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Earth ellipsoid due to rotation 
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#1
Oct1312, 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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#2
Oct1412, 07:18 AM

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#3
Oct1412, 08:11 AM

PF Gold
P: 5,458

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 a_{1} and b_{1}, due to the same forces, Although likely, I'm not sure that ratio a/b is the same as a_{1}/b_{1}, but lets assume that it is. Now obviously a_{1}a > b_{1}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. 


#4
Oct1412, 08:23 AM

Mentor
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Earth ellipsoid due to rotation
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 nonspherical. You need to account for the nonspherical 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. 


#5
Oct1412, 11:41 AM

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#6
Oct1412, 11:49 AM

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#7
Oct1412, 12:22 PM

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#8
Oct1412, 01:07 PM

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#9
Oct1412, 02:57 PM

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#10
Oct1412, 03:21 PM

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#11
Oct1412, 03:50 PM

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#12
Oct1412, 04:38 PM

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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. 


#13
Oct1412, 05:18 PM

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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 nonspherical 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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