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Mercury's Perihelion Precession
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20 years 4 months ago #10274
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />What I find rather questionable in these works is that they use heliocentric coordinates throughout when the sun is in fact not an inertial system as it orbits the barycenter of the solar system (with a period corresponding to mainly the orbital period of Jupiter and a radius of about 7*10^5 km).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The barycenter is an arbitrary reference point, not an observable place or an origin point for any forces. For example, if we added Alpha Centauri to the dynamical system, nothing would change about the motions of solar system bodies because the gravity of Alpha Centauri is too weak. But the barycenter would shift half way to Alpha Centauri, which is 4 lightyears away (the closest bright star).
So the equations of motion actually used do in fact define just such an arbitrary barycenter to serve as a fixed reference point, whose location depends on the particular set of masses to be included. And the Sun's motion is then a part of the dynamical process; i.e., the Sun is <i>not</i> being taken as fixed. The equations of motion then consist of a direct part (e.g., forces on Earth's motion relative to the adopted barycenter) and an indirect part (forces on the Sun's motion relative to the barycenter). The difference of these two then gives the equations of motion of Earth relative to the Sun, which can be compared against observations. (Motion with respect to the barycenter cannot be observed.)
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In the barycentric system there is consequently an additional centrifugal force which should be taken into account as the positional observations of the planets as well as any time corrections are referred to the fixed stars.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a common fallacy. As Earth orbits the Sun, the Sun produces a centripetal force accelerating the Earth towards itself, but <i>there is no centrifugal force acting on the Earth</i>. If there were, the two forces would cancel and Earth's motion would remain a straight line. Instead, the reaction force is Earth pulling back on the Sun with a force equal to that of the Sun on the Earth. And that contributes to the Sun's motion relative to the barycenter.
In general, bodies in freefall do not experience centrifugal force. Centrifugal force is a pseudo-force (not a real one) used to "explain" (for example) why a revolving body on a tether applies tension to the tether. You can "feel" centrifugal force in a car making a turn, but what you really feel is parts of your body in contact with the car through friction applying a force of cohesion to free parts of your body still trying to move ahead on a straight line. This cohesion force prevents your body from coming apart when some parts of it are pulled in a different direction.
In space under normal circumstances, the only relevant force determining orbital motion is that of gravity. No forces of any kind are associated with the barycenter. And everything is in free fall (small tidal forces being the largest exception), so there are no centrifugal forces operating. -|Tom|-
<br />What I find rather questionable in these works is that they use heliocentric coordinates throughout when the sun is in fact not an inertial system as it orbits the barycenter of the solar system (with a period corresponding to mainly the orbital period of Jupiter and a radius of about 7*10^5 km).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The barycenter is an arbitrary reference point, not an observable place or an origin point for any forces. For example, if we added Alpha Centauri to the dynamical system, nothing would change about the motions of solar system bodies because the gravity of Alpha Centauri is too weak. But the barycenter would shift half way to Alpha Centauri, which is 4 lightyears away (the closest bright star).
So the equations of motion actually used do in fact define just such an arbitrary barycenter to serve as a fixed reference point, whose location depends on the particular set of masses to be included. And the Sun's motion is then a part of the dynamical process; i.e., the Sun is <i>not</i> being taken as fixed. The equations of motion then consist of a direct part (e.g., forces on Earth's motion relative to the adopted barycenter) and an indirect part (forces on the Sun's motion relative to the barycenter). The difference of these two then gives the equations of motion of Earth relative to the Sun, which can be compared against observations. (Motion with respect to the barycenter cannot be observed.)
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In the barycentric system there is consequently an additional centrifugal force which should be taken into account as the positional observations of the planets as well as any time corrections are referred to the fixed stars.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a common fallacy. As Earth orbits the Sun, the Sun produces a centripetal force accelerating the Earth towards itself, but <i>there is no centrifugal force acting on the Earth</i>. If there were, the two forces would cancel and Earth's motion would remain a straight line. Instead, the reaction force is Earth pulling back on the Sun with a force equal to that of the Sun on the Earth. And that contributes to the Sun's motion relative to the barycenter.
