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Effect of lunar/solar gravity on shape of earth 
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#73
Sep1311, 03:19 PM

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Personally, I found it kind of an interesting exercise to convince myself (mathematically) that several of the explanations offered earlier in the thread were indeed equivalent. As a side bonus, it tends to help you understand where people arguing from the various viewpoints are (mutually) confused. 


#74
Sep1311, 05:24 PM

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1. Because including a centrifugal force component has a real/nonzero effect on the acceleration of a dropped object, we can say that both gravity and rotation play a role in causing the measured acceleration. One causes positive acceleration, one causes negative acceleration, but both together form the cause of the observed experimental result. The corollary of this is what I am trying to convey in this thread: 2. Because including a centrifugal force component has no effect on the calculated tidal forces, we can say that the centrifugal force does not play any role in the cause the tidal force. 


#75
Sep1311, 06:52 PM

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Starting with the first sentence. The tideraising forces at the earth's surface thus result from a combination of basic forces: (1) the force of gravitation exerted by the moon (and sun) upon the earth; and (2) centrifugal forces produced by the revolutions of the earth and moon (and earth and sun) around their common centerofgravity (mass) or barycenter.Right of the bat, this is just wrong. First off, what centrifugal force? To have a centrifugal force you have to have a rotating frame. There is no centrifugal force in an inertial frame in which the moon orbits the earth. There is a centrifugal force in a frame that rotates with the earthmoon system, but in that frame the earth and the moon are stationary. They are not orbiting one another in that frame. A centrifugal force explanation of orbits can be useful at times. A geosynchronous satellite sits in a fixed position in the sky from our earthbound perspective. From our perspective, such a satellite indeed does have gravitational and centrifugal forces that are in balance, and per Newton's first law (extended to noninertial frames), the satellite remains stationary. A rotating frame in which the earth and moon do not move can be useful at times. Such a frame is very useful for describing the Lagrange points, for example, and also is very nice for depicting the trajectory of a satellite in transit from the earth to the moon. Continuing with the next paragraph, With respect to the center of mass of the earth or the center of mass of the moon, the above two forces always remain in balance (i.e., equal and opposite). In consequence, the moon revolves in a closed orbit around the earth, without either escaping from, or falling into the earth  and the earth likewise does not collide with the moon.That is the silly centrifugal force description of an orbit. And it is wrong. Look at it from the perspective of Newton's first law. If the two forces truly are in balance (i.e., equal and opposite), there is no closed orbit. There is only straight line motion or no motion at all. Now let's look at the image that follows. You will see a similar image in most descriptions that invoke centrifugal force. The moon and earth are clearly shown as orbiting one another here. There is no centrifugal force in this diagram. Let's go on to the next diagram. And this is flat out wrong. What's wrong is that this site (and almost every text that invokes centrifugal force) has the centrifugal force identical at the center of the earth, at the sublunar point, and at its antipode. The centrifugal force at a point removed from the axis of rotation by a distance r is rΩ^{2}. The sublunar point is only 1068 miles from the barycenter; the center of the earth, 2895 miles, and the antipodal point, 6858 miles. The centrifugal force is not the same at these three points. Do the math right and you do not get the tidal forces as shown in the diagram from the perspective of this rotating frame. That's okay, though. Force is a frame dependent quantity when one allows fictitious forces to enter the picture. We don't want the tidal forces as observed from the perspective of this frame. We want the tidal forces as observed on the earth. To get a correct explanation we need to move the origin to the center of the earth  and we need to get rid of that monthly rotation. The centrifugal forces that we added in need to be subtracted out. What we're left with is a nonrotating with origin at the center of the Earth. This frame is the earthcentered inertial frame. Despite its name, this isn't really an inertial frame; it is accelerating toward the moon and the sun. This acceleration of the origin results in a fictitious force, and it is a uniform fictitious force. This fictitious force is the inertial force, not the centrifugal force. There is little need to invoke the earthmoon barycenter here and there is absolutely no need to invoke the centrifugal force. 


#76
Sep1411, 03:59 AM

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#77
Sep1411, 10:35 AM

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Is centrifugal force really this confusing? And finally, to respond to this oft repeated point: 


#78
Sep1411, 03:00 PM

P: 4,049

 The gravitational gradient is the same in every reference frame  The tidal effect computed in every reference frame must be the the same Therefore:  The sum of all inertial forces (which might arise from the choice of the reference frame) must not affect the tidal effect. 


#79
Sep1411, 03:21 PM

P: 4,049

2) If the centrifugal force vector is the same for A & B & C (as shown in the picture) then it cannot cause/affect the tides, so why show/mention it in an explanation of tides? 


#80
Sep1411, 03:51 PM

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#81
Sep1411, 04:11 PM

P: 4,049

In which rotating frame would the centrifugal force vector be the same for A & B (as shown in the picture)? 


