Re: Today's jaw dropper

[brian whatcott <inet@intellisys.net>]

At 11:19 10/5/97 -0700, you wrote:
> On Sun, 5 Oct 1997, John Mallinckrodt wrote:
>
> > "travel completely around the Moon and reach the astronaut at a point 10.0
> > cm below its initial height."
>
> So the moon must have an atmosphere?  If perfect vacuum, wouldn't the
> bullet be in orbit, and therefor NOT fall 10cm after a single orbit?
>
> .....................uuuu / oo \ uuuu........,.............................
> William Beaty 

I fancy the indicated treatment was to separate the horizontal and vertical
motion in the Newtonian manner.

In this way, we are to consider the time needed for an object to drop
0.1m in the Lunar gravity. Let's see, x = 1/2 a.t^2 
where a = 9.8 x 0.16 m.s^-2

Or...if we are more sophisticated, the time for it to drop two moon diameters
plus 0.1m

(This would possibly allow it to travel even slower than 0.1c?)

Sincerely,

brian whatcott <inet@intellisys.net>
Altus OK