Re: a surprising escape speed problem

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

> -----Original Message-----
> From: Carl E. Mungan [mailto:mungan@USNA.EDU]
> Sent: Sunday, October 27, 2002 3:05 PM
> [snip]
> Note that if you apply energy and momentum conservation in this same
> way, it clears up the apparent violation of energy conservation in
> the Appendix of my previous message about a rock dropped 5 m on earth
> as viewed by someone traveling upward at 5 m/s:
>
> cons. of E => v^2 + 2vu = 2gh + M/m*(u^2 - u'^2)
>
> where v = 10 m/s of rock relative to earth after dropping 5 m, u = 5
> m/s of earth initially relative to the observer, and u' = speed of
> earth finally relative to the observer.
>
> cons. of p => v = M/m*(u - u')
>
> Substitute the second into the first to find v = sqrt(2*g*h). Sorry,
> no Nobel prize.

I thought the problem with your initial appendix was relatively simple: h
was measured relative to the earth frame while v was measured relative to
the elevator.  I figured that was the point you were making.  Now I'm
getting lost.  Can someone simply explain the point?

____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

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