If Earth Stopped Spinning Would Equatorial Gravity Increase?

[common_sense_seeker]

Due to loss of centrifugal force, would it then be greater than at the poles? The Moon isn't spinning like the Earth, and it still has flattening and a lower equatorial surface gravity. Maybe it would remain (less due to irregularities of the gravitational field of the inner core)?

[ElectricFetus]

Due to loss of centrifugal force, would it then be greater than at the poles? The Moon isn't spinning like the Earth, and it still has flattening and a lower equatorial surface gravity. Maybe it would remain (less due to irregularities of the gravitational field of the inner core)?

Again the moon has lower equatorial gravity on the near side because it being pull on by the earth. If the earth stopped spinning equatorial gravity would not get any stronger rather it would feel slightly stronger simple because there is not centripetal force negating a little bit of it anymore.

[D H]

The Moon isn't spinning like the Earth
Yes, it is. Evidence:
and it still has flattening and a lower equatorial surface gravity.
More evidence: From the perspective of a moon-fixed observer, stars rise and set. It is the rotation rate with respect to inertial that counts when computing perceived gravitation force.
Maybe it would remain (less due to irregularities of the gravitational field of the inner core)?
What irregularities in the inner core? The Moon's gravity field is incredibly lumpy because the Moon's surface is quite irregular, not the interior.

[Ophiolite]

What irregularities in the inner core? The inner core is not a perfect sphere, or indeed, not even a perfect oblate spheroid.

[D H]

The inner core is not a perfect sphere, or indeed, not even a perfect oblate spheroid.
I did not say that Moon's core is a perfect sphere, or even a perfect oblate spheroid. What I said that the Moon's highly irregular gravity field results largely from the irregularities in the Moon's surface. The Moon is only slightly oblate; the lunar terrain has features with much greater vertical relief than the 2.2 km variation attributable to oblateness: Craters 13 km deep, for example. The surface of the near side of the Moon has several large mass concentrations. The slight oblateness, the mascons, and the irregular terrain are what lead to the Moon's incredibly lumpy gravity field.

If you read the OP's posts, you will see he has some poorly-formed concept of directional gravity based on some poorly-formed concepts of dark matter in a planet's core. The Moon's core is not responsible for the Moon's irregular gravity field. It might contribute a little; but the lion's share is attributable to oblateness, roughness, and mascons.

[CheskiChips]

Due to loss of centrifugal force, would it then be greater than at the poles? The Moon isn't spinning like the Earth, and it still has flattening and a lower equatorial surface gravity. Maybe it would remain (less due to irregularities of the gravitational field of the inner core)?

Why don't you guys answer the question he intended to ask?

Effective Gravity is the term used when taking into account centrifugal force.

The effective gravity at the poles would not change at all in either case (if ignoring relativistic effects).

The gravity at the equator would be lower than the poles, as it's further from the center of mass than the gravitational poles are, if it was a perfect sphere...they would be equal.

D H ~ is correct, the moon is spinning.

[amark317]

one thing is for certain, we would all fly like, 5 miles east!

[Billy T]

...If the earth stopped spinning equatorial gravity would not get any stronger rather it would feel slightly stronger simple because there is not centripetal force negating a little bit of it anymore.This is essentially correct, but not completely as the dirt and rocks we call the "solid" earth are not really solid. Those near the Equator would all compress a little and mover slightly closer to the center of mass and then the inverse square fall off would be less, so true gravity force acting on you would slightly incease as well as the much larger increase in your weight due to the absence of centrifugal force.

I am not certain, but possibly the full Military precession of GPS and averaging over many measurements and correlating with the moon's position can actualy measure the tides in the "solid" earth. I think the peak to valley earth tide is approximately 1 cm in amplituded by calculation. - I do not know but bet that is observable by above procedure. (That is the sort of thing they keep secrete. - God forbid that the enemy should know if our error in knowing were the flag pole in front of his office is 50cm or only 0.5cm - as if that make any difference!)

Also if one neglects the effect of winds (and without spin of the earth, the Corelois effect is absent) the sea surface should be almost a equal-gravitational-potiential surface (Density of water varies with temperature so it is not exactly.)* Thus on deck of a boat the gravity at equator and in polar oceans is essentailly the same and on the boat deck you can not weigh less at the equator than near the pole, as you certainly could now for two reasons. (Greater distance for CoM makes less centripetal force and more centrafugal force)

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*off hand, I am not sure what the net effect is as polar water also has less salt so that partically (or fully?) compenstates for the fact colder temperature increases the density.

[ElectricFetus]

This is essentially correct, but not completely as the dirt and rocks we call the "solid" earth are not really solid. Those near the Equator would all compress a little and mover slightly closer to the center of mass and then the inverse square fall off would be less, so true gravity force acting on you would slightly incease as well as the much larger increase in your weight due to the absence of centrifugal force.

