RE: Centripetal Force at Equator

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From: J. G. Locke (lockejg@redshift.com)
Date: Sun Nov 07 1999 - 06:47:23 PST 

From: "J. G. Locke" <lockejg@redshift.com>
Subject: RE: Centripetal Force at Equator
Date: Sun, 7 Nov 1999 06:47:23 -0800
Message-ID: <000201bf292f$003c0b40$4b30c8d8@daddys-machine>

||
|| Subject: Centripetal Force at Equator
|| From: "David Porter" <dpotasnik@earthlink.net>
|| Date: Sat, 06 Nov 1999 08:11:14 -0700
||
|| A problem asks why, if the earth spins at such a great speed,
|| don't all of us fly off. After calculating the centripetal force
|| due to gravity that is necessary to hold us down, one comes up
|| with around 2.3 N for the equator. That explains any object, like
|| a person, that is heavier than 2.3 N. It obviously doesn't
|| explain why objects that weigh less don't fly off. This question came
|| up with one of my students, but I didn't have an explanation.
|| Is there anyone out there who does? Thanks, david.
||
||

The centripetal force and the gravitational force are both proportional to
the mass of the object. [ centripetal: F=mv^2/2 where v is linear velocity
of object on Earth's surface; gravity: F=GMm/r^2 where G is universal grav.
constant, M and r are mass and radius of Earth. The m is the mass of the
object. ] More information on these relationships can be found in most
physics text books; look under 'circular motion' and 'gravitation'.

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