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Calculate how long would a day be if the Earth were rotating so fast that objects at the equator were apparently weightless?

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Hint: in order to be weightless, acceleration must be equal to g.
Now, We know that acceleration due to gravity varies with the angle such as, \[g'=g-w ^{2}r \cos \Theta\] At, Equator \[\Theta=0^{0}\] Therefore g'=0 to solve this problem !
\[w ^{2} = a/R\] \[w=v/r = 2\pi/t\] \[t =\sqrt(4\pi ^{2} R/a) \] \[t \approx 161 \sec \approx 2.6 \min\] that is awesome result we will no longer inertial frame.

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A quick online search shows this problem to have been solved sometime ago at 17x faster then the normal spin. But this I believe is a trick question. And despite their very strong mathematics in solving it. It was always incorrect. This is a question that can never be solved it is beyond improbabilities. In order to be weightless there also needs to be a lack of gravity. In order for Earth to no longer have gravity it would need to have no mass. Spin has no effect on mass. Trick question that appeared to have been solved 1000x wrong on the internet. There is no speed the Earth to obtain for objects to become weightless.
Weightlessness and absence of gravity are two different things. Your organs are weightless relative to your body, and you feel weightless every time you are in free fall, ie acceleration of your body is equal to gravity. Just drop a pair of scissors in a bin, and the 2 blades are weightless relative to each other. Throw upwards a plastic water bottle with tiny holes in it and, as soon as it leaves your hands, the water is weightless with respect to the bottle: not a drop of water will leak out of the bottle anymore.
Interesting proposition. However if the Earth was spinning at 17x the speed needed for centrifugal force to momentarily cancel out the earths gravitational force weightlessness would only appear to be achieved. Regardless of the rotation speed of the earth objects would still fall at the same rate. Consider this if you were to fly from the North Pole to the Equator. Once you arrived above the equator and stepped out of the plane in midair. You would still fall at the same rate regardless of earth rotational spin. If you were of actual weightlessness you would not fall. The plane would not actually be flying but floating. This may actually be more of a problem of semantics still if this proposition of things floating out into space due to the the spin of the earth 90 year old men could throw 70 ton boulders at the moon. Gravity is the common name given to the gravitational field. Weight is defined as the force on a mass due to the gravitational field of another object. The spinning of a planet never removes this force. Even if we were floating temporarily due to the increased spin our weight would still remain the same. If we stood on the equator of Jupiter our weight would increase and likewise decrease from standing on the moon.
Weight is a frame-dependant quantity. Weight in frame R is defined as the sum of gravitational forces + inertial forces due to the motion of R relative to an inertial frame of reference. In other words, weight is the opposite of the force needed to keep a body still in a given frame of reference, when no other forces than gravity are acting. These two definitions are equivalent.

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