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Effect of lunar/solar gravity on shape of earth 
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#55
Aug2511, 05:05 AM

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#56
Aug2511, 05:14 AM

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The computation I showed gives you a differential in forces acting on the front and far sides of the planet or other object. All it says is that the object will be stretched out. Id est, there will be some sort of bulging. Whether that bulging is symmetric or not is a matter of higher order effects which we have not yet discussed.
If you want to use Mathematica to graph what the Earth's surface would look like if it was a perfect fluid, taking into account Earth's gravity, Earth's rotation, and Moon's gravity, I can help you out with this. But maybe you should send me a PM, so that we don't clutter this topic. Edit: Unless there are other people who want to see it. I suppose, I can just set it up in Mathematica and upload the notebook. 


#57
Aug2511, 06:25 AM

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#58
Aug2611, 01:18 PM

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#59
Sep1311, 06:40 AM

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1. Gamow as per the OP. 2. Feynman, and [extrinsic to this post] [ crackpot link deleted ] daniel6874, the link you cited is not a good link for this site.. And that F. himself saw problems with his idea. 3. A quaint older text suggesting that the pull of the moon is stronger on the earth than the water, so the earth shifts like an eggyolk toward the moon, leaving a bulge on the far side. 4. A reasonable note by xts that water for high tides is shifted from lowtide areas, and glosses from others [outside this post] on the effect of resonant land formations and separation of oceans to explain tidal height discrepancies. 5. An series of notes by K^2 that suggest the explanation of forces depends on choice of coordinate system. 6. Cited by Russ, http://www.lhup.edu/~dsimanek/scenario/tides.htm, an author who calculates the "tidegenerating force" as the gradient of the moon's gravitational potential [I mistakenly said "force"], with a picture that seems to explain tidal bulges in terms of gravitation alone. The author then proceeds to outline what appears to be K^2's approachthe use of a rotating coordinate system with barycenter at the origin, noting that the mathematical origin of the "fictitious" forces is often omitted in this approach. 7. A gloss by harrylin on essentially the eggyolk theory of 3 above, but updated I think to account for discussion in the post. 8. A.T. posts http://www.vialattea.net/maree/eng/index.htm, which seems to agree with K^2 but like Russ emphasizes the nonuse of fictitious forces. I am tempted to conclude that Russ, K^2, Gamow, and Feynman are all correct, taking into account simplifications and assumptions. After reading the author's homepage in 2 above I no longer think it bears pursuing. Thanks to all who helped with this apparently notsoeasy question. 


#60
Sep1311, 08:09 AM

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I'll start off by noting that I am not a fan of the centrifugal force explanation of the tides.
That said, one cannot say that a centrifugal force explanation is incorrect. K^2 was spoton correct way back on page 2: The explanations that result from a nonrotating perspective and a barycentric frame rotating with the Moon's orbital rate are valid with respect to the tidal forcing functions. Both are correct with regard to the shape of the Earth itself in response to these forcing functions. The Earth itself undergoes tidal deformations. See http://en.wikipedia.org/wiki/Earth_tide, for example. Neither explanation is correct with regard to oceanic tides. There is no tidal bulge. There are two huge land masses, the Americas and Eurasia+Africa, that prevent this from arising. To get a correct picture of the oceans' responses to the tidal forces you need to go to yet another perspective, one based on the rotating Earth. There you'll get a picture of amphidromic points, points with virtually no tides, about which the tides appear to rotate. Picture: http://svs.gsfc.nasa.gov/stories/top...erns_hires.tif Movie: http://svs.gsfc.nasa.gov/vis/a000000...01332/full.mov Since all frames are equally valid, I suppose one could come up with an explanation of the ocean tides from the perspective of a nonrotating frame or a frame that rotates with the orbit of the Earth and Moon about one another. Nobody does that because such an explanation would be hideously complex. Addendum An example of a hideously bad explanation based on centrifugal force, courtesy of our government: http://coops.nos.noaa.gov/restles3.html. This is so convoluted that I can't see if they've done the math right. There's just too much hand waving and there is no math to check. If they did do the math right they would of course have come up with the same answer as the much, much simpler gravity gradient based explanation. 


#61
Sep1311, 09:24 AM

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No matter how many times I say I understand that the math works out, people keep harping on it, so I must not be making myself clear. Let me try a completely different approach:
I dropped my pen and it fell to the ground. The centrifugal force due to the earth's rotation caused the pen to fall to the ground. True or false? And why or why not? 


#62
Sep1311, 09:52 AM

P: 788

If you think about it carefully, rotating coordinates aren't so counterintuitive for explaining tideproducing forces. The tidal "bulges" (in all the simplified examples here) are stationary relative to the earthmoon line, not to relative to fixed points on a non rotating earth in an inertial frame. 


#63
Sep1311, 09:56 AM

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But physicswise, I'd like to hear why anybody would explain a pen falling toward the ground as being due to the centrifugal force due to the earth's rotation. 


