Guest 2 jags Posted January 8, 2005 Share Posted January 8, 2005 An interesting question popped up in a conversation the other day. Would the centripetal (centrifugal) force experienced near the equator caused by the rotation of the earth slightly counteract gravity and make you and everything else feel lighter? And also make you feel heavier nearer the poles? Its been a long time since I studied physics for my A-levels and I've forgotten most the equations. I know that velocity at the equator is around 1000 mph which is high but angular velocity is relatively low. I cant remember which of these is the more impotant factor. If you are lighter at the equator,does this cause any other interesting effects? such as people living at the equtor developing weaker bones than people living at the poles. Has this been observed? Link to comment Share on other sites More sharing options...
YT2095 Posted January 8, 2005 Share Posted January 8, 2005 the further you get away from the earths mass center, the lighter you become, so On Mt.Everest, you`de weigh a little less. I think by mentioning the Equator you`re considering Spin, (centripetal force)? Link to comment Share on other sites More sharing options...
Guest 2 jags Posted January 8, 2005 Share Posted January 8, 2005 the further you get away from the earths mass center' date=' the lighter you become, so On Mt.Everest, you`de weigh a little less. I think by mentioning the Equator you`re considering Spin, (centripetal force)?[/quote'] Yes its the spin I'm reffering to. I'm thinking that the nearer the equator you are, the greater your velocity is and so the higher the amount of centripetal force which is pushing you away from the earth. On the poles you would be perfectly stationary and so you'd experience the full force of gravity. Sorry I cant put it in simpler terms, I'm not a very good scientist! Link to comment Share on other sites More sharing options...
Martin Posted January 8, 2005 Share Posted January 8, 2005 An interesting question popped up in a conversation the other day. Would the centripetal (centrifugal) force experienced near the equator caused by the rotation of the earth slightly counteract gravity and make you and everything else feel lighter? And also make you feel heavier nearer the poles? Its been a long time since I studied physics for my A-levels and I've forgotten most the equations. I know that velocity at the equator is around 1000 mph which is high but angular velocity is relatively low. I cant remember which of these is the more impotant factor. If you are lighter at the equator,does this cause any other interesting effects? such as people living at the equtor developing weaker bones than people living at the poles. Has this been observed? the earth is not a sphere, it is oblate. the equatorial radius is greater than the polar radius by a few tens of miles so even if the earth were not spinning, gravity would be weaker at the equator, simply because someone standing at sealevel at the equator is farther from the center of the earth than someone standing at the north pole the EFFECTIVE gravity at the equator is also weaker because of the effect of the earth's spin but both these effects only make a fraction of a percent difference I do not know of any biological effect of this tiny difference in the strength of gravity. Temperature would be overwhelmingly more important on how critters evolve in the two regions. the metric value of sealevel gravity 9.80665 meters per sq. second is only an average or arbitrarily chosen value and in a good physics handbook like CRC Handbook of Chem and Phys you can find the real value of gravity tabulated BY LATITUDE and you can see the variation. maybe it is on the web too, try googling "gravity latitude" or "variation of gravity with latitude" or something like that Link to comment Share on other sites More sharing options...
Martin Posted January 8, 2005 Share Posted January 8, 2005 Yes its the spin I'm reffering to.... OK then let's calculate the effect of the spin the acceleration felt by a point on equator is R w2 where R is radius in meters 6378,000 and w is radians per second (2pi/86400) 6378000 X (2pi/86400)2 = 0.0337 meters per sq second that is pretty tiny compared with the rough size of 9.8 meters per sq second it is only THREE TENTHS OF A PERCENT difference due to spin we are talking three tenths of a percent of one gee. you could not feel this btw the equatorial radius of the earth is 6378 km and the polar radius is 6357 km. so the man at the equator is standing on a 20 km high mountain in effect. he is farther from the earth's center the oblateness of the earth, and of the ocean and atmosphere covering the earth is due to the spin. It is only a fraction of one gee, but the ocean feels that and so sealevel bulges out 20 extra km just as the solid earth bulges out-----bulge matches bulge and nobody notices Link to comment Share on other sites More sharing options...
ydoaPs Posted January 8, 2005 Share Posted January 8, 2005 what is g on the equater? Link to comment Share on other sites More sharing options...