In general, bodies in freefall do not experience centrifugal force. Centrifugal force is a pseudo-force (not a real one) used to "explain" (for example) why a revolving body on a tether applies tension to the tether. You can "feel" centrifugal force in a car making a turn, but what you really feel is parts of your body in contact with the car through friction applying a force of cohesion to free parts of your body still trying to move ahead on a straight line. This cohesion force prevents your body from coming apart when some parts of it are pulled in a different direction.
In space under normal circumstances, the only relevant force determining orbital motion is that of gravity. No forces of any kind are associated with the barycenter. And everything is in free fall (small tidal forces being the largest exception), so there are no centrifugal forces operating. -|Tom|-
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20 years 4 months ago #10209
by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
The barycenter is an arbitrary reference point<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The barycenter is not an arbitrary reference point. It is uniquely determined by the positions and masses of all bodies in the solar system. Only other points in a uniform relative motion would be equivalent. The sun is not such a point.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So the equations of motion actually used do in fact define just such an arbitrary barycenter to serve as a fixed reference point, whose location depends on the particular set of masses to be included. And the Sun's motion is then a part of the dynamical process; i.e., the Sun is <i>not</i> being taken as fixed.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">As far as I can see, the sun <i>is</i> assumed as fixed as the observations are not corrected for the motion of the sun relative to the barycenter. Have a look at the paper of Morrison and Ward or the numerical simulation by Narlikar and Rana: the word 'barycenter' doesn't even appear once there. Everything is done using heliocentric elements and assuming the sun to be an inertial system.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In general, bodies in freefall do not experience centrifugal force. Centrifugal force is a pseudo-force (not a real one)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Of course the centrifugal force is not an active force but merely a passive one (or pseudo- force if you want) due to the inertia of masses. Nevertheless it has observational consequences: consider for instance a satellite orbiting the earth at a certain height. If you would assume now that the earth is not rotating (i.e. assuming that it is an inertial frame) you would interprete the apparent frequency of revolution of the satellite to the effect that the force of gravity has a lower value (for a satellite in a geostationary orbit you would even have to conclude that there is no gravity at all as otherwise it should be falling straight down).
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
The barycenter is an arbitrary reference point<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The barycenter is not an arbitrary reference point. It is uniquely determined by the positions and masses of all bodies in the solar system. Only other points in a uniform relative motion would be equivalent. The sun is not such a point.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So the equations of motion actually used do in fact define just such an arbitrary barycenter to serve as a fixed reference point, whose location depends on the particular set of masses to be included. And the Sun's motion is then a part of the dynamical process; i.e., the Sun is <i>not</i> being taken as fixed.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">As far as I can see, the sun <i>is</i> assumed as fixed as the observations are not corrected for the motion of the sun relative to the barycenter. Have a look at the paper of Morrison and Ward or the numerical simulation by Narlikar and Rana: the word 'barycenter' doesn't even appear once there. Everything is done using heliocentric elements and assuming the sun to be an inertial system.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">In general, bodies in freefall do not experience centrifugal force. Centrifugal force is a pseudo-force (not a real one)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Of course the centrifugal force is not an active force but merely a passive one (or pseudo- force if you want) due to the inertia of masses. Nevertheless it has observational consequences: consider for instance a satellite orbiting the earth at a certain height. If you would assume now that the earth is not rotating (i.e. assuming that it is an inertial frame) you would interprete the apparent frequency of revolution of the satellite to the effect that the force of gravity has a lower value (for a satellite in a geostationary orbit you would even have to conclude that there is no gravity at all as otherwise it should be falling straight down).