I am not certain, but possibly the full Military precession of GPS and averaging over many measurements and correlating with the moon's position can actualy measure the tides in the "solid" earth. I think the peak to valley earth tide is approximately 1 cm in amplituded by calculation. - I do not know but bet that is observable by above procedure. (That is the sort of thing they keep secrete. - God forbid that the enemy should know if our error in knowing were the flag pole in front of his office is 50cm or only 0.5cm - as if that make any difference!)

Also if one neglects the effect of winds (and without spin of the earth, the Corelois effect is absent) the sea surface should be almost a equal-gravitational-potiential surface (Density of water varies with temperature so it is not exactly.)* Thus on deck of a boat the gravity at equator and in polar oceans is essentailly the same and on the boat deck you can not weigh less at the equator than near the pole, as you certainly could now for two reasons. (Greater distance for CoM makes less centripetal force and more centrafugal force)

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*off hand, I am not sure what the net effect is as polar water also has less salt so that partically (or fully?) compenstates for the fact colder temperature increases the density.

Hey if we had gps accurate to .5cm we could have smart "bombs" the size of pencils capable of taking out parts of equipment, so we could drop a smart bomb that will take out just the machine gun (or BAM! straight through the head of the machine gunner), not the machine gun nest, or the building the machine gun nest is on.

[D H]

I am not certain, but possibly the full Military precession of GPS and averaging over many measurements and correlating with the moon's position can actualy measure the tides in the "solid" earth. I think the peak to valley earth tide is approximately 1 cm in amplituded by calculation.
You're off by a bit, Billy. Earth tides have an amplitude of up to 55 cm or so at the equator. The underlying theory has been known for over 100 years.

Also if one neglects the effect of winds (and without spin of the earth, the Corelois effect is absent) the sea surface should be almost a equal-gravitational-potiential surface (Density of water varies with temperature so it is not exactly.)
Mean see surface is an equipotential surface -- in the combined gravitational+centrifugal potential, that is.

Thus on deck of a boat the gravity at equator and in polar oceans is essentailly the same and on the boat deck you can not weigh less at the equator than near the pole, as you certainly could now for two reasons.
One problem here: If the Earth stopped spinning and if it maintained its oblate spheroid shape, there would be no longer be any oceans anywhere near the equator. The oceans would migrate toward the poles, leaving the equator literally very high and very dry.

[Billy T]

You're off by a bit, Billy. Earth tides have an amplitude of up to 55 cm or so at the equator. The underlying theory has been known for over 100 years. Thanks. I work from memory. I actually though it was more like 10cm but wanted not to err on the high side.


Mean see surface is an equipotential surface -- in the combined gravitational+centrifugal potential, that is.Not sure we agree here. I think that making the average density lower all the way to the ocean floor in area "A" would elevate the surface. Consider a vertical U shaped glass tube filled with water. The top surface in both sides are at the same potential we agree and same altitude from the room's floor. Now warm the water in one side. The top surface of that side will become more distant from the floor, will it not? It certainly is not at the same gravitational potential then as the top surface of the still cold side.* (and the centrifugal forces are essentially unchagned, even for the spinning Earth.)

I was just trying to be fully accurate when I allowed that the ocean surface might not be exactly at same gravitation potential even if the Earth were not spinning due to the fact salt and temperature change the density.

Likewise if one side of the U tube is strong salt solution and other is pure H2O then the top of the salt side will be lower.
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*In fact with tiny paddle wheel and inter connection I would have a candidate for the world's least efficient heat engine contest. :)

[D H]

Mean see surface is an equipotential surface -- in the combined gravitational+centrifugal potential, that is.Not sure we agree here. I think that making the average density lower all the way to the ocean floor in area "A" would elevate the surface.
We are talking different things. The mean surface of a fluid body, whether or not the fluid is of uniform density, will be an equipotential surface. Think about it in terms of Hamilton's principle.

[Billy T]

We are talking different things. The mean surface of a fluid body, whether or not the fluid is of uniform density, will be an equipotential surface. Think about it in terms of Hamilton's principle.It has been a while (~45 Years) since I applied that*, but you are correct - the surface is where it is as that is the leased action or in other words, any virtual displacment of it would require work.

We really were speaking of different potentials - Mine, clearly stated in post 8 and now made bold so it is easy to find, was ONLY the gravitational potential. Your was the full potential energy. -I just am not confortable with centrifugal force contributing to potential but that is correct - you can do work against it, even though I do not like to even consider it a force.
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*I still have my copy of Goldstein's Classica Mechanics as I did most of the problems in it once. I am sure we applied Hamilton's principle to the simple harmonic oscillator - Think that (or perhaps the Lagrangian approach?) was when Goldstein noted it was: "Like opening peanuts with a sledge hammer." That and his intrductory admonishment: "Be ye doers of the word and not hearers only." (Translated for the original Hebrew) is all I remember well now. The class met for only and hour each week and had no lectures. - You might be called upon to put your solution to one of the problems on the black board as classed worked thru the book. It is a great book but doubt they still use it.