Martin Posted January 8, 2005 Share Posted January 8, 2005 what is g on the equater? dammit yourda my CRC handbook is upstairs! why cant you look these things up you are a smart kid why dont you google for yourself OK, I looked it up in the handbook, it is 9.78 meters per sq second (compare that with the conventional metric value of 9.80665) Link to comment Share on other sites More sharing options...
ydoaPs Posted January 8, 2005 Share Posted January 8, 2005 i like to make your life difficult. Link to comment Share on other sites More sharing options...
Martin Posted January 8, 2005 Share Posted January 8, 2005 i like to make your life difficult. all right I will reciprocate and ask you a similar question what is the sealevel gravity at the north pole? it will be MORE than the conventional metric value of 9.80665 see if you can find it out Link to comment Share on other sites More sharing options...
Externet Posted January 10, 2005 Share Posted January 10, 2005 Yes, you are lighter at equator; by two reasons; the increased distance from the center of earth gravity and the centrifugal force. http://www.madsci.org/posts/archives/mar97/857469722.Ph.q.html http://www.madsci.org/posts/archives/mar97/857469722.Ph.r.html What I don't get is if an object on top of a scale at equator determines the force of gravity there or the force of gravity less the centrifugal force. By figures from a post above, 9.78 = G(eq) minus 0.0337 In other words, the weight of a body at the equator does not obey only to the gravity constant there but also the centrifugal force; then the pure gravity constant figure measured at the equator is inexact because the centrifugal force was not taken in account.... I think... Miguel Link to comment Share on other sites More sharing options...
Martin Posted January 10, 2005 Share Posted January 10, 2005 Yes' date=' you are lighter at equator; by two reasons; the increased distance from the center of earth gravity and the centrifugal force. http://www.madsci.org/posts/archives/mar97/857469722.Ph.q.html http://www.madsci.org/posts/archives/mar97/857469722.Ph.r.html What I don't get is if an object on top of a scale at equator determines the force of gravity there or the force of gravity less the centrifugal force. By figures from a post above, 9.78 = G(eq) minus 0.0337 In other words, the weight of a body at the equator does not obey only to the gravity constant there but also the centrifugal force; then the pure gravity constant figure measured at the equator is inexact because the centrifugal force was not taken in account.... I think... Miguel[/quote'] Yes in the handbook table (CRC handbook) I used it said that the effective gravity at equator (which includes centrifugal effect) is 9.78 and also the calculation showed that the centrif effect is 0.0337 So now one can add these two and say 9.78... + 0.03... = 9.81... SO IF ONE COULD STOP THE EARTH from turning just for a moment and measure the gravity at the equator it would be 9.81... and also in France where it is often 9.80665 it would be more, because there is also the centrifugal effect which would be temporarily turned off. Everything would be screwed up and the water of the ocean would no longer want to be at the equator and the ocean would flow north towards the pole and south towards the south pole. holy shit! quickly get the earth to start turning again! what you call the "pure gravity" (that is, without including the centrif. effect) is not usually tabulated in the handbook. but we can say that at the equator it would be around 9.81 and using trigonometry we could calculate it for other latitudes also, like France. But maybe this is not interesting. Or maybe it is interesting because it is the pure gravity out in space that determines the orbits of satellites Link to comment Share on other sites More sharing options...
Externet Posted January 11, 2005 Share Posted January 11, 2005 I have deleted my own poorly expressed post, where I should had said the atomic clocks are less related to earth's rotation now after the Ninetyeast ridge quake. Miguel Link to comment Share on other sites More sharing options...
swansont Posted January 11, 2005 Share Posted January 11, 2005 Hi.From NASA calculations' date=' bye bye to atomic clocks, as the north pole just shifted and rose 25mm from the earthquake at the ninetyeast fault, making the globe less flat at the poles thus spinning faster. Now days are shorter some microseconds. To be confirmed with actual readings... ¿Would that have changed the sea level too? Miguel[/quote'] Why say bye bye to atomic clocks? The atom has defined the second for some time now, not the earth's rotation. The slowdown due to the quake is small on the scale of many other perturbations. We add leap seconds when they accumulate enough phase difference. And the north pole wanders around, too, even in the absence of quakes. Link to comment Share on other sites More sharing options...
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