[Phys-l] (no subject)

[Jack Uretsky <jlu@hep.anl.gov>]

Good point about the 60 km/hr being tiny.  The physics is that the extra 
speed goes into moving #2 into a new orbit, slightly outside tho old one, 
at slightly lower orbital speed.  So I agree with JM's answer.

The way to catch up with #1 is to move into a new, and lower (hence, 
higher speed) orbit orbit.  So the thrusters whould be fired pointing 
toward #1, in order to reduce the energy of the orbit (and increase the 
speed).
				Regards,
					Jack



On Tue, 25 Jul 2006, Folkerts, Timothy J wrote:

> John Denker wrote:
>
> > Here's the scenario:  Spacecraft #1 is launched into a nice, circular,
> > low-earth orbit.  Shortly thereafter, spacecraft #2 is launched into
> > the _same_ orbit.  At this point in the story, spacecraft #2 is
> > keeping
> > pace with #1, and is simply 90 km in trail.
> >
> > The flight plan calls for the two craft to rendezvous.  At time t,
> > spacecraft #2 fires its thruster, thereby giving itself a delta v
> > of 60 km/h, directed toward spacecraft #1 in the obvious way.
> >
> > Here's the question:  Within a km or so, how far apart are the
> > spacecraft at time t + 90 minutes?
> >
> > Vis vobiscum.
>
>
> Here is my thinking - no calculations.
>
>
> To first order, the original space craft will make one complete orbit in 90 min.
>
>
>
> To first order, the extra 90 km/hr is trivial compared to the orbital speed, so it is a minor perturbation.  After 90 min it will also have completed one orbit and will be back where it started in 90 min.  Since the second rocket is in a slightly larger orbit, a second order correction implies it will take slightly longer to complete the orbit, so it will be slightly farther away than it had been.  (After 45 min = half an orbit, the second rock would be almost straight "above" the first).  360 km (from John M) seems a little more than I would have guessed, by I could believe it.
>
>
>
> Tim F

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