Nonspherical earth (split from centrifugal forces)

[imatfaal]

 

 

I didn't realise that was your original contention.

That bit is most definitely wrong.

 

Is it? - I thought it was quirky but correct. Planets settle into the lowest energy configuration - a non-rotating spheroid would be a perfect sphere as any variation to that sphere would be at a higher energy configuration and would tend over time to the lower. A rotating spheroid has more stuff on the equator where it weighs less because any configuration with equal amounts at poles and equators is a high energy solution - matter will gravitate to where gpe is lower cos it weighs less

[Strange]

Is it? - I thought it was quirky but correct.

 

Same here. Which is why I think it is an interesting way of looking at the two different effects: the forces flattening the Earth considered as a reduction in weight; and the extra distance from the centre considered as a reduction in weight.

 

But it is important to realise they are different. (which is why the answer that started all this is not a non-answer!)

[studiot]

 

imatfaal

 

gpe is lower cos it weighs less

 

 

 

imatfaal

 

Is it?

 

gpe = mass x g x separation distance.

 

Mass is constant, although g is smaller distance is larger.

 

My objection to MigL's sentence is the same as someone who earlier said "cart before the horse"

 

An object whether solid or liquid or some combination tends to a spherical ball in the absence of other disturbing forces to minimise its free surface surface energy (surface tension).

I think you will find that there are other configurations that minimise gravity by balancing opposing pulls on opposite sides of the COG.

 

Anyway having taken up the minimum free surface configuration, or even before, some disturbing mass impacts the ball and adds its mass and momentum to the pot. This in general will add moment of momentum to the ball and if enough such impacts happen, and also if the impacts arrive from a preferential direction and so are not cancelled out by randomness, will set the ball rotating.

 

If, once the ball is rotating, parts of it are capable of movement relative to other parts, yes a belly will develop.

 

The equatorial material that is lifted to form the belly will actually gain rotational energy compared to its slower polar counterpart.

Alternatively the polar material will loose gpe by subsidence (flattening).

Most likely some combination of both will occur.

 

So horses and carts, which is it?

Edited April 15, 2015 by studiot

[Robittybob1]

 

Is it? - I thought it was quirky but correct. Planets settle into the lowest energy configuration - a non-rotating spheroid would be a perfect sphere as any variation to that sphere would be at a higher energy configuration and would tend over time to the lower. A rotating spheroid has more stuff on the equator where it weighs less because any configuration with equal amounts at poles and equators is a high energy solution - matter will gravitate to where gpe is lower cos it weighs less

Matter will gravitate to where gpe is lower cos it weighs [less] more not less.

Edited April 15, 2015 by Robittybob1

[pavelcherepan]

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.

 

I agree with studiot here, I still think that the underlying cause for things weighing less at the equator is the rotation and angular momentum of the Earth and hence the shape of the Earth and hence the lower g value at the equator.

 

Matter will gravitate to where gpe is lower cos it weighs [less] more.

 

This makes no sense to me.

[Robittybob1]

 

This makes no sense to me.

Closer to the center of the mass things will experience a stronger force and hence weigh more. GPE being a negative energy less is more negative or deeper in the gravity well.

 

I agree with studiot here, I still think that the underlying cause for things weighing less at the equator is the rotation and angular momentum of the Earth and hence the shape of the Earth and hence the lower g value at the equator.

 

But that lower G value is not causing the shape so Studiot is wrong.

[pavelcherepan]

But that lower G value is not causing the shape so Studiot is wrong.

 

You have misread what I wrote and what studiot has written. Not the lower G-value. The angular momentum of the Earth should be the cause. Lower g value should be the result of deformed spherical shape and not the other way around.

 

Closer to the center of the mass things will experience a stronger force and hence weigh more. GPE being a negative energy less is more negative or deeper in the gravity well.

 

OK. Makes slightly more sense now. I can only assume imatfaal was referring to the lower absolute value.

 

On the other hand if you use a simple mechanics formula for total energy then your point doesn't hold:

 

[latex]E_T = \frac{mv^2}{2} + mgh[/latex]

Edited April 16, 2015 by pavelcherepan

[Robittybob1]

 

You have misread what I wrote and what studiot has written. Not the lower G-value. The angular momentum of the Earth should be the cause. Lower g value should be the result of deformed spherical shape and not the other way around.

 

 

OK. Makes slightly more sense now. I can only assume imatfaal was referring to the lower absolute value.

 

On the other hand if you use a simple mechanics formula for total energy then your point doesn't hold:

 

[latex]E_T = \frac{mv^2}{2} + mgh[/latex]

That's OK mgh is positive if something rises but will be negative if the mass falls lower.

