Earth ellipsoid due to rotation


by Ulrich
Tags: earth, ellipsoid, rotation
Ulrich
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#1
Oct13-12, 12:02 PM
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I recently saw a documentary, which claimed that if the earth rotation slows down the water of the oceans will flood to the north and south because the centripetal force at the equator diminishes. In fact, earth’s radius is about 20 km longer at the equator than at the poles. However, I doubt that any water would be displaced because the Earth is not solid in the interior and the crust is by far too thin to hold the earth’s original ellipsoidal form. So I think that all the water would stay at the same place with less rotation.

On the other hand, I calculated the equatorial radius for the actual angular velocity. It should only be about 10 km longer than the polar radius, which means, if the calculation is right, that the solid crust is effectively slow in adapting its form to a decrease in angular velocity ω. However, my calculation may be wrong. Here is what I did: assuming that the earth is covered with water, there are three forces acting on a small volume of water with mass m at the surface. The first is the support force S perpendicular to the surface and therefore to the derivative of a curve y(x) to be determined delimiting the surface in Cartesian coordinates. The second is the gravitational force G pointing to earth’s center and the third is the centripetal force Z perpendicular to the rotation axis. All vector forces are supposed to be in equilibrium (no flow of water) so we have S + G + Z = 0, which yields the differential equation [itex]\left(x^2+y^2\right)^{3/2}=\frac{G M}{ω^2}\left(\frac{y' y}{x}+1\right)[/itex], where G is the gravitational constant and M the mass of the Earth. It’s solution is a curve near to an ellipse for small ω. You can see the maths in the Mathematica file in the attachement.

I am sure that I did the maths correctly but something may be wrong with the equilibrium of the forces. I would greatly appreciate any comments.
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File Type: nb Meeresspiegelerhöhung durch Erdrotation.nb (65.9 KB, 9 views)
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ImaLooser
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#2
Oct14-12, 07:18 AM
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Quote Quote by Ulrich View Post
I recently saw a documentary, which claimed that if the earth rotation slows down the water of the oceans will flood to the north and south because the centripetal force at the equator diminishes. In fact, earth’s radius is about 20 km longer at the equator than at the poles. However, I doubt that any water would be displaced because the Earth is not solid in the interior and the crust is by far too thin to hold the earth’s original ellipsoidal form. So I think that all the water would stay at the same place with less rotation.

On the other hand, I calculated the equatorial radius for the actual angular velocity. It should only be about 10 km longer than the polar radius, which means, if the calculation is right, that the solid crust is effectively slow in adapting its form to a decrease in angular velocity ω. However, my calculation may be wrong. Here is what I did: assuming that the earth is covered with water, there are three forces acting on a small volume of water with mass m at the surface. The first is the support force S perpendicular to the surface and therefore to the derivative of a curve y(x) to be determined delimiting the surface in Cartesian coordinates. The second is the gravitational force G pointing to earth’s center and the third is the centripetal force Z perpendicular to the rotation axis. All vector forces are supposed to be in equilibrium (no flow of water) so we have S + G + Z = 0, which yields the differential equation [itex]\left(x^2+y^2\right)^{3/2}=\frac{G M}{ω^2}\left(\frac{y' y}{x}+1\right)[/itex], where G is the gravitational constant and M the mass of the Earth. It’s solution is a curve near to an ellipse for small ω. You can see the maths in the Mathematica file in the attachement.

I am sure that I did the maths correctly but something may be wrong with the equilibrium of the forces. I would greatly appreciate any comments.
I'd say the water would flow to the poles. The Earth would eventually change shape, but the solid portions would change much more slowly than the flow of water.
Andre
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#3
Oct14-12, 08:11 AM
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Picture Earth as a (pseudo) solid rotating oblate spheroid, with a the major axis and b the minor (spin) axis. Now if we add a shell of water on the surface, it will also assume the oblate spheroid shape with a1 and b1, due to the same forces, Although likely, I'm not sure that ratio a/b is the same as a1/b1, but lets assume that it is. Now obviously a1-a > b1-b since a>b, so the ocean depth at the equator is more than at the poles.

So when the rotation stops both the solid Earth and the water assume a spherical shape eventually with equal ocean depth anywhere. However we started with deeper oceans at the equator so the water has to flow to the poles. And mind there is a lot more equator area than polar area.

D H
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Oct14-12, 08:23 AM
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Earth ellipsoid due to rotation


Quote Quote by Ulrich View Post
I recently saw a documentary, which claimed that if the earth rotation slows down the water of the oceans will flood to the north and south because the centripetal force at the equator diminishes. In fact, earth’s radius is about 20 km longer at the equator than at the poles. However, I doubt that any water would be displaced because the Earth is not solid in the interior and the crust is by far too thin to hold the earth’s original ellipsoidal form. So I think that all the water would stay at the same place with less rotation.
It depends on how quickly the Earth's rotation stops. Slow it down gradually and the Earth will have time to relax to maintain an equilibrium shape; the Earth and oceans will relax toward a spherical shape in unison. Slow the rotation down suddenly and the Earth won't be able to respond in time, but the oceans will. The poles will be flooded.


