The Atmosphere, wind

So if the Earth was round, wouldn't the centrifugal force of its immense spinning create tremendous winds which would make it totally uninhabitable? After all, the air isn't tied down by anything except "gravity" and inertia, and centrifugal force can overcome both of these.

so you're saying centrifugal force is stronger than the gravitational pull of the earth?
oh and if you up a few thousand feet in a plane or something you'll find those "tremendous" winds you mentioned.

so you're saying centrifugal force is stronger than the gravitational pull of the earth?
oh and if you up a few thousand feet in a plane or something you'll find those "tremendous" winds you mentioned.

They aren't consistent at all.

Well, let's figure out the centrifugal contribution to acceleration: it's ω × (r × ω).  Here ω = qz is the Earth's angular velocity where q = π/43200 rad/sec, r is the position vector of an object on the surface of the Earth, and z is the unit vector pointing towards the north star.

Some quick calculations reveal that the acceleration vector is q²r which points directly away from the Earth's surface and has a magnitude of 0.034 m/s² (taking the radius of the Earth to be 6400 km).  This is obviously dwarfed by the much greater gravitational acceleration of 9.8 m/s².

In other words, Earth's gravity is about 289 times stronger than its centrifugal force.  Just in case you're curious, if the Earth were spinning about 17 times per day (once every 5078 seconds), the centrifugal force would be 17² = 289 times stronger than it is now and would just about equal the gravitation force.

Okay, that takes care of the atmosphere getting thrown off by a spinning Earth.

As for the high winds issue, that might be a problem if the Earth suddenly started spinning once per day from a state of non-rotation.  However, air is viscous; eventually, friction would cause the air to be rotating with the Earth.  It's just like when you spin a bucket of water: initially the water stays where it is (i.e. fast currents from the perspective of an insect on the inner surface of the bucket) but eventually the water picks up some of the rotational energy and spins with the bucket.   Fortunately, according to RE scientists, this has never happened because the Earth and its atmosphere were rotating with the same angular velocity even as it was forming (in fact this is required by conservation of momentum).

so you're saying centrifugal force is stronger than the gravitational pull of the earth?
oh and if you up a few thousand feet in a plane or something you'll find those "tremendous" winds you mentioned.

So how is aviation even possible? It happens, I've been on an aeroplane.

Also, why aren't we buffeted by insane winds when we stand on top of a hill or mountain?

oh and if you up a few thousand feet in a plane or something you'll find those "tremendous" winds you mentioned.

If you're thinking of the Jet Stream, it's moving in the wrong direction to be an example of those tremendous winds (it moves west to east, and the Earth spins counterclockwise or west to east).

maths

What is your source for the Earth's rotation speed? or, for that matter, its centrifugal force?

If you're thinking of the Jet Stream, it's moving in the wrong direction to be an example of those tremendous winds (it moves west to east, and the Earth spins counterclockwise or west to east).

Isn't it true though that winds would pick up a lot the further the air itself was from the Earth, since the Earth's gravity would be weaker? Why don't they?

What is your source for the Earth's rotation speed? or, for that matter, its centrifugal force?

Well, you were attacking a feature of the RE description of the world, so I just used the RE description of the world, which has the Earth rotating one full circumference per day or 2π radians every 86400 seconds.

Centrifugal acceleration can be derived (and is done so in this link) by transforming the time-derivative operator in a rotating reference frame into an inertial reference frame and plugging in subsequent derivates of the position vector (r in both my and Wikipedia's example).  In other words, you can compute the centrifugal acceleration once you know the rotational speed and the distance from the axis of rotation.

By the way, I should point out that I assumed that the centrifugal acceleration is being considered at the equator.  It gets weaker as you get closer to the poles, because you're getting closer to the axis of rotation.  Basically you'd multiply the value I got by the cosine of your latitude.

Isn't it true though that winds would pick up a lot the further the air itself was from the Earth, since the Earth's gravity would be weaker? Why don't they?

Again, the atmosphere is, on average, rotating with the same angular velocity as the Earth.  If it weren't, then after a period of time it would be (bucket of water analogy -- just try this experiment; it's fun).  In other words, if you planted a fifty-mile-long pole vertically in the ground, then not taking into account spurious high-altitude air currents the air would be more or less stationary at all points along the pole.

