Nonspherical earth (split from centrifugal forces)

[MigL]

Sorry Strange, I thought I made myself clearer, but I guess not.

Things farther away actually weigh less because weight is a force due to gravity and changes with the inverse of the radius squared. In effect, the equatorial surface is feeling less gravitational force

Mass does not change, and the two are not the same.

Edited April 14, 2015 by MigL

[Strange]

Sorry Strange, I thought I made myself clearer, but I guess not.

Things farther away actually weigh less because weight is a force due to gravity and changes with the inverse of the radius squared. In effect, the equatorial surface is feeling less gravitational force

 

Exactly. So I don't understand why you think that is a non-answer. It seems a really good answer (to me).

[MigL]

I really don't think this is that important an issue to warrant splitting and continued discussion, but I'll explain my reasoning...

 

The distribution of mass/matter, due to gravity, for a non spinning object, tends to a perfect sphere, as gravity is radially symmetric.

The force felt by the mass/matter is the only arbiter of the shape, so if we modify the force by imparting an axial spin, the equatorial region will feel a modified ( lessened )force. This is because, at the equator, the inertia of the mass/matter resists the gravitational or centripetal force to a greater extent than at the poles where angular momentum is trivial.

 

We can measure gravitational or modified gravitational ( due to spin, not MOND ) force directly by weighing things. What we find is that things weigh less at the equator because they feel less net centripetal force compared to the poles. Since they feel less net force there is a deviation from the spherical distribution of the mass/matter such that the radius at the poles is smaller than the equatorial radius.

So, using weight as a measure of force, the radius of the Earth is larger than the polar radius, because mass/matter, or anything else, weighs less in that radial direction.

 

So having established that the surface of the Earth is farther away from the center because the constituent mass/matter weighs less. it makes no sense to say ( as DrP did ) that things weigh less because they are farther away from the center.

While not technically wrong, it is circular reasoning at best.

[Strange]

But that's the point: it isn't circular. There are two different effects: the spin, which causes the shape of the Earth and, separate from that, the lower gravity at the equator because it is further from the centre. So, for example, things would still weigh less at the equator, even if the Earth weren't spinning.

[MigL]

No. If it wasn't spinning the earth would be spherical and there would be no difference in radius between equator and poles.

It is the spin, which alters the force ( also called weight ), which then alters the shape.

Saying that the shape then alters the force, is redundant..

Edited April 14, 2015 by MigL

[Mike Smith Cosmos]

I really don't think this is that important an issue to warrant splitting and continued discussion, but I'll explain my reasoning...

 

The distribution of mass/matter, due to gravity, for a non spinning object, tends to a perfect sphere, as gravity is radially symmetric.

The force felt by the mass/matter is the only arbiter of the shape, so if we modify the force by imparting an axial spin, the equatorial region will feel a modified ( lessened )force. This is because, at the equator, the inertia of the mass/matter resists the gravitational or centripetal force to a greater extent than at the poles where angular momentum is trivial.

 

We can measure gravitational or modified gravitational ( due to spin, not MOND ) force directly by weighing things. What we find is that things weigh less at the equator because they feel less net centripetal force compared to the poles. Since they feel less net force there is a deviation from the spherical distribution of the mass/matter such that the radius at the poles is smaller than the equatorial radius.

So, using weight as a measure of force, the radius of the Earth is larger than the polar radius, because mass/matter, or anything else, weighs less in that radial direction.

 

 

 

What you appear to be saying here about " We can measure gravitational or modified gravitational ( due to spin, not MOND ) force directly by weighing things. What we find is that things weigh less at the equator because they feel less net centripetal force compared to the poles. Since they feel less net force there is a deviation from the spherical distribution of the mass/matter ..."

 

Surely this is what I have been bleeding on about " feel less net centripetal force , means centrifugal due to spin has reduced the centripetal by subtraction . IE net centripetal = gross centripetal - centrifugal due to spin of earth . Is this not what you are saying ?

 

Mike

[Strange]

No. If it wasn't spinning the earth would be spherical and there would be no difference in radius between equator and poles.

It is the spin, which alters the force ( also called weight ), which then alters the shape.

Saying that the shape then alters the force, is redundant..

 

Having the shape it has now, it would not rebound to a spherical shape if it were not spinning (I suppose it might, over a very, very long time).

 

I think it is useful / interesting to separate the effects of "centrifugal force" and distance from the centre.

[MigL]

I made no mention of 'centrifugal' force Mike, as that would only be 'apparent' from the Earth's surface accelerated ( spinning ) frame.

