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Cosmologist
2012-Sep-21, 01:16 PM
Even a few grams?

korjik
2012-Sep-21, 01:20 PM
Heavier. By about 9% times the cosine of your latitude if I remember right.

grapes
2012-Sep-21, 01:22 PM
We would feel heavier, the bulge of the earth would be reduced, putting us closer to the center, and we'd lose the lift of centrifictional force. Although the folks at the poles would be slightly lighter, since they would be a little bit farther from the center of the earth--they're already a bit heavier than folk at the equator, and the mid-latitudes.



Heavier. By about 9% times the cosine of your latitude if I remember right.9% ?? impossible :)

It couldn't be any much more heavier than the folks at the poles are now.

Cosmologist
2012-Sep-21, 01:48 PM
Interesting. Ofcourse we'd all be too busy with the ensuing earthquakes to notice.

Swift
2012-Sep-21, 01:50 PM
Interesting. Ofcourse we'd all be too busy with the ensuing earthquakes to notice.
Not to mention that unless you stopped the spin magically, the amount of energy that would be needed would liquify the surface of the Earth. ;)

DonM435
2012-Sep-21, 04:15 PM
If it stopped suddenly, I think that you'd fall down and roll into whichever ocean happened to be nearer.

Ara Pacis
2012-Sep-21, 08:09 PM
It would take time for the post-rotational rebound to kick in, due to lithospheric viscosity, so the atmosphere and hydrosphere would flow downhill from the equator towards the poles, but I'm not sure how much it would affect the geoid since the bulge would still have some extra mass for gravity.

korjik
2012-Sep-22, 05:15 AM
We would feel heavier, the bulge of the earth would be reduced, putting us closer to the center, and we'd lose the lift of centrifictional force. Although the folks at the poles would be slightly lighter, since they would be a little bit farther from the center of the earth--they're already a bit heavier than folk at the equator, and the mid-latitudes.


9% ?? impossible :)

It couldn't be any much more heavier than the folks at the poles are now.

Like I said, if I remember right. Its been about 15 years since I did the calculation, and I seem to remember it was about .9 m/s2 centripetal acceleration at the equator.

Besides, he said 'feel heavier'. If I am right, people at the equator feel a bit lighter than they would at the pole.

Like I said tho, only if I remember right. The industrious student would confirm my calculations

:)

Ara Pacis
2012-Sep-22, 06:29 AM
Like I said, if I remember right. Its been about 15 years since I did the calculation, and I seem to remember it was about .9 m/s2 centripetal acceleration at the equator.

Besides, he said 'feel heavier'. If I am right, people at the equator feel a bit lighter than they would at the pole.

Like I said tho, only if I remember right. The industrious student would confirm my calculations

:)

Wait, which one of you is the student and which one is the professor?

korjik
2012-Sep-22, 01:39 PM
Wait, which one of you is the student and which one is the professor?

I said 'industrious student' first, so I must be the prof

:)

WaxRubiks
2012-Sep-22, 01:57 PM
I think one would weigh about 0.3% more.


edit: I mean more....:doh:

slang
2012-Sep-22, 03:57 PM
If it stopped suddenly, I think that you'd fall down and roll into whichever ocean happened to be nearer.

Unless the one behind you comes rolling over and falling down on you first... :)

cjameshuff
2012-Sep-23, 03:40 PM
If it stopped suddenly, I think that you'd fall down and roll into whichever ocean happened to be nearer.

If everything sitting on the planet (like people and air) wasn't stopped along with it, the ground speed of the wind and those other objects within about 40 degrees of the equator would in fact be supersonic. "Roll" is probably the wrong word, the actual effect would be something much messier. And louder.

DonM435
2012-Sep-23, 03:59 PM
... "Roll" is probably the wrong word, ... .

I.e., there'd be no one around to debate the choice of words. ;)

SkepticJ
2012-Sep-23, 07:32 PM
The upside is no force in the universe could just stop a planet's rotation in zero time.

