[Starlight (OP)]

If the Earth is spherical and spins at roughly 1000 mph,  how do the oceans ignore the equator centrifugal force?

[Colin2B]

.....how do the oceans ignore the equator centrifuge force ?

They don’t. There is a bulge which is max at the equator giving the earth the appearance of a slightly flattened sphere. Worth noting that it’s not just the oceans, the land bulges as well.

[Starlight (OP)]

.....how do the oceans ignore the equator centrifuge force ?

They don’t. There is a bulge which is max at the equator giving the earth the appearance of a slightly flattened sphere. Worth noting that it’s not just the oceans, the land bulges as well.

Then how could it be possible for the ocean water above and below the equator to ignore the centrifuge force , retaining position ?

From my understanding of the nature of  forces , water will ''flirt'' off ANY rotating object !

Link courtesy of youtube .

[Kryptid]

Then how could it be possible for the ocean water above and below the equator to ignore the centrifuge force , retaining position ?

It doesn't. The effect is just less above and below the equator because the velocity (and therefore centrifugal force) is lower.

From my understanding of the nature of  forces , water will ''flirt'' off ANY rotating object !

Not if the force holding the water to the object (in this case, gravity) is stronger than the centrifugal force.

[alancalverd]

how do the oceans ignore the equator centrifuge force ?

Same way as everything else does. Or doesn't. That's why the Sahara desert is empty - all the sand has flown off into space. It also explains why there are no flying fish in the North Sea.

[Kryptid]

To put some math to this question, I'll use the centrifugal force equation:

F_c = (mv²)/r, where

"F_c" is the force in newtons,
"m" is the mass in kilograms,
"v" is the local rotational velocity in meters per second, and
"r" is the distance from the center of rotation in meters.

For a 70 kilogram human standing along the equator, the resulting centrifugal force acting on them is:

F_c = (mv²)/r
F_c = ((70 kg)(461 m/s)²)/6,334,080 m
F_c = 2.3486 newtons

For comparison, gravity pulls on a 70 kilogram human with a force of 70 kg x 9.807 m/s² = 686.49 newtons. The centrifugal force is only 0.34% of the gravitational force at the equator. The human (or the ocean) is in no danger of being flung off of the Earth.

[Starlight (OP)]

To put some math to this question, I'll use the centrifugal force equation:

F_c = (mv²)/r, where

"F_c" is the force in newtons,
"m" is the mass in kilograms,
"v" is the local rotational velocity in meters per second, and
"r" is the distance from the center of rotation in meters.

For a 70 kilogram human standing along the equator, the resulting centrifugal force acting on them is:

F_c = (mv²)/r
F_c = ((70 kg)(461 m/s)²)/6,334,080 m
F_c = 2.3486 newtons

For comparison, gravity pulls on a 70 kilogram human with a force of 70 kg x 9.807 m/s² = 686.49 newtons. The centrifugal force is only 0.34% of the gravitational force at the equator. The human (or the ocean) is in no danger of being flung off of the Earth.

How much centrifuge force would it take to displace 1g of mass of water carefully placed onto a disk shape that was about to be spun?

A centrifuge force is actually a linear force as I suspect you must already know .  In fluid dynamics , water has little to no way of 'gripping'' a surface . 

It was explained earlier that the centrifuge force of the earth has sufficient magnitude to alter the shape of the earth. 

Am I to believe that this force isn't enough to bulge all the water ?

Why would some water bulge and the above and under remain in location when as mentioned it has no ''grip'' ?

 

[Janus]

.....how do the oceans ignore the equator centrifuge force ?

They don’t. There is a bulge which is max at the equator giving the earth the appearance of a slightly flattened sphere. Worth noting that it’s not just the oceans, the land bulges as well.

Then how could it be possible for the ocean water above and below the equator to ignore the centrifuge force , retaining position ?

From my understanding of the nature of  forces , water will ''flirt'' off ANY rotating object !

Link courtesy of youtube .

You have to take all the forces involved and their relative magnitudes into account.
Gravity pulls in toward the center of the Earth. Centrifugal effect acts outward away from the axis of rotation.  The centrifugal effect at the Equator  found by v^2/r and is equivalent to an acceleration of 0.0334 m/s^2, much, much smaller than the 9.8 m/s^2 inward acceleration due to gravity.
If we move North to the 45th parallel.  we are Moving in a smaller circle (~71% that at the Equator) in the same amount of time, so you are moving 71% as fast as at the equator. You are also 71% the distance from the axis of rotation.
The result is that the centrifugal effect is equivalent to an acceleration of just 0.038 m/s^2.  Gravity is still 9.8 m/s^2.  However, only a percentage of that (again, ~71%) is acting directly opposite the centrifugal effect. But 71% of 9.8 is still much much greater than 0.038.   This holds everywhere on the surface. the component of the gravity force that directly counter acts the centrifugal effect always is larger.

Below is a diagram illustrating the effects.

