On Sat, 22 Jun 2002, Alex Fraser wrote:
> Since this rock wasn't seen till it passed the earth and was effected by the
> earth's (and moon's) gravity then how close can you come to calculating it's
> original orbit?
Very close, if they got good data on its post-encounter trajectory. The
effects of the Earth-Moon encounter are quite precisely predictable --
forward or backward in time -- if the data is good enough. The effect of
a close encounter with the Earth and the Moon isn't mysterious and arcane;
it just amplifies the fuzziness in the data, because a small difference in
(e.g.) encounter distance turns into a bigger difference in the trajectory
on the far side of the encounter.
The hard part, in general, is taking a short stretch of trajectory data
obtained today, and turning that into a good prediction of what the orbit
will be after the *next* Earth encounter. A short optical-tracking
sequence doesn't give you that precise an idea of the orbit, which
translates into quite a bit of fuzziness about the details of the next
encounter, which gets amplified into a whole lot of uncertainty about the
orbit after that.
(What *can* give you a very good idea of the orbit very quickly is radar
tracking, because it fills in the third dimension and does it very
precisely. I haven't heard whether we got any radar data on this one;
probably not, unless somebody moved very quickly.)
> There could be many solutions of incoming orbits to produce
> the outgoing orbit. We can't know the original orbit.
No, the process is exactly reversible: just push the rewind button :-)
and the physics all work the same. In the absence of complications like
tidal energy dissipation, one outgoing orbit corresponds to exactly one
incoming orbit. The only question is, *how precisely* can the original
orbit be determined?
Henry Spencer
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