Gravity of the Earth decreases if it stops spinning?

[BillyT]

Ok, it sounded like you meant that the 40km bulge would still be there. How elastic is rock, if the 40km equatorial bulge can disappear, without increasing the distance between the poles?

It is 20Km bulge in the radius.* Yes part of it would still be there for centuries. Only example I know about is the Norwegian Fiords. Their walls are still slightly rising from a below sea level height and have been since the end of the last ice age. I. e. it takes a long time for elastic rocks to find their new equlibrium shape when a major weight stress has been removed.

Also note that the movement of rocks at the equator would almost entirely vertical so the the integrated mass from Earth's center to surface would not change. It like the fiords would be moving up due to water mass going to the poles and down due to absence of centrifugal force.
Note the centralfugal force is very very small. Same as at edge of a merrry-go-round with 4000 miles radius very slowly turning (360 degrees in 24 hours)! Much too small for a human to even feel; but a diver in ocean could feel 10 meters of water above him removed quickly or the pressure increase by going only 5 meters deeper - harder to expand his lungs to breath.

* I'm accepting your 40Km diameter change number, but think it is high without searching for the facts. Where did you get that value?

@mfb: can you demonstrate that the contraction at equator with removal of centrfugal force is large compared to the expansion due to a large mass of water going towards the poles (less compressive force). Or, like my opposite belief, that is just your belief?

BTW, with no centrifugal force, I agree the surface of the oceans would have the same same gravitational potential everywhere, but with a spherically inhomagenous interior, like Earth has, that is not a spherical surface.

 

[A.T.]

that is not a spherical surface.

Threre are no perfect spheres in nature, but it would be more spherical than now.

 

[A.T.]

'More distance means less gravitational pull' can't imho be universally true though, but only within certain bounds. Because if you go to the extreme, then you can make an ellipsoid that thin that it approaches a thin disc. Now if you are standing on a pole of it, it means that you are really close to the center of mass, but have almost no mass below you. The way biggest part of the mass will be almost around you, so the gravity vectors going almost horizontal, cancelling each other mostly out at your position, only the small vertical component being left over and summing up. That way you can make the gravitational force arbitrarily small,,while still standing closer and closer to the center.

That's at least what I assume without doing the math. In the end you'd have to do the math and integrate and then maybe determine the ellipsoid shape with maximum gravity on a pole or something.

This is a good point. And also without doing the math, it seems that the ellipsoid shape with maximum gravity on both poles would have similar proportions to the drop shape that maximizes gravity on one of the poles, shown in post #35.

 

[BillyT]

Threre are no perfect spheres in nature, ...

A small drop of water inside an air current shielding glass container in orbit about the Earth, is a sphere.

 

[A.T.]

A small drop of water inside an air current shielding glass container in orbit about the Earth, is a sphere.

It isn't a perfect sphere.

 

[BillyT]

Another point I need to make:The Earth's bulge is NOT due to the current very very weak centrifugal force. The entire Earth was once moltent and spinning much more rapidly before the moon existed, carring much of the original angular monementum away from the earth. The Earth's bulge when its surface solified, was larger than now, and is still adjusting (becoming less). I. e. it is NOT yet back to an equlibrium bulge. Many 100s of thousands of year from now, we still will have more bulge than the current very very weak centrifugal force would make.

Again I note: The current centrifugal force is the same as an 8000+ mile diameter merry-go-round turning very very slowly* would have at its edge.

* One turn takes 24 hours!

 

[BillyT]

It isn't a perfect sphere.

If the container is spherical and with unform wall thickness then at its center, or anywhwere inside, there is no "micro-gravity" from the container mass. Make container orbit / drift in space at least a 1000AU from the sun* and as it has opaque copper walls, with small amount of decaying isotope in the copper, so even in deep space the temperature inside is more than 0C and with no gradients; Then there is no sunlight pressure on one side of the drop and no convection air currents to distort it.

* The container is in "free fall" with path very slightly influenced by sun and other cosmic masses but that path removes all gravity effects from remote objects. Also the inside walls are "hydrophobic," so even if the small water drop should briefly touch a wall, it is repelled and becomes spherical immediately again.

Why would the drop not be a perfect sphere?

 

[A.T.]

The Earth's bulge when its surface solified, was larger than now, and is still adjusting (becoming less). I. e. it is NOT yet back to an equlibrium bulge. Many 100s of thousands of year from now, we still will have more bulge than the current very very weak centrifugal force would make.

What does "more" mean? How many km more? According to

https://en.wikipedia.org/wiki/Geoid

the equipotential surface (based on current centrifugal potential) deviates only −106 to +85 m from the reference Ellipsoid with a 21km radius difference (current Earth shape approximation).

 

[A.T.]

Why would the drop not be a perfect sphere?

Because it’s made of molecules.

