How does the Earth's rotation affect time and weight at different latitudes?

[RandallB]

With a speed of about 460m/s for the Earth's surface at the Equator as it spins around its axis.

Using 2 Atomic Clocks tested to be accurate and match each other and both weighing exactly 100 lbs at an arctic lab site pretty much at the north pole.
Moving one clock to Ecuador pretty much on the Equator and the same altitude (as measured from the center of the earth) for one year.

How much “centripetal acceleration” would it feel from the Earth's rotation?

How much will the clock weigh in Ecuador?

After bringing the two clocks back together how will the time on the Ecuador clock compare to the Polar Clock? How much and why for any difference.

 

[dextercioby]

With a speed of about 460m/s for the Earth's surface at the Equator as it spins around its axis.

That's about 465m/s.

Using 2 Atomic Clocks tested to be accurate and match each other and both weighing exactly 100 lbs at an arctic lab site pretty much at the north pole.

You know there are much more people on this planet using the SI-MKgs unit sytem.Or do you really hate the French that much as not to use their units...?? :tongue2:That's round 45.3Kg.And for the length,the unit,one inch,coming from the little finger of one of the hands of king Henry the VIII-th...God,that's sick! :yuck: What if king Henry had lost his hands in a battle?They woulkd have changed the unit? :rofl: :tongue2:

Moving one clock to Ecuador pretty much on the Equator and the same altitude (as measured from the center of the earth) for one year.

The altitude in Ecuador varies in the range 0->6310 (the frozen peak of Chimborazo volcano).So taking the Arctic lab at see level,the difference in heights between the labs would vary in the domain (21-27) Km.So the part with the same altitude measured from the center of the Earth makes no sense to me.Perhaps u'd like to explain...

How much “centripetal acceleration” would it feel from the Earth's rotation?

Why ask us??Don't u know to compute $a_{cp}=\omega^{2}R_{eq}$ ?

How much will the clock weigh in Ecuador?

Gravity force:$F=G\frac{M_{clock}M_{Earth}}{(R_{eq}+h)^{2}}$Compute.Earth's mass round $6\cdot 10^{24}Kg$.Cavendish constant can be taken from a table of fundamental constants.The Earth's radius at equator is approximately 6378000m,"h" goes from 0->6310 m (see above for explanation).I assumed no relativistic effects on the mass.No effects from SR or GR.

After bringing the two clocks back together how will the time on the Ecuador clock compare to the Polar Clock? How much and why for any difference.

Times goes differently at Equator than it goes at the Pole according to both SR and GR.i don't feel like making this sorts of calcultions.Do them yourself... :tongue2:

Daniel.

PS.Where do you bring the 2 clocks back together,at what speed (or even acceleration) do you bring them?

 

[RandallB]

That's about 465m/s.

Got to give you that one Danny - 465 is about 460 only 'about' 1% diff
OR maybe you'd prefer the aprox. speed and % to two decimals. With m being so long and inexact would you prefer a unit a little shorter like foot or inch?

You know there are much more people on this planet using the SI-MKgs unit sytem.Or do you really hate the French that much as not to use their units...?? That's round 45.3Kg.And for the length,the unit,one inch,coming from the little ...

Oops – guess inch is out eh.
How can anyone so anal get the Units for Force and weight wrong?
( Or do you think you can answer the question “In grams, how much does a 100 gram weight weigh on the moon – ‘about’)
And since you were first to bring up inch – WHY? Did you think this was a history thread. And what’s up with ‘French’ – you have some kind of England France conflict going on in your head?

The altitude in Ecuador varies in the range 0->6310 (the frozen peak of Chimborazo volcano).So taking the Arctic lab at see level,the difference in heights between the labs would vary in the domain (21-27) Km.So the part with the same altitude measured from the center of the Earth makes no sense to me.Perhaps u'd like to explain...

Let see maybe to keep the same altitude! You may construct a tower at the pole – I’d suggest using your imagination.

Why ask us??Don't u know to compute $a_{cp}=\omega^{2}R_{eq}$ ?

Not so much for you but for those that might want enjoy working on a problem.If I really had clocks and dog sled ready to go, I’d have posted in another area.
But I do approve of shareing Formulas with everyone - that's nice of you.

Gravity force:$F=G\frac{M_{clock}M_{Earth}}{(R_{eq}+h)^{2}}$Compute.Earth's mass round $6\cdot 10^{24}Kg$.Cavendish constant can be taken from a table of fundamental constants.The Earth's radius at equator is approximately 6378000m,"h" goes from 0->6310 m (see above for explanation).I assumed no relativistic effects on the mass.No effects from SR or GR.

Gee Danny I think I got to mark you down a point on that one. You missed a conversion or translation – try looking up “the same altitude”.

Times goes differently at Equator than it goes at the Pole according to both SR and GR.i don't feel like making this sorts of calcultions.Do them yourself...