An inertial observer at rest, high above one of the poles, would note the angular momentum at the rotating equator, and attribute the centripetal ( gravitational ) force modification to inertial effects.

 

And as Earth is mostly in a liquid form, surrounded by solid plates, I think it would take a spherical form very quickly, Strange; although with world-wide, extreme earthquakes. Again I make no mention of centrifugal or other frame dependent forces.

[DrP]

No. If it wasn't spinning the earth would be spherical and there would be no difference in radius between equator and poles.

It is the spin, which alters the force ( also called weight ), which then alters the shape.

Saying that the shape then alters the force, is redundant..

 

If you stopped it spinning, assuming it doesn't 'pop' back into a sphere straight away, you would weigh less at the equator than at the poles due to your distance from the centre of mass/gravity of the earth, no? Therefore it is not redundant at all. Unless I am completely misunderstanding what you are saying again. ;-)

Which Strange has already said above... sorry.

We are going round in circles.. ;-) lol.

Edited April 14, 2015 by DrP

[Robittybob1]

 

If you stopped it spinning, assuming it doesn't 'pop' back into a sphere straight away, you would weigh less at the equator than at the poles due to your distance from the centre of mass/gravity of the earth, no? Therefore it is not redundant at all. Unless I am completely misunderstanding what you are saying again. ;-)

Which Strange has already said above... sorry.

I was attempting to show why I think you are both wrong but the idea got split off from the thread.

I am tempted to think if the Earth is an oblate spheroid you would be wise to consider where and what the additional dimension is composed of.

Edited April 14, 2015 by Robittybob1

[DrP]

I was attempting to show why I think you are both wrong but the idea got split off from the thread.

I am tempted to think if the Earth is an oblate spheroid you would be wise to consider where and what the extra dimension is composed of.

Actually, I think we are both right. lol One about the centripetal force and the other about the the force being related to distance to centre of mass. ;-)

[pavelcherepan]

And as Earth is mostly in a liquid form, surrounded by solid plates, I think it would take a spherical form very quickly, Strange; although with world-wide, extreme earthquakes. Again I make no mention of centrifugal or other frame dependent forces.

Where did you get that from? The crust is solid, mantle is solid, inner core is solid too. The only liquid bits are the outer core, bits and pieces within astenosphere and some minor amount of melt within the crust. That's far off from being mostly liquid.

 

On the other hand the mantle is known to exhibit solid flow within timeframes of tens and hundreds of millions of years, so my guess would be that should the rotation stop it should take a spherical shape but after a relatively long time.

[Robittybob1]

Where did you get that from? The crust is solid, mantle is solid, inner core is solid too. The only liquid bits are the outer core, bits and pieces within astenosphere and some minor amount of melt within the crust. That's far off from being mostly liquid.

 

On the other hand the mantle is known to exhibit solid flow within timeframes of tens and hundreds of millions of years, so my guess would be that should the rotation stop it should take a spherical shape but after a relatively long time.

The Earth flexes under the influence of the tidal forces of the Sun and the Moon. If it does this twice a day and don't see why the oblateness should be much of a problem to get rid of. As the Earth's slows its rotation the oblateness would disappear.

[pavelcherepan]

Not straight away. It will take time. A long time. Solid rock flows slowly.

[MigL]

If I remember correctly Pavel, the Earths two largest layers are the low viscosity outer core, and the high viscosity mantle, accounting for approx. 5000 km of the radius almost evenly split. Now a high viscosity liquid like toffee, may not seem to flow very fast, but you'd be surprised how fast it flows when it has billions of pounds pressing on it.

Do you call lava from a volcano a solid or a liquid; it certainly seems to flow very readily.

 

And no, you can't consider what would happen if you suddenly stopped the Earth as then you'd have to ignore angular momentum, which is the cause of the effect in the first place.

Look, I'm not saying DrP is wrong, what he said is certainly valid. But as an explanation he has put 'the cart before the horse'.

[Robittybob1]

Actually, I think we are both right. lol One about the centripetal force and the other about the the force being related to distance to centre of mass. ;-)

I'm not saying the reasons are wrong in themselves for they are not but the reason you get this lowered gravity at the equator, instead of what should happen. Could this be due to the lower density material that is separating you from the main source of the Earth's gravity? That is what I'm thinking at least, but I haven't worked it out as yet.

Edited April 15, 2015 by Robittybob1

[pavelcherepan]

If you want to talk viscosity, then consider this. Astenosphere, the less viscous part of either crust or mantle has an average viscosity of around 10^24 pa*s (can't remember the exact number of the top of my head) which is like 8 orders of magnitude more than solid steel. The rest of the mantle is in 10^27 pa*s region so whatever the pressure it won't flow very fast.

