Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/140

The quadratic equation (24) gives two values of the correction to be applied to the position of the body. When they are not too large, they will belong to two different solutions of the problem, generally to the two least removed from the values assumed. But a very large value of $\Delta q_{2}$ must not be regarded as affording any trustworthy indication of a solution of the problem. In the majority of cases we only care for one of the roots of the equation, which is distinguished by being very small, and which will be most easily calculated by a small correction to the value which we get by neglecting the quadratic term.^[1]

When a comet is somewhat near the earth we may make use of the fact that the earth's orbit is one solution of the problem, i.e., that $-\rho _{2}$ is one value of $\Delta q_{2},$ to save the trifling labor of computing the value of $({\mathfrak {T}}''{\mathfrak {S}}'''{\mathfrak {S}}').$ For it is evident from the theory of equations that if $-\rho _{2}$ and $z$ are the two roots,

$\rho _{2}-z={\frac {({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}''')}{{\tfrac {3}{5}}({\mathfrak {T}}''{\mathfrak {S}}'''{\mathfrak {S}}')}}$⁠$-\rho _{2}z={\frac {({\mathfrak {S}}{\mathfrak {S}}'''{\mathfrak {S}}')}{{\tfrac {3}{5}}({\mathfrak {T}}''{\mathfrak {S}}'''{\mathfrak {S}}')}}\cdot$

Eliminating $({\mathfrak {T}}''{\mathfrak {S}}'''{\mathfrak {S}}'),$ we have

$(\rho _{2}-z)({\mathfrak {S}}{\mathfrak {S}}'''{\mathfrak {S}}')=-\rho _{2}z({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}'''),$

whence

${\frac {1}{z}}={\frac {1}{\rho _{2}}}-{\frac {({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}''')}{({\mathfrak {S}}{\mathfrak {S}}'''{\mathfrak {S}}')}}\cdot$

Now $-{\frac {({\mathfrak {S}}{\mathfrak {S}}'''{\mathfrak {S}}')}{({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}''')}}$ is the value of $\Delta q_{2},$ which we obtain if we neglect the quadratic term in equation (24). If we call this value $[\Delta q_{2}],$ we have for the more exact value^[2]

$\Delta q_{2}={\frac {[\Delta q_{2}]}{1+{\frac {[\Delta q_{2}]}{\rho _{2}}}}}\cdot$

(26)

The quantities $\Delta q_{1}$ and $\Delta _{2}$ might be calculated by the equations

$-({\mathfrak {S}}{\mathfrak {S}}''{\mathfrak {S}}''')=$	$({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}''')\Delta q_{1}+$	${\tfrac {6}{10}}({\mathfrak {T}}''{\mathfrak {S}}''{\mathfrak {S}}''')\Delta q_{2}^{2}$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$ (27)
$-({\mathfrak {S}}{\mathfrak {S}}'{\mathfrak {S}}'')=$	$({\mathfrak {S}}'{\mathfrak {S}}''{\mathfrak {S}}''')\Delta q_{3}+$	${\tfrac {6}{10}}({\mathfrak {T}}''{\mathfrak {S}}'{\mathfrak {S}}'')\Delta q_{2}^{2}$

1. In the case of Swift's comet (V, 1880), the writer found by the quadratic equation –.247 and –.116 for corrections of the assumed geocentric distance 1250. The first of these numbers gives an approximation to the position of the earth; the second to that of the comet, viz., the geocentric distance .134 instead of the true value .1333. The coefficient ${\tfrac {6}{10}}$ was used in the quadratic equation; with the coefficient the approximations would not be quite so good. The value of the correction obtained by neglecting the quadratic term was .079, which indicates that the approximations (in this very critical case) would be quite tedious without the use of the quadratic term.
2. In the case mentioned in the preceding footnote, from $[\Delta q_{2}]=.079$ and $\rho _{2}=.25,$ we get $\Delta q_{2}=-.1155,$ which is sensibly the same value as that obtained by calculating the quadratic term.