What would and does happen is that the air would be more likely to escape into space as you get farther from the Earth, at least in part because gravity is weaker and the centrifugal force is stronger.  I haven't done a calculation to determine the altitude at which gravity and the centrifugal force are equal but it might be interesting.

So are we done here?

There aren't any tremendous winds because Earth's atmosphere isn't glued to the Earth. The lower atmosphere just rotates at roughly the same speed as the Earth due to friction.

There aren't any tremendous winds because Earth's atmosphere isn't glued to the Earth.

I don't think this implies what you think it implies.  If there atmosphere was glued to the Earth, there still wouldn't be tremendous winds, because we're glued to the Earth too (effectively).

I've always been curious about what role the centrifugal force contributed by the earth's orbit around the sun would contribute to the measurement of weight on earth. Intuitively it seems that someone would weigh less at night, and more during the day, but I've never been determined enough to go through with all the calculations.

... but I've never been determined enough to go through with all the calculations.

Is that a request? :)

Could be.

Well now that I've done it once, it's pretty simple.  Centrifugal acceleration is just q²r, where q is rotational speed and r is distance from the axis of rotation.

If you assume the centre of the Earth makes it around the sun once every 365.256 days, then its rotational speed is q = 2π/3.1558e7 rad/sec = 1.991e-7 rad/sec.

1 AU is 149,598,000,000 m (google) and the Earth's radius is rEarth = 6,400,000 m.  We get two interesting distances-from-axis-of-rotation: rNight = 1 AU + rEarth = 149,604,400,000 m and rDay = 1 AU - rEarth = 149,591,600,000 m.

So the centrigugal force during the day is fDay = q²rDay = 0.0059299 m/s², and fNight = q²rNight = 0.0059304 m/s².

Thus total acceleration toward the center of the Earth is aDay = g + fDay = 9.8125798641 m/s² and aNight = g - fNight = 9.8007196285 m/s².

Therefore you are 1.0012101393 times as heavy -- or about 0.121% heavier -- during the day than you are during the night, due to the Earth's motion around the sun.

If we also take into account centrifugal acceleration due to the Earth's rotation, calculated earlier to be 0.034 m/s², then we get aDay' = aDay - 0.034 = 9.7786 m/s² and aNight' = aNight - 0.034 = 9.7667.  The ratio comes out to be about the same, within four decimal places.

Probably you could test this if you found a heavy enough object whose weight wouldn't change between day and night (i.e. not an animal)...

I've always been curious about what role the centrifugal force contributed by the earth's orbit around the sun would contribute to the measurement of weight on earth. Intuitively it seems that someone would weigh less at night, and more during the day, but I've never been determined enough to go through with all the calculations.

Interesting idea, similar to a story by Larry Niven (Nuetron Star).
He has a ship go around a Nuetron Star, and the difference in the gravitation pull due to the differing distances from the gravitational Center, and the ship whipping around the orbit with it's nose pointing "Down".

Quote from: "Unimportant"

... but I've never been determined enough to go through with all the calculations.

Is that a request?

Well if I just came out and asked it would make me look lazy. This way, you just look friendly and helpful... everyone wins!

Therefore you are 1.0012101393 times as heavy -- or about 0.121% heavier -- during the day than you are during the night, due to the Earth's motion around the sun.

Probably you could test this if you found a heavy enough object whose weight wouldn't change between day and night (i.e. not an animal)...

That's pretty interesting, and a more significant difference than I was expecting.

Even with something not especially huge - like, say, 100 lbs of weightlifting plates - you're looking at a difference of a couple ounces, which is certainly a significant enough weight to be measured. Now if only someone had access to an accurate enough and reliable scale to perform an experiment for us...

(See I'm doing it again.)