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
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20 years 4 months ago #10331
by Thomas
Replied by Thomas on topic Reply from Thomas Smid
I did now a rough mathematical analysis of the problem:
The distance r of a planet with mass m in a circular orbit around the sun (mass M) is given by the equation
G*(M+m)/r² = ω²*r,
where ω is the relative angular frequency of revolution and G the gravitational constant.
ω²*r is the centrifugal acceleration here which cancels the gravitational acceleration at any instant such as to maintain a constant distance r (r=5.8*10^7 km for Mercury).
According to General Relativity, this equation has to be corrected to
G*(M+m)/r² = ω²*[r-3ρ] ,
where ρ is the sun's Schwarzschild Radius GM/c² = 1.5 km (this results in the precession rate of 43"/100y for Mercury's perihelion if one takes into account the actual eccentricity of the orbit).
However, if one re-writes the original Newtonion equation and superimposes the movement of the sun around the barycenter, one gets in fact a very similar offset of the centrifugal acceleration, because now the equation has to be written as
G*(M+m)/r² = (ω +Δω*|Δr|/r)² *(r+Δr) ,
where Δω is the angular frequency of the sun around the barycenter and Δr the distance of the sun from the barycenter
(see www.physicsmyths.org.uk/imgs/orbit.gif for a schematic illustration; the order of the orbits has been reversed here)
If one evaluates the above expression and leaves out the terms linear in Δr and Δω (as these change sign (depending on the positions of Mercury and Jupiter) they lead only to periodical changes) one obtains
G*(M+m)/r² = ω²*[r + (Δω/ω)²*(Δr²/r)]
With ω and Δω given b the orbital periods of Mercury and the Sun (i.e. Jupiter) respectively and r and Δr by the corresponding orbital radii (Δr=7.4*10^5 km) this leads to an offset in the angular bracket of 3.9 km, so rather close to the relativistic theory of 3*1.5=4.5 km (the sign is different, but apparently this drops out anyway as the precession depends only on the square of this offset (see www.mathpages.com/rr/s6-02/6-02.htm ).
I am not necessarily saying that this effect due to the neglection of the motion of the sun around the barycenter could exactly explain all of the observed precession, but it is of a similar magnitude as the alleged relativistic effect and, if taken into account, would obviously invalidate GR unless further effects are found that re-establish the 43" discrepancy.
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
The distance r of a planet with mass m in a circular orbit around the sun (mass M) is given by the equation
G*(M+m)/r² = ω²*r,
where ω is the relative angular frequency of revolution and G the gravitational constant.
ω²*r is the centrifugal acceleration here which cancels the gravitational acceleration at any instant such as to maintain a constant distance r (r=5.8*10^7 km for Mercury).
According to General Relativity, this equation has to be corrected to
G*(M+m)/r² = ω²*[r-3ρ] ,
where ρ is the sun's Schwarzschild Radius GM/c² = 1.5 km (this results in the precession rate of 43"/100y for Mercury's perihelion if one takes into account the actual eccentricity of the orbit).
However, if one re-writes the original Newtonion equation and superimposes the movement of the sun around the barycenter, one gets in fact a very similar offset of the centrifugal acceleration, because now the equation has to be written as
G*(M+m)/r² = (ω +Δω*|Δr|/r)² *(r+Δr) ,
where Δω is the angular frequency of the sun around the barycenter and Δr the distance of the sun from the barycenter
(see www.physicsmyths.org.uk/imgs/orbit.gif for a schematic illustration; the order of the orbits has been reversed here)
If one evaluates the above expression and leaves out the terms linear in Δr and Δω (as these change sign (depending on the positions of Mercury and Jupiter) they lead only to periodical changes) one obtains
G*(M+m)/r² = ω²*[r + (Δω/ω)²*(Δr²/r)]
With ω and Δω given b the orbital periods of Mercury and the Sun (i.e. Jupiter) respectively and r and Δr by the corresponding orbital radii (Δr=7.4*10^5 km) this leads to an offset in the angular bracket of 3.9 km, so rather close to the relativistic theory of 3*1.5=4.5 km (the sign is different, but apparently this drops out anyway as the precession depends only on the square of this offset (see www.mathpages.com/rr/s6-02/6-02.htm ).
