speed at the poles?
- Thread starter ColJack
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so here's a follow up question..
why don't things at the pole get crushed?
presumably at the equator there is a certain amount of centrifugal force trying to send us flying into space, but as you're not moving at as great a velocity at or near the pole then why doesn't gravity cause you to get much heavier..?
why don't things at the pole get crushed?
presumably at the equator there is a certain amount of centrifugal force trying to send us flying into space, but as you're not moving at as great a velocity at or near the pole then why doesn't gravity cause you to get much heavier..?
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If the earth was not spinning, and was a perfect sphere, you would weigh the same where ever you were.
As the earth is spinning, it creates a bulge at the equator, which means you are slightly further away from the centre, thus you weigh slightly less at the equator.
You are also subject to a centripetal force at the equator, but this is much much smaller than the gravitational force holding you down.
It is quite difficult to measure the weight difference at the pole/equator, as any weighing machine is subject to the same forces as you, dependant on where you are.
EDIT: I have just realised that as the platform of your weighing machine is subject to a centripetal force, making it appear slightly lighter than it really is, you're all fatter than you think you are............
D
dextrous
Er, what centripetal force at the equator?
And gravitational pull decreases by an inverse proportion in relation to the square of the distance between the objects.
And gravitational pull decreases by an inverse proportion in relation to the square of the distance between the objects.
Centripetal acceleration at the equator is 4*pi2*r/T2.
The period of rotation is 86400 seconds and the radius of the Earth is about 6400 km. This means that the centripetal acceleration at the equator is about 0.03 m/s2.
Gravity is about 10 m/s2. therefore you will weigh about 0.3% less at the equator than at the poles.
Your point being?
D
dextrous
Not doubting you here, but where did you get all this from, as it's been a while since I studied this sort of stuff and was under the impression that mv2/r was the centripetal force? (can't immediately see how this relates to 4pi2r/t2)
Just providing clarification
You are right, mv2/r is used to calculate the centripetal FORCE, when you have the mass of a body.
4pi2r/t2, however is used to calculate centripetal ACCELERATION, and I used this to compare it to gravitational acceleration of 10 m/s2.
I think I am on the right lines with my calculations, but I'm open to correction..........
If you stood on a frictionless turntable, yes.
As does a pendulum.....
http://en.wikipedia.org/wiki/Foucault_pendulum
D
dextrous
anything with mass will be affected by gravity and also create it's own gravitational pull. So any temptation to spew out is counterbalanced by other molecules' (and its own) gravitational pull.
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