Edited April 15, 2015 by pavelcherepan

[Robittybob1]

If you want to talk viscosity, then consider this. Astenosphere, the less viscous part of rather crust or mantle has an average viscosity of around 10^24 pa*s (can't remember the exact number of the top of my head) which is like 8 orders of magnitude more than solid steel. The rest of the mantle is in 10^27 pa*s region so whatever the pressure it won't flow very fast.

What is viscosity measured in? http://www.hydramotion.com/pdf/Website_Viscosity_Units_V2.pdf

 

 

Pascal-second (symbol: Pa·s)

This is the SI unit of viscosity,

equivalent to newton-second per square

metre (N·s m–2). It is sometimes referred

to as the “poiseuille” (symbol Pl).

One poise is exactly 0.1 Pa·s. One

poiseuille is 10 poise or 1000 cP, while

1 cP = 1 mPa·s (one millipascal-second).

[pavelcherepan]

Yeah, that's what I wrote Pascal-seconds.

[Robittybob1]

Yeah, that's what I wrote Pascal-seconds.

I have never worked with those units before.

[pavelcherepan]

If you want to talk viscosity, then consider this. Astenosphere, the less viscous part of either crust or mantle has an average viscosity of around 10^24 pa*s (can't remember the exact number of the top of my head) which is like 8 orders of magnitude more than solid steel. The rest of the mantle is in 10^27 pa*s region so whatever the pressure it won't flow very fast.

 

Correction to my earlier post (I was on the plane so couldn't look the numbers up properly). Viscosity of astenosphere is actually 10¹⁹ Pa*s and the rest of the mantle is between 1-2*10²¹ Pa*s. Compare that with calculated viscosity for steel (10³-10⁵ Pa*s) and aluminium (3-4*10⁴ Pa*s). Although the numbers i originally posted were wrong the difference now is even greater - over 14 orders of magnitude.

 

Do you call lava from a volcano a solid or a liquid; it certainly seems to flow very readily.

 

Lavas are viscous liquids of course, the paper below gives viscosity numbers for Hawaiian lavas which are generally around 4-5*10⁴ Pa*s. Most other lavas would be less than that.

 

http://www.higp.hawaii.edu/~scott/Nichols_articles/Nichols_lava_viscosity.pdf

[swansont]

Not straight away. It will take time. A long time. Solid rock flows slowly.

You're both right. The solid earth has tides, so it flexes every day, but north america is still rebounding from the effect of having been covered in glaciers hower many thousands of years ago.

[studiot]

Shape of Earth, nonspherical or what?

 

Well I suppose all this ballyhoo really depends on what you take the shape of.

 

This the prime decision to make before debating the shape.

The second one is what do you mean by centre?

 

Geologically speaking the most abundant rock at the surface is water, mostly liquid.

The surface of this rock is decidedly non spherical.

 

So perhaps you mean the geoid?

 

The average density of the non hydrous rock at the surface is just under 3 times that of water.

But we can deduce from astronomic studies that the average density of the equivalent globe is around 5.5 times the density of water.

 

So we can say for certain that the variation of gravity with radius deviates significanly from the inverse square law since the density is not constant.

Edited April 15, 2015 by studiot

[MigL]

I have a feeling this thread is going to be split off again. Now wiith regards to Earth's composition and fluidity.

I'm no expert in this field, I'm just going by memory of things I've read or been told.

But back to the original point...

 

You cannot just eliminate Earth's angular momentum by snapping your fingers. If you could dissipate it and 'brake' the Earth, it would take time, and in the end, you'd find that the Earth would be close, if not spherical.

Just like, as it was coming together 4.5 Bi yrs ago, it was probably extremely flattened ( accretion disc ), and slowly rotating. It then speeded up ( conservation of ang. momentum ) and reached an equilibrium of forces in the somewhat flattened spherical shape it has today. The very fact that the surface of the Earth sits appreciably higher in the gravity well, at the equator, is because it feels less net force. In effect, it weighs less. Even rock feels the effects of buoyancy, given several billion years to react, and so the surface has 'floated' up higher at the equator than the poles.

 

I stand by my original assertion...

The earth's shape ( flattened sphere ) is due to the fact that things weigh less at the equator.

And while not wrong, saying that things weigh less at the equator because of Earth's shape, is redundant.

 

We could argue about it till the cows come home, but its probably just semantics anyway.

[studiot]

 

MigL

The earth's shape ( flattened sphere ) is due to the fact that things weigh less at the equator.

 

 

I didn't realise that was your original contention.

That bit is most definitely wrong.