FORCES.png (24.78 kB . 600x600 - viewed 19503 times)
Green arrow is gravity
Red arrow is centrifugal effect
Light blue arrow is the gravity component acting directly against the centrifugal effect
Yellow arrow is gravity component acting parallel to axis of rotation.
The blue ellipse is the shape the water takes as a result of these combined effects. The purple resultant force will always be perpendicular to the surface of the water at that point.
The purple arrow is the resultant combined effect of all these influences.

[Starlight (OP)]

You have to take all the forces involved and their relative magnitudes into account.

I did !

Centrifugal effect acts outward away from the axis of rotation.

A centrifuge force is a linear force and it is for the same reason of orbiting planets .  As you must be aware the Earth wants to travel a linear path but the suns gravitational hold creates the curvature path the earth travels .  An object that is spinning throws things off in a linear path +v(x) .  The centrifuge force is a linear ejection !


Drawing (1).png (61.44 kB . 461x404 - viewed 20000 times)

Your answers this far have not explained how some of the water has enough inertia to retain location north and south of the equator whilst some of the water forms an ocean bulge at the equator !

[evan_au]

Your answers this far have not explained how some of the water has enough inertia to retain location north and south of the equator whilst some of the water forms an ocean bulge at the equator !

In his diagram above, Janus showed how water away from the equator is subject to:
- A dominant force pulling it towards the center of the Earth
- A small force pulling it towards the equator

The answer to your question is that all of the water on Earth has already been pulled towards the equator by a small amount.
- This forms the equatorial bulge
- And the corresponding polar "dip"
- Now, the centrifugal force pulling the water away from the poles towards the equator is exactly counterbalanced by the force of gravity pulling on this bulge, trying to move it towards the poles.
- In scientific terms, we would call this an "equilibrium" - a stable state.
- While the Earth's rate of rotation and amount of polar ice remains unchanged*, no more water will move towards the equatorial bulge, as it has to climb "uphill" to get up the equatorial bulge.

*In fact, polar ice is melting, and Earth's rate of rotation is changing (slightly, by microseconds per day), and the Sun and Moon pull do on the oceans, so there is always some sloshing around in the oceans - but by a matter of a meter, not the kilometers of the equatorial bulge.
See: https://en.wikipedia.org/wiki/Leap_second#Slowing_rotation_of_the_Earth

[Kryptid]

How much centrifuge force would it take to displace 1g of mass of water carefully placed onto a disk shape that was about to be spun?

Since water is a liquid, I presume that any amount of force that is strong enough to break the weak, molecular adhesion forces sticking it to the disk would work. It would just be a very slow progression for small forces. In either case, that's not a good analogy for the Earth because gravity is holding the water to the surface. If you wanted the Earth to spin fast enough to spin all of the water off into space, then you would need to make the centrifugal force equal to the gravitational force. Since centrifugal force increases with the square of the velocity, you'd need to make the Earth spin about 17.1 times faster than it currently does to make that happen.

Am I to believe that this force isn't enough to bulge all the water ?

It depends on what you mean when you say "all" the water. There is a finite volume of water on the Earth, so any increase in water depth at the equator due to centrifugal force must be compensated for by a corresponding decrease in water depth near the poles. So all of the water is indeed moved by the centrifugal force.

Why would some water bulge and the above and under remain in location when as mentioned it has no ''grip'' ?

Water is a liquid. It doesn't "remain in location". Currents would mix the water around the equator with water above and below the equator. A molecule of water at the equator now could be near Greenland next year.

Your answers this far have not explained how some of the water has enough inertia to retain location north and south of the equator whilst some of the water forms an ocean bulge at the equator !

The bulge isn't just at the equator. It's just largest at the equator.

[Janus]

You have to take all the forces involved and their relative magnitudes into account.

I did !

Centrifugal effect acts outward away from the axis of rotation.

A centrifuge force is a linear force and it is for the same reason of orbiting planets .  As you must be aware the Earth wants to travel a linear path but the suns gravitational hold creates the curvature path the earth travels .  An object that is spinning throws things off in a linear path +v(x) .  The centrifuge force is a linear ejection !


Drawing (1).png (61.44 kB . 461x404 - viewed 20000 times)

Your answers this far have not explained how some of the water has enough inertia to retain location north and south of the equator whilst some of the water forms an ocean bulge at the equator !

As water North and South of the equator tries to move toward the equator it can only do so by displacing water that is already there upward.  Since gravity is still the predominate effect at the equator, This water has to lifted against gravity.  As more water builds up at the equator, the bulge gets taller and contains more water.  Gravity acting on this bulge wants to flatten it, resisting the growth of the bulge. The larger the bulge gets, the more "push back" is created.   Eventually, the force of the "push back" equals the force from the water trying to move toward the equator  and a balance is reached.
This will result in a oblate spheroid shape.   The rate of spin of the Earth is so slow that the amount of bulge isn't even noticeable to the bare eye.

[Starlight (OP)]

As water North and South of the equator tries to move toward the equator it can only do so by displacing water that is already there upward.  Since gravity is still the predominate effect at the equator, This water has to lifted against gravity.  As more water builds up at the equator, the bulge gets taller and contains more water.  Gravity acting on this bulge wants to flatten it, resisting the growth of the bulge. The larger the bulge gets, the more "push back" is created.   Eventually, the force of the "push back" equals the force from the water trying to move toward the equator  and a balance is reached.
This will result in a oblate spheroid shape.   The rate of spin of the Earth is so slow that the amount of bulge isn't even noticeable to the bare eye.

