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Equivalence Principle and Gravity 
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#1
Dec2812, 04:04 PM

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I'm reading a book wherein the earth's rotation is supposedly slowing down. The author claims that a ball thrown in the air would fall faster and harder... But if the rotation slows, wouldn't the equivalence principle say that the smaller acceleration could also be interpreted as a smaller force of gravity?



#2
Dec2812, 04:22 PM

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hi schaefera!
the force of gravity doesn't depend on whether either body is moving (ok, technically, there will be an undetectably slight difference, caused by the undetectably slightly slower mass of a slowerrotating earth ) 


#3
Dec2812, 04:28 PM

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Centrifugal acceleration (or the requirement of centripetal acceleration if you like inertial frames) would be reduced. As a result and if we neglect other effects, objects would fall a bit quicker (about 0.3% at the equator) and not perfectly vertical.
One of those other effects: earth would change its shape a bit, too, and become more spherical. This changes the distance between its surface and the center, reducing gravitational acceleration at the poles and increasing it at the equator even more. 


#4
Dec2812, 04:30 PM

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Equivalence Principle and Gravity
I was thinking that there would be a smaller centripetal acceleration from us being pulled along with the rotations. And while this would increase apparent weight (normal forces on us would slightly increase due to the lower centripetal acceleration), it wouldn't make gravity stronger for anything else...



#5
Dec2812, 04:36 PM

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but the author was talking about a falling ball, so there's no normal forces



#6
Dec2812, 04:58 PM

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Gravity (as fundamental force) does not get stronger, but the acceleration towards the bottom does.
@tinytim: Earth is not a perfect sphere. 


#7
Dec2812, 06:10 PM

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Why does the acceleration toward the bottom get bigger what equations could I consider?



#8
Dec2912, 08:49 AM

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In a frame relative to the surface of earth, and neglecting coriolis force and the nonspherical shape of earth: $$\vec{a}= \frac{MG}{r^3} \vec{r}  \omega \times (\omega \times \vec{r})$$
where r is the vector between surface and center of earth. I hope I got the sign right. The second part always points away from the surface (perpendicular on the equator, in other directions elsewhere), if you remove it (by setting ω=0), the acceleration towards the ground gets reduced. 


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