MATHEMATICAL SECTION. 95 
ing to look at the problem from another point of view, and one 
that will lead us to consider more closely its geometry. 
Imagine a set of rectangular axes in space, the origin being at 
the center of the planet, and denote by X, Y, Zthe points on the 
celestial sphere made by the intersections of these axes. Let S be 
the point where the prolongation of the radius vector of the satel- 
lite strikes the sphere; then we have for the co-ordinates of the 
satellite 
z=r. cos SX 
y= r.cos SY 
z= r. cos SZ 
We can express these cosines by means of six auxiliary quanti- 
ties similar to those that Gauss has used for computing the position 
of a planet. Take the prolongation of the right line drawn from 
the earth to the planet as the axis of Z, the axis of Y in the plane of 
the declination circle that passes through Z, and the axis of X at 
right angles to this plane and in the direction of increasing right 
ascensions. Let O be the pole of the equator and T' the positive 
pole of the orbit of the satellite. Introduce the following notation, 
which is the same as Bessel’s : 
arc TX =f, angle OTX=F 
Re A ete So WR cae 
Ny Ore ar 
Since the are TS = 90°, the spherical triangles STX, STY, and 
STZ give 
cos SX = sin f cos STX 
cos SY = sin gcos STY 
cos SZ = sin h cos STZ 
The distance of the satellite in its orbit from the node being u, 
and the angle OTN being 90°, we have 
STX = 90° — (F + u) 
STY = 90° — (G+ u) 
STZ = 90° — (H+ u) 
And the values of the co-ordinates are therefore: 
z=r.sinfsin(F + wu) 
y =r. singsin(G+ u) (4) 
z=r.sinhsin (H + u) 
