi), \]
observing to add to the logarithm of \( T \) the constant logarithm 1-7644179 (the compliment of 8-2355821), in order that \( T \) may be expressed in mean solar days.
Computing from this formula, there results in the present case,
\[ T = 26-35895 \]
Epoch of first observation, October...22-68488
Passage through perihelion, November 18-04883.
To determine the position of the orbit, we must first compute the heliocentric latitudes and longitudes. Let \( \alpha \), \( \alpha' \) be the heliocentric longitudes of the points \( C \) and \( C' \); \( \beta \), \( \beta' \) their heliocentric latitudes; then, denoting for an instant by \( \sigma \) the comet's distance from the sun, we shall have \( r \sin \beta = \sigma \sin b \); but \( \sigma = \frac{\varepsilon}{r} \sec b \), therefore
\[ r \sin \beta = \frac{\varepsilon}{r} \tan b', \quad \text{whence} \]
\[ \sin \beta' = \frac{\varepsilon}{r} \tan b', \quad \sin \beta = \frac{\varepsilon}{r} \tan b. \]
These two formulas, which are very easily computed, give the heliocentric latitudes at the extreme observations. In the present case they become
\[ \sin \beta' = (9-4155979), \quad \sin \beta = (9-4341627). \]
Now, to compute the heliocentric longitudes, let us suppose \( K \) to be the projection of \( C \) on the plane of the ecliptic. In the rectilinear triangle \( ESK \), formed by joining the positions of the earth, the sun, and the projected place of the comet, we have \( EK = \varepsilon \); \( SK = r \cos \beta \); the angle \( ESK \), opposite to \( EK \), is \( \alpha - A \), the difference between the heliocentric longitude of the comet and the longitude of the sun; and the angle \( KES \), opposite to \( SK \), is \( \alpha - A \), the difference between the geocentric longitude of the comet and the longitude of the sun; therefore
\[ \frac{\varepsilon}{r} \cos \beta : \sin (\alpha - A) : \sin (\alpha - A), \]
from which, since \( \frac{\varepsilon}{r} = \frac{\sin \beta}{\tan b'} \), we deduce for the first and third observations,
\[ \sin (\alpha - A) = \tan \beta' \sin (\alpha - A), \] \[ \sin (\alpha' - A') = \tan \beta' \sin (\alpha' - A'). \]
In the present case these formulas become
\[ \sin (\alpha - A) = (9-6598871), \quad \sin (\alpha' - A') = (9-8155230) \] \[ \alpha - A = 27^\circ 11' 18'', \quad \alpha' - A' = 40^\circ 50' 15'' 45' \] \[ \alpha = 29^\circ 19' 50'', \quad A' = 41^\circ 22' 26'' \] \[ \alpha = 59^\circ 31' 8'', \quad \alpha' = 82^\circ 12' 41'' 45''. \]
Let \( N \) be the ascending node, \( \gamma \) its longitude, and \( I \) the inclination of the orbit. In the spherical triangle \( NCX \), the side \( NK = \alpha - \gamma \), \( CK = \beta \), and the angle \( CNK = I \). Therefore, by Napier's analogies, cot \( I = \cot \beta \sin (\alpha - \gamma) \), which, on substituting the longitudes \( \alpha \) and \( \alpha' \), gives the double equation
\[ \cot I = \cot \beta' \sin (\alpha - \gamma) = \cot \beta \sin (\alpha - \gamma); \]
from which we have
\[ \frac{\sin (\alpha - \gamma)}{\sin (\alpha' - \gamma)} = \frac{\tan \beta'}{\tan \beta}, \]
whence
\[ \frac{\sin (\alpha - \gamma) + \sin (\alpha' - \gamma)}{\sin (\alpha - \gamma) - \sin (\alpha' - \gamma)} = \frac{\tan \beta' + \tan \beta}{\tan \beta' - \tan \beta}. \]
But if \( A \) and \( B \) represent two arcs, then, by the trigonometrical formulae,