#82
Sep1411, 04:39 PM

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#83
Sep1411, 05:16 PM

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< Moon's gravity is pulling to the left with a r^{2} fall off. > Uniform centrifugal component pulls to the right  In the Earth's center they cancel  In A Moon's gravity wins and creates a bulge to the left  In B Uniform centrifugal component wins and creates a bulge to the right 


#84
Sep1511, 03:30 AM

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Here's a web site where an oceanographer does it right: http://oceanworld.tamu.edu/resources...apter17_04.htm. Emphasis mine: Note that many oceanographic books state that the tide is produced by two processes: i) the centripetal acceleration at earth's surface as earth and moon circle around a common center of mass, and ii) the gravitational attraction of mass on earth and moon. However, the derivation of the tidal potential does not involve centripatal acceleration, and the concept is not used by the astronomical or geodetic communities.This is a very nice description of the tides. Stewart starts off in terms of potential, develops the horizontal component of the tidal force, and then proceeds to dive deep, eventually reaching the concept of Doodson numbers. Let's see what that constant [itex]r\omega^2[/itex] term of yours really is. The angular velocity of a pair of objects orbiting one another circularly due to gravity is given by [tex]\omega^2= \frac{G(m_1+m_2)}{R^3}[/tex] where ω is the angular velocity, G is the universal gravitational constant, m_{1} and m_{2} are the masses of the two bodies, and R is the distance between the two bodies. (This generalizes very nicely to elliptical orbits; it is Kepler's third law.) Multiply both sides of the above by r_{1}, the distance between the center of mass and m_{1}, and we have the desired rω^{2} term on the left hand side: [tex]r_1\omega^2 = \frac{G(m_1+m_2)r_1}{R^3}[/tex] The relationship between r_{1} and R is given by m_{1}r_{1}=m_{2}r_{2}=m_{2}(Rr_{1}) or (m_{1}+m_{2})r_{1}=m_{2}R. Substituting this into the right hand side of the equation for r_{1}ω^{2} yields [tex]r_1\omega^2 = \frac{Gm_2}{R^2}[/tex] Recognizing that our generic body 1 is the Earth and that body 2 is the Moon, we have [tex]r\omega^2 = \frac{Gm_{\text{M}}}{R^2}[/tex] The right hand side is the translational acceleration of the Earth toward the Moon. This is what you really want because the answer you really want is in a frame with origin at the center of the Earth and rotating with the Earth. You can use a centrifugal force concept to develop those forces in that frame with origin at the barycenter and rotating at the EarthMoon orbital rate. Do it right and you will get some goofy answers. That's okay though because those answers are to a question other than the question we are trying to answer. To answer the question that was asked, you are going to have to transfer those forces from that frame to the Earthcentered, Earthfixed frame. All of those centrifugal terms you so painstakingly added in have to be painstakingly subtracted out. A barycentric POV misses one fictitious force, that due to the translational acceleration of the Earth toward the Moon. This missed term also needs to be taken into account when transforming those forces from the rotating barycentric frame to an Earthcentered frame. Do the math right and you will get the same answer as you would have gotten by starting in an Earthcentered frame in the first place. So why not start in an Earthcentered frame in the first place? For one thing, the math is a lot easier, and for another, it avoids all this going around your elbow to get to your thumb business. Just because the fictitious centrifugal force in the rotating barycentric frame at the center of the Earth has the same value as the fictitious inertial force in an Earthcentered at the center of the Earth does not mean these are the same fictitious force. They are very different fictitious forces. 


#85
Sep1511, 04:22 AM

P: 4,049

The decomposition of the centrifugal force in the barycentric rest frame is explained here in Fig 12 http://www.vialattea.net/maree/eng/index.htm http://www.vialattea.net/maree/eng/index.htm This uniform inertial force field turns this: into this: So I think it is OK to use that uniform inertial force field to explain why the water on the opposite side is accelerated away from the fixed Earth's center. What is misleading, is calling that uniform inertial force field "centrifugal force". 


#86
Sep1511, 06:07 AM

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There is one, and only one, fictitious force in Newtonian mechanics that is everywhere uniform: The fictitious force that results from choosing an origin that is accelerating with respect to an inertial frame. The simple answer, that the tidal acceleration is the difference between the gravitational acceleration of the Earth as a whole toward the Moon from the gravitational acceleration at some point of interest toward the Moon, looks just as handwavy as getting rid of those "residual vectors". What justifies this subtraction? What justifies it is that the question implicitly asks us to look at things from the perspective of an accelerating reference frame (accelerating from a Newtonian perspective, that is). Such a perspective results in a d'Alembert force, the inertial force resulting from the acceleration of the origin of the reference frame.  You referenced this website, http://www.vialattea.net/maree/eng/index.htm, in your post. The author opens with At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of the second tidal bulge, the one opposite to the Moon. Often they invoke the centrifugal force as an "explanation". But centrifugal force is a fictitious force, and we can't justify a real effect with a fictitious force, can we?That is perhaps a bit too dismissive. Moreover, a physicist who comes from a GR background would most likely say that it is wrong. Let me parody Sirtoli's opening paragraph to demonstrate why (these are [b]not[b] Sirtoli's words): At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of two tidal bulges. Often they invoke the gravitational force as an "explanation". But gravitational force is a fictitious force, and we can't justify a real effect with a fictitious force, can we? 


#87
Sep1511, 06:39 AM

P: 4,049




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