A good answer that I can not at this time find a counter argument for so perhaps I could discuss your answer a little more deeper and ask about tensors .

Does the North and South magnetic poles also apply tensors on the oceans to create the stretching of the oceans ?

I ask this as you have now in so many words defined the equator relative to water as being down bank !


Drawing.png (26.89 kB . 390x449 - viewed 17943 times)














You have to take all the forces involved and their relative magnitudes into account. [/quote]

I did !

Centrifugal effect acts outward away from the axis of rotation.

A centrifuge force is a linear force and it is for the same reason of orbiting planets .  As you must be aware the Earth wants to travel a linear path but the suns gravitational hold creates the curvature path the earth travels .  An object that is spinning throws things off in a linear path +v(x) .  The centrifuge force is a linear ejection !


Drawing (1).png (61.44 kB . 461x404 - viewed 20000 times)

Your answers this far have not explained how some of the water has enough inertia to retain location north and south of the equator whilst some of the water forms an ocean bulge at the equator !













[/quote]
As water North and South of the equator tries to move toward the equator it can only do so by displacing water that is already there upward.  Since gravity is still the predominate effect at the equator, This water has to lifted against gravity.  As more water builds up at the equator, the bulge gets taller and contains more water.  Gravity acting on this bulge wants to flatten it, resisting the growth of the bulge. The larger the bulge gets, the more "push back" is created.   Eventually, the force of the "push back" equals the force from the water trying to move toward the equator  and a balance is reached.
This will result in a oblate spheroid shape.   The rate of spin of the Earth is so slow that the amount of bulge isn't even noticeable to the bare eye.
[/quote]

[chiralSPO]

How much centrifuge force would it take to displace 1g of mass of water carefully placed onto a disk shape that was about to be spun?

A centrifuge force is actually a linear force as I suspect you must already know .  In fluid dynamics , water has little to no way of 'gripping'' a surface . 

It was explained earlier that the centrifuge force of the earth has sufficient magnitude to alter the shape of the earth. 

Am I to believe that this force isn't enough to bulge all the water ?

Why would some water bulge and the above and under remain in location when as mentioned it has no ''grip'' ?

The spinning dish analogy fails to reproduce what we observe precisely because the earth has a gripping force that the dish/water doesn't: gravity.

A simple tweak to the analogy can work though. Put the water in a large conical dish and spin it--the water will experience a force away from the axis of rotation, but it has to flow uphill to move outwards. There will be an equilibrium established between those two opposing forces, and the equilibrium will be set by the angle of the cone (slope of the side of the dish, ie how far up does it have to go up to go 1 cm out?) and the rate of the rotation (assuming gravity is fixed). If the dish is spun quickly enough all the water would fly up and out, but for any sufficiently slow spinning, the equilibrium will allow all of the water to remain in the dish.

[Kryptid]

Does the North and South magnetic poles also apply tensors on the oceans to create the stretching of the oceans ?

Such an effect would be very, very small if it does exist. Water is not particularly magnetic (and Earth's magnetic field is weak).

[Starlight (OP)]

The spinning dish analogy fails to reproduce what we observe precisely because the earth has a gripping force that the dish/water doesn't: gravity.

I do not think your statement is true .  All objects have mass , this includes a spinning  disk . The disk also has the force of gravity .
A centrifuge force is stronger than the weaker force of gravity !

[Starlight (OP)]

Does the North and South magnetic poles also apply tensors on the oceans to create the stretching of the oceans ?

Such an effect would be very, very small if it does exist. Water is not particularly magnetic (and Earth's magnetic field is weak).

If that is so , then I still do not feel the answers given this far have given a satisfactory explanation !

As water North and South of the equator tries to move toward the equator it can only do so by displacing water that is already there upward.

Water mixes with water !


It seems very odd that some water ignores the centrifuge force that apparently has the magnitude to reform a planets shape but not enough magnitude to displace less dense water ! 


Drawing.png (22.65 kB . 270x477 - viewed 10660 times)

I see no physics of why all the water wouldn't run downwards to the equator by centrifuge force! The linear centripetal force of gravity not being a tensor for the latitude of water  .


9a3d6cae-40f7-49e3-8d14-dba7e6ee5a9c.png (13.04 kB . 270x477 - viewed 10588 times)

[alancalverd]

A centrifuge force is stronger than the weaker force of gravity !

Which, presumably, is why your car always ends up in the ditch instead of going round a corner. Fortunately, most vehicles have a different opinion.

[Bored chemist]

A centrifuge force is stronger than the weaker force of gravity !

Yes, but only because it spins very fast.

A centrifuge that only spun at one revolution per day wouldn't achieve much.

[Kryptid]

If that is so , then I still do not feel the answers given this far have given a satisfactory explanation !

Why...?

A centrifuge force is stronger than the weaker force of gravity !

The centrifugal force of the Earth's rotation is significantly weaker than Earth's gravity. Please refer to reply #5 for the math that supports this.