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Gravity at the equator 
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#1
Dec2208, 09:54 PM

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Assuming a perfect sphere,
mg  w' = m (v^2/r) g' = g  (v^2/r) I understand there's a constant radial acceleration because we're not flying off the planet. Theoretically, the difference should be 0.0337 m/s^2, but isn't because the Earth "bulges" at the equator. Why is there a difference on a rotating perfect sphere? It's not quite connecting intuitively or physically. 


#2
Dec2208, 11:02 PM

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It's because of centrifugal force:
http://en.wikipedia.org/wiki/Centrifugal_force The centrifugal force exists when there is rotation in the observer's frame of reference, so the exact shape doesn't really matter, except to play a role in the magnitude of the effect, which is negligible for most purposes on earth. The bulge at the equator means that the mass of the earth is not spread uniformly (topography affects this too), so this creates variations in the local gravity. Here is something else interesting to consider. Suppose your at some latitude. As the earth rotates on its axis, your motion traces out the circumference of the circle within a plane that is perpendicular to the earth's rotational axis. Now the centrifugal force vector lies in the same plane and is directed radially outward. Now consider the acceleration vector for gravity. It points toward the center of the earth. So when you drop an object, it actually deviates by a small angle from the direction of the gravitational acceleration vectorthe result is negligible but it is there. Draw a picture and construct the resultant vector from the two aforementioned vectors and you can see this result. 


#3
Dec2208, 11:04 PM

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On a rotating perfect sphere on uniform density, it's pretty easy to calculate the difference between g at the poles and g at the equator. Just subtract the centripetal acceleration. If the planet deforms, it's not. I'm not sure what's confusing you. The bulge is going to change the gravitational field and the radius. How depends on the rigidity of the material.



#4
Dec2208, 11:05 PM

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Gravity at the equator



#5
Dec2208, 11:08 PM

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It's fictitious in the sense that it's not a 'force', it's an acceleration. It's still there. You still have to factor it into F=ma, because that contains both forces and accelerations.



#6
Dec2208, 11:11 PM

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#7
Dec2208, 11:12 PM

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#8
Dec2208, 11:14 PM

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And to echo someone else's input, say, above the equation, an object's downward resultant vector would be slightly higher than it would be without the rotational consideration. 


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Dec2208, 11:24 PM

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#10
Dec2208, 11:25 PM

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#11
Dec2208, 11:27 PM

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