[MigL]

OK, this is my last attempt. And only because studiot said I was wrong ( wouldn't want him to think less of me ).

 

Take a pail with outward sloping sides and fill it with water, or if you prefer sand or mud to more closely resmble the Earth's crust.

Now spin it about its vertical axis and see what happens to the contents of the pail.

We note that the liquid ( or flowable solid ) goes from a flat surface to one that is concave. The edges close to the pail sides rise wwhile the center sinks. It is still at its lowest energy state, but that has now been modified by the angular momentum ( and the angle of the pail's sides ). The force of gravity acting on the contents of the pail has been modified also by the angular momentum ( as Pavel has also agreed ) such that the NET force is lessened close to the edges, and this leads to the deformation of the pail's contents.

Now weight is nothing more than the measure of the force of gravity. It has been conveniently been given units which are equal to inertial mass on earth, but while mass is invariant, weight will change on the moon, or any othe place where gravity has changed.

So when we say that the deformation of the pail's contents is due to a modification to the force of gravity ( by angular momentum ), it is equivalent to saying the deformation is due to a change ( lessening ) of its weight. Not mass mind you, the contents have the same mass throughout, but its weight has changed, and it 'floats' higher along the sides of the pail.

 

This is the exact same effect as the Earth spinning about its axis.

And the deformation to the lowest energy state is caused by the angular momentum changing the NET force acting on the Earth's composition. This NET force is also known as weight ( sometimes given in Newtons, go figure, a unit of force, not mass ).

In effect, the, the deformation is caused by the change in weight of the composition of the Earth.

 

So, again, saying that the deformation causes a change in weight is redundant.

Edited April 16, 2015 by MigL

[Robittybob1]

MigL - I'm sure you should have used the word pressure somewhere in your explanation.

[MigL]

Why ?

How does the pressure change from the pail at rest and the spinning pail ?

 

It doesn't . What does change is the angular momentum, and as a consequence, the net force, or weight.

[Robittybob1]

Why ?

How does the pressure change from the pail at rest and the spinning pail ?

 

It doesn't . What does change is the angular momentum, and as a consequence, the net force, or weight.

Have you heard of centrifugal pumps? What you have done is a very cursory centrifugal pump.

How does a Centrifugal pump work ?

[MigL]

And do you think a centrifugal pump that si pumping a liquid or flowable solid actually works by compressing it ?

(as opposed to a centrifugal compressor on turbomachinery )

[Strange]

So, again, saying that the deformation causes a change in weight is redundant.

But to go back to my point, if you were to have a non-rotating planet (not Earth, because we don't want to get sidetracked by the physical effects of stopping rotation) with the same mass, size and shape as Earth then things would still weigh less at the Equator because they are further from the centre, even with no rotation.

[Robittybob1]

But to go back to my point, if you were to have a non-rotating planet (not Earth, because we don't want to get sidetracked by the physical effects of stopping rotation) with the same mass, size and shape as Earth then things would still weigh less at the Equator because they are further from the centre, even with no rotation.

It does but just saying it is further from the center is not the whole answer for if you were able to drop a mass down a hole drilled into the Earth its weight is not going to go up simply because it is getting closer to the center.

[pavelcherepan]

But to go back to my point, if you were to have a non-rotating planet (not Earth, because we don't want to get sidetracked by the physical effects of stopping rotation) with the same mass, size and shape as Earth then things would still weigh less at the Equator because they are further from the centre, even with no rotation.

 

Not sure I follow that thought. Why would on a non-rotating body equator be further away from center of mass than poles? Non-rotating planet should be ideally spherical (or very close to it).

 

EDIT: Sorry, didn't notice the 'shape' bit initially. In that case you're right, but then there's the question of why the shape is the same.

Edited April 16, 2015 by pavelcherepan

[studiot]

 

robittybob

MigL - I'm sure you should have used the word pressure somewhere in your explanation.

 

Yes indeed.

There is a big difference between a constrained (by the bucket) low viscoscity fluid with next to ne shear stress and an unconstraned (except by surface tension) ultra high viscoscity fluo-solid.

[Robittybob1]

 

Not sure I follow that thought. Why would on a non-rotating body equator be further away from center of mass than poles? Non-rotating planet should be ideally spherical (or very close to it).

 

EDIT: Sorry, didn't notice the 'shape' bit initially. In that case you're right, but then there's the question of why the shape is the same.

The planet might have cooled and froze solid in that shape and then slowed its rotation (somehow).

[studiot]

Strange offered another gem of perceptual insight.

 

pavelchaperan

Not sure I follow that thought. Why would on a non-rotating body equator be further away from center of mass than poles? Non-rotating planet should be ideally spherical (or very close to it).

robbitybob

The planet might have cooled and froze solid in that shape and then slowed its rotation (somehow).