On the other hand, I calculated the equatorial radius for the actual angular velocity. It should only be about 10 km longer than the polar radius, which means, if the calculation is right, that the solid crust is effectively slow in adapting its form to a decrease in angular velocity ω. However, my calculation may be wrong. Here is what I did: assuming that the earth is covered with water, there are three forces acting on a small volume of water with mass m at the surface. The first is the support force S perpendicular to the surface and therefore to the derivative of a curve y(x) to be determined delimiting the surface in Cartesian coordinates. The second is the gravitational force G pointing to earth’s center and the third is the centripetal force Z perpendicular to the rotation axis. All vector forces are supposed to be in equilibrium (no flow of water) so we have S + G + Z = 0, which yields the differential equation [itex]\left(x^2+y^2\right)^{3/2}=\frac{G M}{ω^2}\left(\frac{y' y}{x}+1\right)[/itex], where G is the gravitational constant and M the mass of the Earth. It’s solution is a curve near to an ellipse for small ω. You can see the maths in the Mathematica file in the attachement.

I am sure that I did the maths correctly but something may be wrong with the equilibrium of the forces. I would greatly appreciate any comments.
What I see is mostly gobbledygook of the form Cell[BoxData[{RowBox[{RowBox[{...}]}]...]. And lots and lots of magic numbers such as CellChangeTimes->{{3.5589800338894997`*^9, 3.5589801006420164`*^9}}. Is there any way you can write that for those of us who don't have Mathematica on our home computers?

One thing that I did see that is definitely wrong is a [itex]GM/r^2[/itex] term for gravitation, and nothing else. That simple form of Newton's law of gravitation is only valid for point masses and objects with a spherical mass distribution. The Earth is not a spherical body. The equatorial bulge makes the Earth's gravity field non-spherical. You need to account for the non-spherical nature of the gravity field in your computations. One way to do this is via spherical harmonics. There are no gravitational dipoles (mass is always positive), so the [itex]Y_1^m[/itex] terms vanish with the origin at the Earth's center of mass. Adding just the [itex]Y_2^0[/itex] term, the degree 2 zonal harmonic, to the [itex]Y_0^0[/itex] term ([itex]GM/r^2[/itex]) should come close to doing the trick.
Ulrich
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Oct14-12, 11:41 AM
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Quote Quote by D H View Post
Is there any way you can write that for those of us who don't have Mathematica on our home computers?.
The basic calculation is in the first attachement, the Mathematica code in the second as pdf file.
One thing that I did see that is definitely wrong is a [itex]GM/r^2[/itex] term for gravitation, and nothing else. That simple form of Newton's law of gravitation is only valid for point masses and objects with a spherical mass distribution. The Earth is not a spherical body. The equatorial bulge makes the Earth's gravity field non-spherical. You need to account for the non-spherical nature of the gravity field in your computations.
Yeah, this is right, I didn't think to that but for slow rotation I think it's a good approximation. That for such cases my calculation is correct can also be seen in another way: if the earth is non-rotating it has radius r and the gravitationnal force on a mass m on the surface at the equator is [itex]G M m/r^2[/itex]. Wenn the earth rotates, the mass m is elevated to a hight h where it has nearly the force [itex]G M m/(r+h)^2[/itex]. If h is little, the term [itex]1/(r+h)^2[/itex] can be approached by [itex]1/r^2-2 h/r^3[/itex]. The difference between these forces must be equal to the centripetal force so we get [itex]2 G M h/r^3 ≈ (r + h) w^2[/itex] and [itex]h≈r^4 w^2/(2 G M - r^3 w^2) [/itex]. This also yields about 10 km. That in reality it is 20 km may effectively be due to the crust that is upholding a shape from a time when the earth rotated faster. The atmospheric pression may also play a role. In any case I applied the same calculation for Jupiter that has no crust. There, the results fit better to reality.
One way to do this is via spherical harmonics. There are no gravitational dipoles (mass is always positive), so the [itex]Y_1^m[/itex] terms vanish with the origin at the Earth's center of mass. Adding just the [itex]Y_2^0[/itex] term, the degree 2 zonal harmonic, to the [itex]Y_0^0[/itex] term ([itex]GM/r^2[/itex]) should come close to doing the trick.
This may be true but I don't think that Mathematica will be able to solve such a differential equation...
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Ulrich
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#6
Oct14-12, 11:49 AM
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Quote Quote by Andre View Post
However we started with deeper oceans at the equator so the water has to flow to the poles. And mind there is a lot more equator area than polar area.
You are right, this could explain the flow showed in the documentary.
Ulrich
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#7
Oct14-12, 12:22 PM
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Quote Quote by Ulrich View Post
[itex]h≈(r^4 w^2)/(2 G M - r^3 w^2) [/itex]. This also yields about 10km.
Since this is the difference from the mean radius, the polar radius must have about 10km less such that the difference between equatorial and polar radius is 20km. So there must something be wrong with my calculation of the shape since it yields 10km for this last difference, but what?
Andre
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Oct14-12, 01:07 PM
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Quote Quote by Ulrich View Post
... That in reality it is 20 km may effectively be due to the crust that is ubholding a form from a time when the earth rotated faster...
From maybe "daily" coral rings it is assumed that the year had 400 days in the Devonian 400 million years ago. Would that add up?
D H
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Oct14-12, 02:57 PM
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Quote Quote by Ulrich View Post
Wenn the earth rotates, the mass m is elevated to a hight h where it has nearly the force [itex]G M m/(r+h)^2[/itex].
That's still invalid. You are using still using GM/r^2; all you've done is replaced r with r+h. It's just not valid. The difference in acceleration due to gravity (vector sum of gravitational acceleration and centrifugal acceleration) on our real Earth is 0.052 m/s^2. (9.832186 m/s^2 at the poles vs 9.780327 m/s^2 at the equator). Treating the Earth as a point mass results in a 0.066 m/s^2 difference for gravitational alone. Add in the 0.034 m/s^2 difference due to centrifugal acceleration and you're off by a factor of 2. This is why you are off by a factor of 2 in your estimation of the size of the bulge.
Ulrich
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Oct14-12, 03:21 PM
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Quote Quote by D H View Post
Treating the Earth as a point mass results in a 0.066 m/s^2 difference for gravitational alone.
If the matter density is a function of the radius alone, that is, if it is constant everywhere at the same distance from the gravity center of a mass distribution, the entire mass can be considered as a point mass. We made an exercise on that in classical mechanics in the second year. h=10km is very little compared to the 6378km of the radius (0.157%) so a spherical treatment should be a good approximation.
D H
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Oct14-12, 03:50 PM
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Quote Quote by Ulrich View Post
If the matter density is a function of the radius alone, that is, if it is constant everywhere at the same distance from the gravity center of a mass distribution, the entire mass can be considered as a point mass.
And that's not the case here. Think about it. The polar radius is 6356.752 km. The equatorial radius is 6378.137 km. At a distance of 6358 km from the center of the Earth, the density is nearly zero at the poles; air at 1.4 km up isn't very dense. The density is about 2.9 gm/cc at the same distance from the center of the Earth at the equator. Here 6358 km is 20 km under the surface.