#64
Sep1311, 10:22 AM

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#65
Sep1311, 10:30 AM

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#66
Sep1311, 10:43 AM

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You're missing my point: I don't think it works that way, but that is what I think I am seeing from others.
What I'm seeing from others in this thread is this: 1. Since we can construct a mathematical model of the tides that incorporates a centrifugal force component, we can say that the centrifugal force causes the tides. So why can't we then say this: 2. Since we can construct a mathematical model of the motion of a falling pen that incorporates a centrifugal force component, we can say that the centrifugal force causes the tides. Why is one correct and the other not? 


#67
Sep1311, 11:36 AM

P: 3,187

'the effect of the external gravitational force produced by another astronomical body may be different at different positions on the earth because the magnitude of the gravitational force exerted varies with the distance of the attracting body.' (but why "may be"?!) Also (and happily!) they do point out that 'It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory.' PS: note that I fully agree with your criticism on the way they phrase things. 


#68
Sep1311, 12:19 PM

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Let g' be the local gravitational acceleration vector due to the Earth's mass only. From the perspective of a frame rotating with the Earth, your pen accelerates Earthward at g' due to the Earth's mass and accelerates outward at r×Ω^{2} due to the Earth's rotation. Here r is the vector from the Earth's rotation axis to the pen and Ω is the Earth's rotation vector, 1 revolution per sidereal day. The pen's net acceleration is g=g'r×Ω^{2}. Assuming that gravity and centrifugal acceleration change very little over the short fall, the time the pen takes to fall is given by [itex]\sqrt{2h/g}[/itex]. This calculation becomes even easier if you just use g in the first place rather than g'. For example, the standard value for Earth gravity, g_{0}=9.80665 m/s^{2} includes both the gravitational and centrifugal forces at sea level and at a latitude of 45.5 degrees. From the perspective of an inertial observer, the only force acting on the pen is gravity. The pen has an initial horizontal velocity due to the Earth's rotation that puts it on a curved trajectory. That initial horizontal velocity coupled with the Earth's curvature means the pen will fall for a distance slightly greater than the initial height h above the surface of the Earth. Account for that increase in the vertical distance traveled by the pen and make a simple second order assumption (the first order correction is zero) and you will get the exact same answer as above, [itex]\sqrt{2h/g}[/itex]. Just not as easily. This is one of those occasions where working in a rotating frame is easier than an inertial one. 


#69
Sep1311, 12:59 PM

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Consider a moon the same mass as ours that is in an orthogonal orbit about some planet the same size as the Earth. (Orthogonal orbit: An orbit with a zero angular momentum. When the moon hits the planet, the impact angle will be orthogonal to the ground.) At the instant that this hypothetical moon is one lunar distance from the planet, the tidal forces exerted by this moon on that planet will be exactly the same as the tidal forces exerted by our orbiting Moon on the Earth. Explaining that this is the case is a piece of cake if you use a gravity gradient based explanation. Now try from explaining using a centrifugal force based explanation. The reason that centrifugal forces due to the Earth's rotation must be disregarded is because those forces have already been taken into account. The starting point for an Earthfixed based explanation of the tides is the geoid. The geoid is an equipotential surface of the gravitational plus centrifugal forces. Accounting for those centrifugal force again would be a double booking. 


#70
Sep1311, 01:28 PM

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#71
Sep1311, 02:21 PM

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Compare that to a gravity gradient based approach. The desired outcome is the gravitational acceleration due to the Moon (or Sun) relative to the Earth as a whole. The Earth as a whole is accelerating toward the Moon by [tex]{\boldsymbol a}_e = \frac {GM_m}{{\boldsymbol R}^3}\,{\boldsymbol R}[/tex] where R is the vector from the center of the Earth to the center of the Moon. The gravitational acceleration toward the Moon at some point r away from the center of the Earth is [tex]{\boldsymbol a}_p = \frac {GM_m}{{\boldsymbol R}{\boldsymbol r}^3}\,({\boldsymbol R}{\boldsymbol r})[/tex] The relative acceleration is the difference between these two vectors: [tex]{\boldsymbol a}_{\mbox{tidal}} = {\boldsymbol a}_p  {\boldsymbol a}_e[/tex] The vector R can always be expressed in terms of a unit vector [itex]\hat{\boldsymbol r}[/itex] directed along [itex]\boldsymbol r[/itex] and some other unit vector [itex]\hat{\boldsymbol{\theta}}[/itex] normal to [itex]\hat{\boldsymbol r}[/itex]: [tex]{\boldsymbol R} = R(\cos\theta \,\hat{\boldsymbol r} + \sin\theta\,\hat{\boldsymbol{\theta}})[/tex] Using this notation and making a first order approximation for r<<R yields [tex]{\boldsymbol a}_{\mbox{tidal}} \approx \frac {GM_m r}{R^3} \left((3\cos^2\theta1)\,\hat{\boldsymbol r} + 3\cos\theta\sin\theta\,\hat{\boldsymbol {\theta}}\right)[/tex] Note that this covers the entire globe, not just the sublunar point. Try to do the same with a centrifugalbased approach. This is one of those cases where the inertial explanation is much easier, and much more general, than a rotating frame explanation. 


#72
Sep1311, 02:24 PM

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