I am not necessarily saying that this effect due to the neglection of the motion of the sun around the barycenter could exactly explain all of the observed precession, but it is of a similar magnitude as the alleged relativistic effect and, if taken into account, would obviously invalidate GR unless further effects are found that re-establish the 43" discrepancy.
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
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20 years 4 months ago #10168
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Thomas</i>
<br />The barycenter is not an arbitrary reference point.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I gave you an example involving Alpha Centauri that showed that it is.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It is uniquely determined by the positions and masses of all bodies in the solar system.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">And suppose the solar system has an undiscovered "Planet X" with a mass of 0.001 that of the Sun (same as Jupiter) orbiting at a distance of 2000 au. Then the barycenter shifts outside the orbit of Mars with no detectable consequences for motions of the inner planets.
Indeed, one can move the barycenter to anyplace in the universe by including a distant galaxy of suitable mass in the dynamical system under consideration. That is what I mean by "arbitrary".
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Only other points in a uniform relative motion would be equivalent. The sun is not such a point.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I agree the Sun cannot serve as a barycenter. If you thought I was arguing for that, you missed my point altogether. But there are an infinite number of points that can serve as a barycenter or a "fixed" origin of coordinates. (Naturally, any such origin point is neither fixed nor truly inertial because it is moving rapidly and accelerating in a Galactic orbit. But it serves as a suitable anchor point for dynamical purposes.) My point is that the choice of barycenter from among the infinity of eligible points has no detectible consequences.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">As far as I can see, the sun <i>is</i> assumed as fixed as the observations are not corrected for the motion of the sun relative to the barycenter.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">On the contrary, the motion of the Sun is always considered. That is what the indirect portion of the equations of motion accomplishes. Do you really think that all dynamicists are that stupid?
The only time one could neglect solar motion is when orbits are taken as fixed ellipses, which basically means massless planets. But no dynamicist could be so foolish as to calculate only the perturbations of Jupiter on the Earth and neglect the perturbations of Jupiter on the Sun. Because the two perturbations are so similar (Earth and Sun are at a similar distance from Jupiter in a similar direction), neglecting Jupiter's perturbations on the Sun would make Earth's predicted orbit <i>worse</i> than assuming no perturbations at all.
Moreover, modern high-precision observations (VLBI, radar and laser randing, spacecraft, etc.) have precisions of meters or better over hundreds of millions of kilometers. Do you think NASA could thread Saturn's rings with the Cassini spacecraft if they neglected an important force?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Have a look at the paper of Morrison and Ward or the numerical simulation by Narlikar and Rana: the word 'barycenter' doesn't even appear once there. Everything is done using heliocentric elements and assuming the sun to be an inertial system.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Both analytic theories and numerical integrations are always (necessarily) done in an inertial reference system. The choice of which inertial frame is a matter of convenience. Once the motions are determined, the coordinates of the Sun are subtracted from the coordinates of each body (including the Sun) to reset the origin to the Sun. That is done purely for purposes of comparing theory with observations because the barycenter is not an observable point and has an arbitrary location.
