Re: Today's jaw dropper

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Sun, 5 Oct 1997, Leigh Palmer wrote:

> > Chapter 4, Problem 19 in Serway's, 4th Edition asks students to calculate
> > the muzzle velocity and time of flight for a bullet which is fired
> > horizontally by an astronaut standing on the Moon and which is then to
> > "travel completely around the Moon and reach the astronaut at a point 10.0
> > cm below its initial height."
> >
> > Making the assumptions that this problem seems to call for, I get
> >
> > time of flight = 0.35 s
> > muzzle velocity = 0.10 c
> >
> > Who knew that the Moon was so nearly a black hole?
>
> What was the dissipative mechanism suggested? Is this a three body
> problem? If not, one would expect energy conservation to return
> the bullet to its initial elevation after one orbit. Perhaps the
> Moon is considered to rotate, and the bullet is in an eccentric
> orbit? Please drop the other shoe. Some of us don't have copies of
> Serway.

Given its position in the 2-d kinematics chapter and *before* introducing
the concept of force, I can only assume that the intention was for
students to model this as a *very* long baseline constant gravity problem
with a drop of 10 cm over a distance equal to the circumference of the
moon.  Doing so yields the absurd answers I proposed above.

BTW, Henry Kuhlman wrote to me privately saying that, in his third
printing copy (mine is the first), the "10 cm" reference has been
exorcised, so I guess someone had already noticed the preposterous nature
of the implied solution method. 

John
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