Lazy ass. :x

Well, I could test it, but Erasmus would have to figure out what the precise times are in my area so that I could get the most weight difference possible. If you want to take the time and help in this experiment Erasmus, I am located at (or at least will be located at tomorrow) 47*29'15.02" N; 117*34'29.03". Cheney, Washington, in case you were wondering. If you calculate the times for those coordinates where the greates change in weight could be recorded (so as to make it easier to see the difference and get rid of any nasty little imperfections in the scale or my reading of it) then I will gladly take the time to weigh an object at those specific times to see how much of a difference it really amounts to.

The times don't need to be exact; 12 noon and 12 midnight would work fine.

Let us know what you're measuring, the type of scale you're using, and what the results are.

There's a little bit of an issue that I made some assumptions (circular orbit, constant orbital speed) and used some insufficiently precise values (6,400 km for the Earth's radius, 0.034 for the centrifugal acceleration due to Earth's rotation).  Now that I think about it, the Earth doesn't even rotate once in exactly 24 hours so that's inaccurate too.  I think in order to get a prediction with enough significant digits I'll need to do most of it over again... except for the derivation of the formula of course.

Good luck.

There's a little bit of an issue that I made some assumptions (circular orbit, constant orbital speed) and used some insufficiently precise values (6,400 km for the Earth's radius, 0.034 for the centrifugal acceleration due to Earth's rotation).  Now that I think about it, the Earth doesn't even rotate once in exactly 24 hours so that's inaccurate too.  I think in order to get a prediction with enough significant digits I'll need to do most of it over again... except for the derivation of the formula of course.

Well I think the idea is that, on a flat earth, and assuming constant acceleration, the two values would be precisely identical.

On a round earth, no matter how "rough" your calculations may be, there would be some not-insignificant difference.

If you really want to go all out on the calculations, though, don't forget to take into account seasons and axis tilt / the longitude of the area the measurements are taking place, since they'll have a substantial effect on what "radius of earth" you use for your equations. Canada won't have as big a radius as Mexico, presumably.

The problem is what kind of scale will you use that is not also affected?
No kind of balance, spring, or hydrolic scale will work.

Yeah, that's why I asked for the type of scale.

Balance scales are obviously out, but why wouldn't a spring scale, like a force meter, work? The spring constant isn't going to change, so I think the measurements will be meaningful.

Well I think the idea is that, on a flat earth, and assuming constant acceleration, the two values would be precisely identical.

On a round earth, no matter how "rough" your calculations may be, there would be some not-insignificant difference.[/quote]

Definitely, definitely, but this experiment has the potential not only to refute FE theory, but also confirm RE theory.  The latter goal, however, requires a more precise prediction from RE.

If you really want to go all out on the calculations, though, don't forget to take into account seasons and axis tilt

That was the plan.

the longitude of the area the measurements are taking place, since they'll have a substantial effect on what "radius of earth" you use for your equations. Canada won't have as big a radius as Mexico, presumably.

I'm planning to ignore longitude, since the measurements will be done at noon and midnight, and take latitude only into account.  It's easy to factor latitude in -- it should be easy to do, and it can just be offset to take axial tilt into account.  Additionally in a few days we'll be at the autumnal equinox, so the offset will be zero.  I'm assuming the Earth is a perfect sphere and that the measurements will be done at the average distance from the centre both times.  I guess one longitude issue is that local midnight/noon are not exactly what's printed on nearby clocks, so I should figure that out too.

Yeah, that's why I asked for the type of scale.

Balance scales are obviously out, but why wouldn't a spring scale, like a force meter, work? The spring constant isn't going to change, so I think the measurements will be meaningful.

Because a spring scale's measurements are affected by the weight of the components of the scale.  

Picture a fish scale.
When you weigh the fish, you are also weighing the spring, the connection from the spring to the hook, and the hook.

I suppose the electric chemistry scales we used in college would work, since you reset balance to zero every time you used it.  I'm not entirely sure what principle was used for them so I just don't know.

Because a spring scale's measurements are affected by the weight of the components of the scale.  

Picture a fish scale.
When you weigh the fish, you are also weighing the spring, the connection from the spring to the hook, and the hook.

Interesting puzzle.  I think things like the weight of the spring don't matter in our case because our experiment involves measuring the difference between two weighings, just to see if there is one.  Of course if we want to use it to test the RE prediction, the spring's weight need to be taken into account.