Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/137

Our equation may therefore be regarded as signifying that the three vectors ${\mathfrak {R}}_{1},{\mathfrak {R}}_{2},{\mathfrak {R}}_{3}$ lie in one plane, and that the three triangles determined each by a pair of these vectors, and usually denoted by $[r_{2}r_{3}],[r_{1}r_{3}],[r_{1}r_{2}],$ are proportional to

$A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right)$⁠$\left(1-{\frac {B_{2}}{r_{2}^{3}}}\right)$⁠$A_{3}\left(1+{\frac {B_{3}}{r_{3}^{3}}}\right)\cdot$

Since this vector equation is equivalent to three ordinary equations, it is evidently sufficient to determine the three positions of the body in connection with the conditions that these positions must lie upon the lines of sight of three observations. To give analytical expression to these conditions, we may write ${\mathfrak {E}}_{1},{\mathfrak {E}}_{2},{\mathfrak {E}}_{3}$ for the vectors drawn from the sun to the three positions of the earth (or, more exactly, of the observatories where the observations have been made), ${\mathfrak {F}}_{1},{\mathfrak {F}}_{2},{\mathfrak {F}}_{3}$ for unit vectors drawn in the directions of the body, as observed, and $\rho _{1},\rho _{2},\rho _{3}$ for the three distances of the body from the places of observation. We have then

${\mathfrak {R}}_{1}={\mathfrak {E}}_{1}+\rho _{1}{\mathfrak {F}}_{1},$⁠${\mathfrak {R}}_{2}={\mathfrak {E}}_{2}+\rho _{2}{\mathfrak {F}}_{2},$⁠${\mathfrak {R}}_{3}={\mathfrak {E}}_{3}+\rho _{3}{\mathfrak {F}}_{3}.$

(6)

By substitution of these values our fundamental equation becomes

$A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right)({\mathfrak {E}}_{1}+\rho _{1}{\mathfrak {F}}_{1})-\left(1-{\frac {B_{2}}{r_{2}^{3}}}\right)({\mathfrak {E}}_{2}+\rho _{2}{\mathfrak {F}}_{2})+A_{3}\left(1+{\frac {B_{3}}{r_{3}^{3}}}\right)({\mathfrak {E}}_{3}+\rho _{3}{\mathfrak {F}}_{3})=0,$

(7)

where $\rho _{1},\rho _{2},\rho _{3},r_{1},r_{2},r_{3}$ (the geocentric and heliocentric distances) are the only unknown quantities. From equations (6) we also get, by squaring both members in each,

$r_{1}^{2}={\mathfrak {E}}_{3}^{2}+2({\mathfrak {E}}_{3}.{\mathfrak {F}}_{3})\rho _{3}+\rho _{3}^{2},\,\,\,r_{2}^{2}={\mathfrak {E}}_{2}^{2}+2({\mathfrak {E}}_{2}.{\mathfrak {F}}_{2})\rho _{2}+\rho _{2}^{2},$		$\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$ (8)
$r_{3}^{2}={\mathfrak {E}}_{3}^{2}+2({\mathfrak {E}}_{3}.{\mathfrak {F}}_{3})\rho _{3}+\rho _{3}^{2},$

by which the values of $r_{1},r_{2},r_{3}$ may be derived from those of $\rho _{1},\rho _{2},\rho _{3},$ or vice versâ. Equations (7) and (8), which are equivalent to six ordinary equations, are sufficient to determine the six quantities $r_{1},r_{2},r_{3},\rho _{1},\rho _{2},\rho _{3};$ or, if we suppose the values of $r_{1},r_{2},r_{3}$ in terms of $\rho _{1},\rho _{2},\rho _{3},$ to be substituted in equation (7), we have a single vector equation, from which we may determine the three geocentric distances $\rho _{1},\rho _{2},\rho _{3}.$

It remains to be shown, first, how the numerical solution of the equation may be performed, and secondly, how such an approximate solution of the actual problem may furnish the basis of a closer approximation.

Solution of the Fundamental Equation.

The relations with which we have to do will be rendered a little more simple if instead of each geocentric distance we introduce the