This is the logical equivalent of predicting the motion of an artificial satellite using a particular observer's coordinates as the origin. No one is assuming the satellite revolves around the observer, or that the observer's location influences the satellite in any way. But the observer's coordinates are the most convenient origin for both predicting where the satellite will be seen, and for comparing the observer's observations with those predictions.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Of course the centrifugal force is not an active force but merely a passive one (or pseudo-force if you want) due to the inertia of masses.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">"Passive" and "pseudo" are not equivalent. The former implies a real force, the latter implies an imaginary one. Psuedo-forces do not exist in nature. They are introduced for observers or origin points that are themselves accelerating, so they can explain the anomalous behavior of target bodies. But there is no such pseudo-force acting on the target body. The non-linear motion of the target body is merely an illusion created by the accelerated motion of the observer or origin reference point.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">ù²*r is the centrifugal acceleration here which cancels the gravitational acceleration at any instant such as to maintain a constant distance r (r=5.8*10^7 km for Mercury).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a mistaken impression about what centrifugal force is. The motion of Mercury would be linear but for the existence of a centripetal acceleration due to gravity that bends its motion into an ellipse (or possibly a circle; but a circle cannot have perihelion motion). There is no centrifugal acceleration. All of the velocity change (the definition of acceleration) is directed toward the Sun.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">According to General Relativity, this equation has to be corrected to
(M+m)/r² = ù²*[r-3ñ] ,
where ñ is the Schwarzschild Radius GM/c² = 1.5 km (this results in the precession rate of 43"/100y for Mercury's perihelion if one takes into account the actual eccentricity of the orbit).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I don't know where you found this notion, but it is wrong. GR has three contributions to perihelion motion, none of which has any association with the Schwarzschild radius.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">However, if one re-writes the original Newtonion equation and superimposes the movement of the sun around the barycenter, one gets in fact a very similar offset of the centrifugal acceleration<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a wrong analogy based on a wrong assumption.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">the orbital periods of Mercury and the Sun (i.e. Jupiter)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Jupiter's period is not a good approximation of the Sun's motion with respect to a barycenter because Saturn's effect is comparable to Jupiter's. So this assumption is also wrong.
I hope the preceding is enough to get you back to the drawing board and interested in learning some basic mechanics, t
<br />The barycenter is not an arbitrary reference point.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I gave you an example involving Alpha Centauri that showed that it is.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It is uniquely determined by the positions and masses of all bodies in the solar system.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">And suppose the solar system has an undiscovered "Planet X" with a mass of 0.001 that of the Sun (same as Jupiter) orbiting at a distance of 2000 au. Then the barycenter shifts outside the orbit of Mars with no detectable consequences for motions of the inner planets.
Indeed, one can move the barycenter to anyplace in the universe by including a distant galaxy of suitable mass in the dynamical system under consideration. That is what I mean by "arbitrary".
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Only other points in a uniform relative motion would be equivalent. The sun is not such a point.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I agree the Sun cannot serve as a barycenter. If you thought I was arguing for that, you missed my point altogether. But there are an infinite number of points that can serve as a barycenter or a "fixed" origin of coordinates. (Naturally, any such origin point is neither fixed nor truly inertial because it is moving rapidly and accelerating in a Galactic orbit. But it serves as a suitable anchor point for dynamical purposes.) My point is that the choice of barycenter from among the infinity of eligible points has no detectible consequences.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">As far as I can see, the sun <i>is</i> assumed as fixed as the observations are not corrected for the motion of the sun relative to the barycenter.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">On the contrary, the motion of the Sun is always considered. That is what the indirect portion of the equations of motion accomplishes. Do you really think that all dynamicists are that stupid?
The only time one could neglect solar motion is when orbits are taken as fixed ellipses, which basically means massless planets. But no dynamicist could be so foolish as to calculate only the perturbations of Jupiter on the Earth and neglect the perturbations of Jupiter on the Sun. Because the two perturbations are so similar (Earth and Sun are at a similar distance from Jupiter in a similar direction), neglecting Jupiter's perturbations on the Sun would make Earth's predicted orbit <i>worse</i> than assuming no perturbations at all.
