Re: a surprising escape speed problem

[John Mallinckrodt <ajm@CSUPOMONA.EDU>]

Interesting problem!

Because it was not immediately evident how to work this out
analytically, I opted for the experimental method and put together an
Interactive Physics simulation.  The initial results do not support
your conclusion.  In my simulation v_esc = v_orbit = 1 for now and I
find that the launch speed (relative to the Earth) required for
escape depends on the direction I launch.  For instance, if I launch
directly upward from the midnight position, the required speed is
~.55.  If I launch directly up from the forward position on Earth
(i.e., in the direction of the orbit), v is ~ .64.  In both cases the
orbits of the rocket are complicated and I'm simply finding the first
speed that leads to a trajectory that doesn't eventually come
crashing back down to Earth.

I need to go to class now, but I'll play around some more with this later on.

John

> I present a simplified version of the problem I have been toying
> with. Suppose we get rid of the sun and instead put the earth at the
> end of a string (whose length equals the usual earth-sun distance)
> and whirl it around so that it matches earth's usual orbital speed of
> v_orbit = 29.8 km/s. Also suppose that the earth does not spin on its
> axis.
>
> RIGHT ANSWER.
>
> I find v = sqrt(v_orbit^2 + v_esc^2) = 31.8 km/s no matter what
> direction you launch it! And yes, this is the speed relative to the
> earth not relative to the sun. Can you confirm this answer? Gene and
> I would really like to know if this is right.


--
   A. John Mallinckrodt        http://www.csupomona.edu/~ajm
   Professor of Physics      mailto:ajm@csupomona.edu
   Physics Department         voice:909-869-4054
   Cal Poly Pomona              fax:909-869-5090
   Pomona, CA 91768-4031     office:Building 8, Room 223

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.