Rotation about the pointing axis, on the other hand, could potentially induce significant curvature. If the satellite has a rotational velocity of $\theta \u02d9$ about the pointing axis, which will be taken as $z$, and the target has velocity components ($vx$, $vy$, $vz$) and coordinates of Display Formula
$x=xo+vxt,y=yo+vyt,z=zo+vzt,$(1)
with respect to the satellite center of mass, then the location of the target in the detector coordinate system is given by Display Formula$x\u2032=(xo\u2032+vx\u2032t)cos(\theta \u02d9t)+(yo\u2032+vy\u2032t)sin(\theta \u02d9t),y\u2032=\u2212(xo\u2032+vx\u2032t)sin(\theta \u02d9t)+(yo\u2032+vy\u2032t)cos(\theta \u02d9t),$(2)
where the primes represent the mapping of object space to pixel space and rotation of the satellite about the $x$-and $y$-axis has been folded into the components $vx$ and $vy$. Note that Mapping the detector coordinate system to equatorial coordinates will be achieved by the astrometric solution to the star field being observed. An equatorial angles-only determination of the target orbit can then be achieved.