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==>I take the scale with me in the shuttle and "stand" on it
==>and get zero. Therefore, I'm weightless.
I was very surprised by Leigh's statement (about 2 days ago
see below), but I was too busy writing tests to reply earlier.
So now I'm left to reply "me too".
I'll be saying "I take the scale with me in the shuttle
and "stand" on it and get zero. Therefore, I'm weightless."
again in my conceptual physics course in about 10 days.
I think it is very clear because everyone has used a
bathroom scale. Conversely they have never used a
triple-beam balance (My lab equipment consists of a set
of children's toys that I bring in from home - don't ask),
do not know what one is or how it works.
Sorry, but I felt a need to put in my $0.02.
On to what Leigh wrote:
==>Calling the shuttle a microgravity (better would be migro-gee) is
entirely
==>appropriate. The gravitational field of the Earth varies by one part in
a
==>million over a radial distance of three meters from the center of the
==>Earth if my quick mental calculation does not err*. That would be a
==>microgee over a scale parameter appropriate to shuttle experiments.
==>
==>scratch misconception 5. Gravity in the shuttle is small, as one might
==>well expect seeing those folks floating around in there.
==>
==>* OK. Mental calculation redone below, partly to show the value of math
==>in doing conceptual problems. Conventional symbols (g is grav. field):
==>
==> G M
==> g = -----
==> 2
==> r
==>
==>
==> -6 dg dr
==> 10 = ---- = -2 ---- ; r = 6600 km (or so)
==> g r
==>
==>
==> r dg
==> dr = --- ---- = (-) 3.3 m
==> 2 g
It appears that Leigh has proved that in the shuttle over
a range of 3.3 meters dg/g varies by only a factor of
1 ppm, so the term microgravity makes sense. But this
proof goes through just fine on the surface of the Earth
(with a reduction in the allowed distance to 3.2 m).
Either I'm missing something here or it was a joke.
Which is it?
Bill Larson
Geneva