Gravity On Earth Dependant On Distance From Center

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SUMMARY

The discussion centers on the relationship between Earth's oblateness and gravitational force, emphasizing that gravity is slightly stronger at the poles than at the equator due to the planet's shape. Participants argue that if Earth were to become more oblate, this gravitational differential would increase until a critical point is reached, where the gravitational forces may equalize. The centrifugal force from Earth's rotation is identified as a key factor in creating this oblateness, affecting gravitational acceleration experienced at different latitudes. The conversation concludes that while the gravitational force at the poles is greater, it may not balance out even with significant flattening of the Earth.

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  • Understanding of gravitational force and its dependence on mass distribution
  • Familiarity with the concept of centrifugal force and its effects on planetary shape
  • Knowledge of basic physics principles, particularly those related to mass and density
  • Ability to interpret mathematical ratios, such as r/R (minor/major axis)
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Darwin
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I happened to come upon the fact that gravity at the Earth's poles is slightly greater than at the equator because the oblate shape of the planet positions the poles closer to the center of Earth's mass. So if the Earth became a bit more oblate this differential would seemingly increase. Yet, there must be a limit to which such a phenomenon would persist. Taken to the extreme, if the Earth became more like a fat pancake there would be very little mass between the poles and quite a bit between any two antipodal points on the equator. So, if this notion holds any water, at what point of flattening (increasing oblateness) would the gravity equalize and the differential begin to reverse itself?
 
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Ideally (that is, ignoring friction/rigidity), that would happen whenever the shape of the Earth deviates from a perfect sphere. There would be sheer stresses tending to shape the Earth to a sphere, because then the gravitational force would be normal to the surface everywhere.

The reason for the Earth's oblateness its rotation. The centrifugal force tends to throw matter outwards (away from the axis of rotation). It is the combined gravitational and centrifugal acceleration that we experience as an effective gravitational acceleration. If this acceleration is everywhere perpendicular to the surface of the earth, there would be no sheer stresses. I`m not sure, but I think the shape of the Earth satisfies that condition to a very high degree.

I have tried to calculate the exact shape of the Earth if it satisfies that condition, but that was some time ago. I can't remember what I got out of it.
 
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Galileo,

I don't think you quite understand the issue. The centrifugal force aside, the mere oblate shape itself engenders a differential in the gravitational force at polar positions than at equatorial positions. If the Earth had no spin what so ever a person standing at one of the poles would weigh more than when standing at the equator. The reason for the oblateness is immaterial.
 
Darwin, nothing in science is irrelevant. Every factor has to be taken into account.
 
My intuition is the increased gravity at the poles due to oblateness would zero out when r/R [minor/major axis] reaches .707 [(2^.5)/2], assuming uniform density and no rotation. It's a very difficult calculation when you toss in these variables.
 
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I think this question is easier to think about if we think about charge and electric field, rather than mass and gravitational field.

In the case of spherically symmetric solutions, the field goes like 1/r^2 simply... Now what happens in a pancake situatuion? Well for an infinite flat plane the field doesn't fall off at all. Obviously there is never such a thing as infinite flat plane of charge, but close to the pancake (like standing on it) the field is constant. Yet near the edge there is negligible field.

Hence my prediction is that they never balance out. Meaning, someone on the poles will always feel a stronger force when the ball is flattened.
 
Chronos said:
My intuition is the increased gravity at the poles due to oblateness would zero out when r/R [minor/major axis] reaches .707 [(2^.5)/2], assuming uniform density and no rotation.
If you have 'intuition' that works on that level, you scare the hell out of me. :biggrin:
 
I agree with Jareds' conclusion, but for slightly different reasons. As Marcus mentioned, the oblateness of the Earth is because of centrifugal force. Even if we set aside the effect this "force" will have on the observer on the ground, we cannot set aside the effect it has on the ground itself. It is because of this force that the Earth is oblate in the first place.

This leads me to conclude that a person standing on the equator does NOT have more mas beneath his feet than a person standing on the pole. The equatorial bulge is a result of centrifugal force spreading out the ground under the equator. This would have the effect of making the ground slightly less dense. So although the observer at the equator would have more volume of Earth beneath his feet, he would not have more mass.
 
LURCH said:
The equatorial bulge is a result of centrifugal force spreading out the ground under the equator. This would have the effect of making the ground slightly less dense. So although the observer at the equator would have more volume of Earth beneath his feet, he would not have more mass.
I am not convinced it would affect the density of the ground. I think the ground moves, but does not change density. If you spun a water balloon, you would get an oblate shape, but the density would be constant anywhere in the balloon. The shape would be from movement of material, not from rarefaction of material.
 

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