Moreover, modern high-precision observations (VLBI, radar and laser randing, spacecraft, etc.) have precisions of meters or better over hundreds of millions of kilometers. Do you think NASA could thread Saturn's rings with the Cassini spacecraft if they neglected an important force?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Have a look at the paper of Morrison and Ward or the numerical simulation by Narlikar and Rana: the word 'barycenter' doesn't even appear once there. Everything is done using heliocentric elements and assuming the sun to be an inertial system.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Both analytic theories and numerical integrations are always (necessarily) done in an inertial reference system. The choice of which inertial frame is a matter of convenience. Once the motions are determined, the coordinates of the Sun are subtracted from the coordinates of each body (including the Sun) to reset the origin to the Sun. That is done purely for purposes of comparing theory with observations because the barycenter is not an observable point and has an arbitrary location.
This is the logical equivalent of predicting the motion of an artificial satellite using a particular observer's coordinates as the origin. No one is assuming the satellite revolves around the observer, or that the observer's location influences the satellite in any way. But the observer's coordinates are the most convenient origin for both predicting where the satellite will be seen, and for comparing the observer's observations with those predictions.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Of course the centrifugal force is not an active force but merely a passive one (or pseudo-force if you want) due to the inertia of masses.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">"Passive" and "pseudo" are not equivalent. The former implies a real force, the latter implies an imaginary one. Psuedo-forces do not exist in nature. They are introduced for observers or origin points that are themselves accelerating, so they can explain the anomalous behavior of target bodies. But there is no such pseudo-force acting on the target body. The non-linear motion of the target body is merely an illusion created by the accelerated motion of the observer or origin reference point.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">ù²*r is the centrifugal acceleration here which cancels the gravitational acceleration at any instant such as to maintain a constant distance r (r=5.8*10^7 km for Mercury).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a mistaken impression about what centrifugal force is. The motion of Mercury would be linear but for the existence of a centripetal acceleration due to gravity that bends its motion into an ellipse (or possibly a circle; but a circle cannot have perihelion motion). There is no centrifugal acceleration. All of the velocity change (the definition of acceleration) is directed toward the Sun.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">According to General Relativity, this equation has to be corrected to
(M+m)/r² = ù²*[r-3ñ] ,
where ñ is the Schwarzschild Radius GM/c² = 1.5 km (this results in the precession rate of 43"/100y for Mercury's perihelion if one takes into account the actual eccentricity of the orbit).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I don't know where you found this notion, but it is wrong. GR has three contributions to perihelion motion, none of which has any association with the Schwarzschild radius.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">However, if one re-writes the original Newtonion equation and superimposes the movement of the sun around the barycenter, one gets in fact a very similar offset of the centrifugal acceleration<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This is a wrong analogy based on a wrong assumption.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">the orbital periods of Mercury and the Sun (i.e. Jupiter)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Jupiter's period is not a good approximation of the Sun's motion with respect to a barycenter because Saturn's effect is comparable to Jupiter's. So this assumption is also wrong.
I hope the preceding is enough to get you back to the drawing board and interested in learning some basic mechanics, t
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20 years 4 months ago #10169
by Jim
Replied by Jim on topic Reply from
Tom, You say above only GR has an explaination for the 43" of the orbit of Mercury being kicked around here. Does MM support GR in this detail?
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20 years 4 months ago #10912
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />You say above only GR has an explaination for the 43" of the orbit of Mercury being kicked around here. Does MM support GR in this detail?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I said "The difference is 43"/cy, the part explained only by GR and later field theories of gravity." MM is one of those "later field theories of gravity". MM makes the same prediction as GR for pericenter advance when only one mass is of significant size (as for Sun-Mercury), and when both masses are equal. It makes a different (smaller) prediction when both masses are significant but unequal. -|Tom|-
<br />You say above only GR has an explaination for the 43" of the orbit of Mercury being kicked around here. Does MM support GR in this detail?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I said "The difference is 43"/cy, the part explained only by GR and later field theories of gravity." MM is one of those "later field theories of gravity". MM makes the same prediction as GR for pericenter advance when only one mass is of significant size (as for Sun-Mercury), and when both masses are equal. It makes a different (smaller) prediction when both masses are significant but unequal. -|Tom|-
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