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Our weight if the earth spins faster... 
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#1
May412, 10:51 AM

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I came up with this problem in a competition. They asked what would be the effect on our weight on the earth if the earth spun any faster.
Thinking deep, I came up to a conclusion that people who live nearer to the equator than to the poles find their weight reduced since any faster spinning of the earth would push us slightly out, due to centrifugal force. Am i right?? would there be anymore effect?? 


#2
May412, 12:39 PM

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Yes, you would weigh slightly less at the poles than at the equator due to centripetal force. Acceleration due to gravity is reduced by about 0.03 m/s^2. You are also slightly closer to the center of the earth at the poles  also due to the earth's rotation [it causes the earth to bulge a bit]. The net effect is a difference of about 0.05 m/s^2 between the equator and poles  meaning a sumo wrestler who weighs 200 kg at the south pole would weigh a mere 199 kg in Padang, Indonesia.



#3
May412, 10:59 PM

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#4
May512, 12:44 AM

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Our weight if the earth spins faster...
Hey  i think we weigh less at equator
As already mentioned  firstly due to the fictitious centrifugal force and also because earth's a bit more outwards at the equator 


#5
May512, 12:48 AM

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So, the reasoning that the faster spinning of earth would lighten us is correct?



#6
May512, 03:06 AM

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A dyslexic moment. Less at the equator, as asserted by the rest of my post.



#7
May512, 07:53 PM

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So there are two factors involved in making us weigh more at the poles.
1. We are closer to earth's center due to the earth's oblateness 2. There is less centripedal force hurling us outward at the poles. Excerpt Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator. http://curious.astro.cornell.edu/que...php?number=310 


#9
May1212, 02:42 PM

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I wonder if the centripetal acceleration actually contributes as much as v^{2}/r? I could see this if Earth were perfectly rigid, and able to hold its spherical shape. But it can't. That's why it bulges. So not only is v^{2}/r happening to a person standing on the equator, but its also happening to the ground they are standing on, negating most or all of the effect.



#10
May1412, 02:55 AM

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#11
May1412, 05:17 AM

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So it's not perfectly spherical, but it does hold its current shape perfectly. 


#12
May1412, 07:45 AM

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If we use the formula [itex]\omega^2 r[/itex] for the centripetal acceleration instead, we can look at it this way: As long as we are considering the Earth as it is, Then the angular velocity stays the same as we move from pole to equator, r increases, and the fact that the Earth bulges at the Equator only has the effect of increasing the maximum value of r. In this case, the formula does give us the full value of the centripetal acceleration. This is because the bulge is not actively changing. On the other hand, assume you start with a rigid spherical Earth, and then left it "soften" so that it forms an equatorial bulge. As it does so, Its moment of inertia increases, and in order to conserve angular momentum, its angular velocity goes down. The decrease in angular velocity has a larger effect than the increase of radius and the centripetal acceleration at the Equator goes down. But this a comparison made between pre and post bulge formation. It is the change in the bulge that causes the decrease. As far as I can tell, this discussion deals with the Earth "as is", in which case, the formula holds. 


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