ShinAce
2012-Sep-23, 08:15 PM
I think one would weigh about 0.3% more.


edit: I mean more....:doh:

Confirmed! I got 0.34%. That's the max(which means for those on the equator).

Hornblower
2012-Sep-23, 11:57 PM
Confirmed! I got 0.34%. That's the max(which means for those on the equator).

I don't think most of us would notice such a small difference. A standard rate turn in an airliner generates more centrifugal effect than that, and when the pilot did it smoothly I never noticed a thing unless I was looking out the window and saw the wingtip drop below the horizon.

novaderrik
2012-Sep-24, 02:07 AM
The upside is no force in the universe could just stop a planet's rotation in zero time.

not even Q?

grapes
2012-Sep-24, 03:27 AM
Confirmed! I got 0.34%. That's the max(which means for those on the equator).
The garden gnome got 0.6% for equator to pole, did you guys just calculate centrifictional force, or also the reduced radius once the bulge subsides?

http://news.sciencemag.org/sciencenow/2012/03/scienceshot-garden-gnome-tests-e.html

WaxRubiks
2012-Sep-24, 08:14 AM
it would be less than that as the poles would bulge out a bit if the Earth had stopped spinning, wouldn't it?

Yeah, I just calculated centrifugal force.:einstein:

grapes
2012-Sep-24, 12:21 PM
it would be less than that as the poles would bulge out a bit if the Earth had stopped spinning, wouldn't it?
Yes, that would be an upper limit

ShinAce
2012-Sep-24, 01:45 PM
The garden gnome got 0.6% for equator to pole, did you guys just calculate centrifictional force, or also the reduced radius once the bulge subsides?

http://news.sciencemag.org/sciencenow/2012/03/scienceshot-garden-gnome-tests-e.html

Using acc = velocity2/radius

Since the earth's radius varies by less than 1 percent from pole to equator, why bother adjusting for it? It wouldn't even show up for 1 significant digit. As for 0.34%, it might be 0.35 or 0.33 after adjustement, but unlikely. We're still nowhere near 0.6%.

I'm not about to adjust for the gravitational field at different radii. The OP is suspended, and the answer seems to be on the order of 0.5%. So in other words, a difference of 1 pound for a 200 pound person. Measurable, but small.

grapes
2012-Sep-24, 03:19 PM
Using acc = velocity2/radius

Since the earth's radius varies by less than 1 percent from pole to equator, why bother adjusting for it? It wouldn't even show up for 1 significant digit. As for 0.34%, it might be 0.35 or 0.33 after adjustement, but unlikely. We're still nowhere near 0.6%.

I'm not about to adjust for the gravitational field at different radii. The OP is suspended, and the answer seems to be on the order of 0.5%. So in other words, a difference of 1 pound for a 200 pound person. Measurable, but small.A 0.6% increase is the difference between 1.000 and 1.006, four significant digits. :)

Still, an oblate spheroid that returns to a spherical shape, gains twice as much radius through the poles, more or less, than it loses at the equator. Using 6,356.7523 km, 6,371.009 km, 6,378.1370 km ( http://en.wikipedia.org/wiki/Earth_radius ), which are each the polar, mean, and equatorial earth radius, the polar gavitational acceleration is less than the mean by 0.45% ( (1/6,356.7523 km)^2/(1/6,371.009 km)^2 - 1 ), and the equatorial is greater by 0.22% ( (1/6,371.009 km)^2/(1/6,378.1370 km)^2 - 1 ).

So, just by dint of going from the equator to the pole, we experience 0.67% ( (1/6,356.7523 km)^2/(1/6,378.1370 km)^2 - 1 ) increase in gravity, irrespective of the centrifictional force

Weird, is that garden gnome lying? :)

ETA: No, not lying. This wiki subsection on latitude dependence of gravity says the difference is about 0.5%
http://en.wikipedia.org/wiki/Gravity_of_Earth#Latitude