 

 

That is if it was ever rotating in the first place.

 

I offered a condition whereby this might happen before.

It will only start rotating if there is a net influx of mass and momentum from a particular direction.

If, on the other hand, it is hit many times at random from all directions then the net turning effect will be zero.

[MigL]

I think Pavel now gets it.

The planet is only squished because of its rotation.

Unless it was a much smaller asteroid, it would tend to spherical. ( is the Moon or Mercury squished ? )

 

And while pressures are changing to some degree, they neither add or subtract from the thought experiment, i.e. inconsequential.

[Strange]

It does but just saying it is further from the center is not the whole answer for if you were able to drop a mass down a hole drilled into the Earth its weight is not going to go up simply because it is getting closer to the center.

 

For things on the surface (which is what we are discussing) it is the whole answer.

 

Actually, no it isn't the whole answer because someone has already mentioned the inhomogeneity of the Earth, for example. But we are dealing with idealisations here, surely.

 

Not sure I follow that thought. Why would on a non-rotating body equator be further away from center of mass than poles? Non-rotating planet should be ideally spherical (or very close to it).

 

EDIT: Sorry, didn't notice the 'shape' bit initially. In that case you're right, but then there's the question of why the shape is the same.

 

Because the Magritheans made it that way for the purpose of this thought experiment. Talk about herding cats! This is worse than trying to manage a team of software engineers.

Edited April 16, 2015 by Strange

[Janus]

It does but just saying it is further from the center is not the whole answer for if you were able to drop a mass down a hole drilled into the Earth its weight is not going to go up simply because it is getting closer to the center.

But is is a great part of the answer.

 

Consider the following diagram. the grey ellipse is the profile of an oblate spheroid with a semi-major axis (equatorial radius) of 2 and a semi-minor axis (polar radius) of 1.

 

 

Let's compare the force of gravity on the two points of the pole and equator.

 

First consider a point on the equator. It will be effected by all the mass inside the sphere closer to the center of the spheroid than it is (shown by the black line). This is the entire mass of the spheroid.

The mass of the spheroid is its density times its volume which makes the mass directly proportional to the volume.

 

The volume of an oblate spheroid is found by:

 

[math]V = \frac{4a^2b \pi}{3}[/math]

where a is the semi-major axis and b is the semimajor axis.

 

For our values, this gives an answer of 16/3 pi

 

Acceleration due to gravity is

 

[math]\frac{GM}{d^2}[/math]

 

In this case the M is 16/3 pi and d=a=2

 

and we get a value of 4/3 G(pi)

 

Now consider a point on the pole.

 

At the very least this is effected by the mass inside the solid black area.

 

This is a sphere with a volume of 4/3 pi

 

Using the same formula for the acceleration of gravity as above and using 4/3 pi for M and 1 for d, we get

 

4/3 G(pi), the same value we got at the equator. Essentially, the mass of this sphere is 1/4 that of the entire oblate spheroid, but the pole is 1/2 the distance from the center.

 

But this only accounts for a part of the gravity felt at the equator. You still have to account for the mass of the bulges that extend beyond the area of the black circle. They will exert a net downward (towards the center) pull on anything resting on the equator, making the total gravitational pull at the pole greater than that at the equator.

Matter will gravitate to where gpe is lower cos it weighs [less] more not less.

GPE is not related to the local weight. If you are on the surface of the Earth and move away, your weight goes down, and your GPE increases (goes less negative). However if you are digging a hole towards the center of the Earth, you would start to way less and your GPE would decrease (go more negative). It is as equally possible to have GPE increase when moving to from low to high gravity as it is to have it increase when moving from high to low.

 

Or potential energy can remain the same when moving from low to high weight or high to low weight.

 

This is the case when it come to the surface of the Earth. Due to its spin and oblate shape, effective gravity is weaker at the equator than it is at the poles. However, the surface of the Earth is at equipotential.

 

There are two potentials to consider: The GPE and the potential due to the rotating frame of the Earth (ignoring gravity, it would take work to move from the equator to the axis of a rotating Earth).

 

It is the balance between these two potentials that determines the shape of the Earth.

[studiot]

Janus, doesn't all this assume a homogenous density function?

 

As we know the actual density/depth function is concave down starting from the surface. It rises quite sharply just down from the surface, with the rate of increase steadily decreasing with depth, finally becoming asymptotic to some max value near the middle.

 

Have you any working to show the effect of this?

[MigL]

I just wanna know one thing...

 

Who are the Magretheans ?

[moth]

Did somebody say "Slartibartfast"