|Nov6-10, 10:40 PM||#1|
Can gravitational field strength equal the centripetal acceleration?
As a homework question, it asks, "...if the earth were rotating so fast that the objects at the equator were apparently weightless?"
Somewhere, someone said that, quote:
In order for the rotation of the earth to cancel weight, the gravitational field strength should equal the centripetal accel. (v^2/R)
Do they mean g=a(centripetal)?
I don't get how that makes sense.
|Nov7-10, 01:41 AM||#2|
The object on the equator moves along a circle of radius of the Earth (R) with the velocity of the equator (v). The centripetal force needed to this motion is provided by gravity Fg=GmM/R^2 (M is the mass of Earth) and the normal force N acting between the object and ground:
The object is weigthless if the ground does not push it upward, and the object does not push the ground, that is N=0. the If the normal force is 0 the centripetal force is equal to gravity at the equator.
The gravitational field strength is Fg/m. Dividing the previous equation by m,
Fg/m = G M/R^2= v^2/R.
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