 

[BillyT]

@mfb: I am still waiting for you to demonstrate that the contraction at equator with removal of centrfugal force is large compared to the expansion due to a large mass of water going towards the poles (less compressive force). Or, like my opposite belief, just admit that is ONLY your belief too?

 

[BillyT]

Because it’s made of molecules.

ok. I admit it is only on the stastical average spherical, but at any time will have a surface roughness of a few atomic diameters.

However, if you have chosen the total number of molecules correctly and the drop is very small and at 1C, then the surface roughness can be less about the diameter of an oxygen atom.

As H2O is a polar molecule with both Hs on the same side, 105 degrees apart, regular stable 3D structures do form that are almost spherical, in the conditions I have described.

 

[BillyT]

What does "more" mean? How many km more? According to

https://en.wikipedia.org/wiki/Geoid

the equipotential surface (based on current centrifugal potential) deviates only −106 to +85 m from the reference Ellipsoid with a 21km radius difference (current Earth shape approximation).

I don't know how many Km of adjustment from the shape the molten, much more rapidly spining Earth took when it solified, still remains to be made. Your reference is not fully correct - even a non-spinning ocean covered Earth without any exposed land mass would not have oceans in an equal gravitational potential because there are thermally driven ocean currents (They would be much larger flows if the Earth were not spinning.)

I worked 30 years at APL/JHU where we made the first orbiting radar altimeter with a precision of about 1 cm. It could see the gulf stream. The Coriolis force on the North flowing gulf steam pushes it eastward. That makes the east edge about a meter higher that the west edge as the GS water is forced into the non-flowing water on its east side.

 

[anorlunda]

The Earth as a whole is quite liquid. It even has tides.

Oh wow, I never thought of that. Do I read you correctly that the crust (not just the oceans) have measurable lunar tidal distortions.

Without rotation, the equatorial bulge would vanish over time, leading to some really nasty earthquakes and to a spherical shape of Earth. Once that shape has been reached, the water will be distributed over the whole surface again.

But what about plate tectonics? Wouldn't that still create mid-ocean ridges and continents independent of tidal forces? Of course that depends on the time scale since eventually the center of the planet will freeze and all internal motion will stop.

 

[A.T.]

I don't know how many Km

Then how do you know it's even relevant, compared to the ~21km radius difference? You are just going off on tangents about irrelevant details.

 

[mfb]

@mfb: can you demonstrate that the contraction at equator with removal of centrfugal force is large compared to the expansion due to a large mass of water going towards the poles (less compressive force). Or, like my opposite belief, that is just your belief?

I did so already. Compare the mass of 20 km of rock to the mass of ~3 km of water. The rock wins by a factor of about 20.

Another point I need to make:The Earth's bulge is NOT due to the current very very weak centrifugal force.

It is. Compare the shape to the geoid.

The entire Earth was once moltent and spinning much more rapidly before the moon existed, carring much of the original angular monementum away from the earth. The Earth's bulge when its surface solified, was larger than now, and is still adjusting (becoming less). I. e. it is NOT yet back to an equlibrium bulge. Many 100s of thousands of year from now, we still will have more bulge than the current very very weak centrifugal force would make.

The continental plates are constantly moving around - plates that are now at the equator were once at the poles and vice versa, multiple times over the last billion years. Earth is not a solid object. It is a liquid with a few shallow blocks of solid matter on top that follow the flow of the liquid interior.

Your reference is not fully correct - even a non-spinning ocean covered Earth without any exposed land mass would not have oceans in an equal gravitational potential because there are thermally driven ocean currents

And we are back at meter-sized effects while you keep ignoring the 20 kilometer effect.

@mfb: I am still waiting for you to demonstrate that the contraction at equator with removal of centrfugal force is large compared to the expansion due to a large mass of water going towards the poles (less compressive force). Or, like my opposite belief, just admit that is ONLY your belief too?

I'm sorry for not being available 24/7 to answer your questions. No wait, I am not. I also do physics to earn money, and that has priority (together with sleep, I'm from Europe and your question was at 1:20 am here).

Do I read you correctly that the crust (not just the oceans) have measurable lunar tidal distortions.

Yes. And the LHC has to take this into account to get the beams around the ring. Story about it

But what about plate tectonics? Wouldn't that still create mid-ocean ridges and continents independent of tidal forces? Of course that depends on the time scale since eventually the center of the planet will freeze and all internal motion will stop.

You still get seas and land masses as today, of course, but their height is small compared to the effect we are talking about. It is also much more local.

 

[ogg]

Wow. Seems to me this tread has jumped the shark. The OP asked about centripital forces and gravitational forces and we're now talking about variations in the g field (not "G"!) due to variations in matter density, temperature, and fluid flow. I mean, really? There's also the claim that centripital force has "nothing" to do with gravitational force. Really? I thought acceleration was "indistinguishable" from a g-field. The FACT is that they DO have something in common (obviously) - although you might argue that their similarities are outside the realm of classical physics...or that discussing their similarities will involve issues of (2nd order) differentials.