Gosh Danny based on the problem as defined this is the only fun part to work on - come back when your more in the mood.QUOTE=dextercioby]
PS.Where do you bring the 2 clocks back together,at what speed (or even acceleration) do you bring them? [/QUOTE]Well just for you I'll give some grownup bonus points if you can show how anywhere on Earth at any speed could affect the clocks to make any significant differance.

 

[dextercioby]

Got to give you that one Danny - 465 is about 460 only 'about' 1% diff.

I see u haven't looked in an atlas for years...If u had,u didn't do it with the glasses on.That's how 460 got on the retina instead of 465. :tongue2:

OR maybe you'd prefer the aprox. speed and % to two decimals.

I guess u missed the point.My figure didn't include decimals.I wouldn't have expected u put the decimals.But at least the (VERY) significant digits...

With 'm' being so long and inexact would you prefer a unit a little shorter like foot or inch? Oops – guess inch is out eh.

Since it relies on the decimal basis and the decimal system,i guess the meter is irreplaceble.

How can anyone so anal get the Units for Force and weight wrong?

To make a bad joke,the word 'anal' has other meanings as well...
Who got the units for Force and Weight wrong??

( Or do you think you can answer the question “In grams, how much does a 100 gram weight weigh on the moon – ‘about’)

I assume this is a joke...A lousy oe,that is... :tongue2:

And since you were first to bring up inch – WHY? Did you think this was a history thread. And what’s up with ‘French’ – you have some kind of England France conflict going on in your head?

:rofl: Apparently u have...Who brought it up??

Let see maybe to keep the same altitude! You may construct a tower at the pole – I’d suggest using your imagination.

Bad suggestion!Since entering Phyics Department at my hometime university,my imagination tends to take into account science.It therefore cannot conceive building a 21->27Km tower at the pole.Or you thought of this as a 'Gedankenexperiment'??You know,Einstein's imagination used to rely to science too...

Gee Danny I think I got to mark you down a point on that one. You missed a conversion or translation – try looking up “the same altitude”.

Have been reading my post without glasses??I guess it seemed like an atlas and u decided it's better if u took out your glasses...

Gosh Danny based on the problem as defined this is the only fun part to work on - come back when your more in the mood

Now that's something really funnny... :rofl: :rofl: Finally u made a good joke... :tongue2:

Well just for you I'll give some grownup bonus points if you can show how anywhere on Earth at any speed could affect the clocks to make any significant differance.

That should have probably offended me,but I'm "different".'Gosh',if it didn't make any 'significant differance',then how come 'this is the only fun part to work on'?

Daniel.For you,and only for you:'Danny'. :tongue2:

 

[RandallB]

Did you look up "About" -- fine use Decameter 46 save it you want 47 I know.

You were first to use anything but metric for Length – need your glasses to read the posts?

You were first to use Grams for Weight – noticed you didn’t try to answer – maybe because using your units for weight it cannot be done.

If you cannot answer any of them – just leave to someone else.
Or did you need a separate Thread area for cracking Wise.

But if you do find the time how about using the Decimal version of time the French tried to make metric.

 

[dextercioby]

Did you look up "About" -- fine use Decameter 46 save it you want 47 I know.

Wow,u keep trying to make jokes... :tongue2:

You were first to use anything but metric for Length – need your glasses to read the posts?

I don't know if you need my glasses.I have mixed astigmatism at both eyes...I gave "inch" as an example of how ridiculous does your unit system look to me.And i gave my reasoning a historical grounding.Besides,u started with "lbs.".

You were first to use Grams for Weight – noticed you didn’t try to answer – maybe because using your units for weight it cannot be done.

Show me that post and i'll delete it by eating it... :tongue2: What cannot be done using my units (i hope you mean:Kilogram and meter) ?

But if you do find the time how about using the Decimal version of time the French tried to make metric.

Good joke,but that missing comma makes it imperfect...

Daniel.

 

[vikasj007]

well, off course, the clock in ecuador, will weigh more due to centripetal acceleration, but i am not in a mood to calculate all that.

but as far as the time difference is concerned, i think that the concept of time dilation is applied only in case the velocity is of the order of 'c' which is not the case here, hence SR cannot be applied here, neither can GR be applied as both the places have no relative acceleration.

 

[dextercioby]

well, off course, the clock in ecuador, will weigh more due to centripetal acceleration, but i am not in a mood to calculate all that.

I'm afraid it's the other way around.It will weigh less,due to the centrifugal effect created by the Eatrth's rotation.It'a acceleration due to Earth's rotation is $a_{cfi}=\omega^{2} R_{equator}$,sot that the weight force will be
$$\vec{G}=m(g_{equator}-\omega^{2} R_{equator})\vec{j}$$

but as far as the time difference is concerned, i think that the concept of time dilation is applied only in case the velocity is of the order of 'c' which is not the case here, hence SR cannot be applied here, neither can GR be applied as both the places have no relative acceleration.

Both theories can be applied.GR can be applied due to the fact that Earth creates different gravitational fields at the poles and at the equator.

Daniel.