\[ \frac{\sin A + \sin B}{\sin A - \sin B} = \frac{\tan \frac{1}{2}(A + B)}{\tan \frac{1}{2}(A - B)}, \]
and
\[ \frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin (A + B)}{\sin (A - B)}; \]
whence the above equation becomes
\[ \frac{\tan \left( \frac{1}{2}(\alpha + \alpha') - \gamma \right)}{\tan \left( \frac{1}{2}(\alpha' - \alpha) \right)} = \frac{\sin (\beta + \beta')}{\sin (\beta - \beta')}, \]
consequently
\[ \tan \left( \frac{1}{2}(\alpha + \alpha') - \gamma \right) = \frac{\sin (\beta + \beta')}{\sin (\beta - \beta')} \tan \frac{\alpha' - \alpha}{2}. \] The tangent now found belongs to two different angles; but this circumstance can occasion no ambiguity; for as the inclination $I$ is supposed to be always less than $90^\circ$, the cotangent of $I$ must be always positive, so that $\sin(\alpha^2 - \gamma)$ must always have the same sign as $\cot \beta$. This consideration will enable the computer to choose between the two values of $\gamma$, which correspond to $\tan(\frac{1}{2}(\alpha^2 + \alpha') - \gamma)$. With regard to the sign of $\beta$, the northern latitudes are supposed positive, and the southern negative; and the motion will be direct or retrograde, according as $\alpha' - \alpha$ is positive or negative.
On computing the above formula, there results
$$\frac{1}{2}(\alpha^2 + \alpha') - \gamma = 84^\circ 14' 30''$$
but
$$\frac{1}{2}(\alpha^2 + \alpha') = 69^\circ 21' 54''$$
therefore
$$-\gamma = 14^\circ 52' 36''$$
and consequently $\gamma = 345^\circ 7' 24''$.
By correcting this value from the double equation given above, Legendre finds
$$-\gamma = 14^\circ 52' 34'' 5$$
$$\gamma = 345^\circ 7' 25'' 5$$
Having found the longitude of the node, the inclination $I$ is given by the formula $\cot I = \cot \beta \cdot \sin(\alpha^2 - \gamma)$, from which we deduce
$$\log \cot I = 0.5458488, I = 15^\circ 53' 0'' 5$$
Lastly, let $H$ be the angular distance of the comet from the node, and $\Pi$ be the longitude of the perihelion. Then
$$\cos H = \cos \beta \cdot \cos(\alpha^2 - \gamma)$$
$$\Pi = H + \gamma + \phi;$$ if the comet is advancing to the perihelion; or
$$\Pi = H + \gamma - \phi;$$ if it has already passed the perihelion.
In the present example,
$$\alpha^2 - \gamma = 71^\circ 23' 42'' 5$$
$$H = 72^\circ 3' 32'' 9$$
$$\cos(\alpha^2 - \gamma) = 9.5038443$$
$$\gamma = 345^\circ 7' 25'' 5$$
$$\cos \beta = 9.9847558$$
$$\phi = 91^\circ 33' 43'' 3$$
$$\cos H = 9.4886003$$
$$\Pi = 148^\circ 44' 41'' 7$$
The elements of the orbit of this comet, therefore, as deduced from the above solution, are as follows:
| Log. $\Pi$ | Passage through the perihelion, November | Inclination | |------------|------------------------------------------|------------| | | | 18-04383 | | | | 15^\circ 53' 0'' 5 | | Longitude of ascending node | 345^\circ 7' 25'' 5 | | Longitude of the perihelion | 148^\circ 44' 41'' 7 | | Motion direct |
The orbit determined according to the method which has here been explained, will always satisfy the first and third observations. In general it will also represent the second; but should it occur that the place of the comet at the second observation, computed from the elements thus found, differs considerably from its observed place, we may then conclude that the true orbit cannot be represented by a parabola. In this case recourse must be had to the ellipse, for computing which, the best method is that which is given by Gauss in his *Theoria Motuum Corporum Coelestium*.
---