We made an exercise on that in classical mechanics in the second year. h=10km is very little compared to the 6378km of the radius (0.157%) so a spherical treatment should be a good approximation.
No, it's not. You are off by a factor of 2, and this spherical approximation is by far the biggest source of your error. A simple accounting of the non-spherical nature of Earth's gravity yields an inverse flattening of [itex]\frac{GM}{a^3\omega^2}[/itex], or 288.9 (a bulge of 22.08 km). The actual value: 298.257223563 (a bulge of 21.384 km).
D H
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Oct14-12, 04:38 PM
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Quote Quote by Andre View Post
From maybe "daily" coral rings it is assumed that the year had 400 days in the Devonian 400 million years ago. Would that add up?
The Earth is a bit slow to respond to changes, but it's not that slow. The response time is in the tens of thousands of years. For example, even though its been about 10,000 years since the end of the last glaciation, the Earth is still experiencing glacial isostatic rebound. That ten to twenty thousand year response time is a blink of the eye compared to 400 million years. On that huge scale, the Earth is almost always in isostatic equilibrium. The equatorial bulge was bigger 400 million years ago.


As far as the "what if the Earth stopped spinning" show on Discovery, I suspect it was based on this article: http://www.esri.com/news/arcuser/0610/files/nospin.pdf.
D H
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Oct14-12, 05:18 PM
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Quote Quote by Ulrich View Post
The formula [itex]h≈r^4 w^2/(2 G M - r^3 w^2) [/itex] is a good approximation for small h.
No, it is not. First, just drop that [itex]r^3\omega^2[/itex] term of yours. It's tiny, less about 0.1% of GM. There's no reason to carry around a 0.1% correction term when you're off by a factor of 2.

And you are off by a factor of 2. A much, much better approximation is [itex]h\approx r^4 \omega^2 / \mu[/itex], where [itex]\mu=GM[/itex] is the Earth's standard gravitational parameter. Aside: The reason for using mu as opposed to GM: The Earth's mu is observable to nine or so decimal places. G is known to a paltry four places, and M is derived from mu and G.)

You are off by a factor of two because you continue to insist that you can use the simple Newtonian approximation. You can't. Newton's law still applies, of course; there's nothing relativistic going on here. However, what you need to do is to calculate the volume integral [itex]\int GM/r^2 dm[/itex], where r is the distance from the point on the surface of the Earth in question to the infinitesimal quantity of mass dm. Ignoring the non-spherical nature of the Earth's gravity field is the key cause of your error.

Another aside: The non spherical nature of the Earth's gravity field is a huge boon to some space scientists. Put a satellite in low Earth orbit in a slightly retrograde orbit, an inclination of about 98 degrees. Do it just right and the equatorial bulge will cause the satellite's orbit to precess by about one degree per day, perfectly in synch with the how the Sun appears to move over the seasons. Some sun synchronous satellites are in orbits where it is always close to local noon / local midnight right below the satellite, others where it is close to sunrise / sunset right below. And shades in between.


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