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#1
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Has anyone read Feynman's "lost lecture" where he uses "plane geometry"
only to prove elliptical orbits of planets. And if so is there a hole in his presentation. Cheers, Don W |
#2
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"Donald Wilgus" wrote in message
. com... Has anyone read Feynman's "lost lecture" where he uses "plane geometry" only to prove elliptical orbits of planets. And if so is there a hole in his presentation. Cheers, Don W Hi Don Why would there be a hole in his presentation? The derivation of elliptic orbits based on Euclidian Geometry has been well tested over the 300 years since Newton fist preformed the derivation. |
#3
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John,
Well, because he used a just "plane geometry" in his unique derivation. No calculus, etc. I had heard that there might be a problem in the approach that he used. Before I really got into it, I wanted to know if there really was a problem. Cheers, Don W "John Zinni" wrote in message .. . "Donald Wilgus" wrote in message . com... Has anyone read Feynman's "lost lecture" where he uses "plane geometry" only to prove elliptical orbits of planets. And if so is there a hole in his presentation. Cheers, Don W Hi Don Why would there be a hole in his presentation? The derivation of elliptic orbits based on Euclidian Geometry has been well tested over the 300 years since Newton fist preformed the derivation. |
#4
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"Donald Wilgus" wrote in message
m... John, Well, because he used a just "plane geometry" in his unique derivation. No calculus, etc. I had heard that there might be a problem in the approach that he used. Before I really got into it, I wanted to know if there really was a problem. Cheers, Don W Sorry, thought you were going somewhere else with this. I'm going to engage in some hand waving here because I don't seem to be able to find the references I want. (As such, I will happily stand corrected if someone knows better) I believe that any purely geometric treatment of the problem must, necessarily, be just an approximation. The error associated with the approximation can be made to be as small as we like, but when we start to do this we are getting close to the realm were The Calculus steps into the picture. |
#5
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"John Zinni" wrote in message
... "Donald Wilgus" wrote in message m... John, Well, because he used a just "plane geometry" in his unique derivation. No calculus, etc. I had heard that there might be a problem in the approach that he used. Before I really got into it, I wanted to know if there really was a problem. Cheers, Don W Sorry, thought you were going somewhere else with this. I'm going to engage in some hand waving here because I don't seem to be able to find the references I want. (As such, I will happily stand corrected if someone knows better) I believe that any purely geometric treatment of the problem must, necessarily, be just an approximation. The error associated with the approximation can be made to be as small as we like, but when we start to do this we are getting close to the realm were The Calculus steps into the picture. John, Well, Feynman boasts about this approach. So I am getting his lecture (the lost one) from local library. Has voice CD and some explanatory text by someone. There is a Java script on the Internet to demo some of this approach .... but I found it incomplete (for my understanding). Cheers, Don W. |
#6
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"John Zinni" wrote in message
... I believe that any purely geometric treatment of the problem must, necessarily, be just an approximation. The error associated with the approximation can be made to be as small as we like, but when we start to do this we are getting close to the realm were The Calculus steps into the picture. Do you mean a geometric treatment of this particular problem, or do you mean that geometric proofs, in general, are necessarily approximations? |
#7
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"Greg Neill" wrote in message
... "John Zinni" wrote in message ... I believe that any purely geometric treatment of the problem must, necessarily, be just an approximation. The error associated with the approximation can be made to be as small as we like, but when we start to do this we are getting close to the realm were The Calculus steps into the picture. Do you mean a geometric treatment of this particular problem, or do you mean that geometric proofs, in general, are necessarily approximations? No no, I mean just the problem at hand (elliptic orbits). |
#8
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"John Zinni" wrote in message
... "Greg Neill" wrote in message ... "John Zinni" wrote in message ... I believe that any purely geometric treatment of the problem must, necessarily, be just an approximation. The error associated with the approximation can be made to be as small as we like, but when we start to do this we are getting close to the realm were The Calculus steps into the picture. Do you mean a geometric treatment of this particular problem, or do you mean that geometric proofs, in general, are necessarily approximations? No no, I mean just the problem at hand (elliptic orbits). Good Lord, I'm not about to throw geometry out the window. Euclid is spinning in his grave. ;-) |
#9
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"Donald Wilgus" wrote in message
. com... John, Well, Feynman boasts about this approach. So I am getting his lecture (the lost one) from local library. Has voice CD and some explanatory text by someone. There is a Java script on the Internet to demo some of this approach .... but I found it incomplete (for my understanding). Hi Don Could you post the link to this please, I'd like to take a look. |
#10
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"John Zinni" wrote in message
... "Donald Wilgus" wrote in message . com... John, Well, Feynman boasts about this approach. So I am getting his lecture (the lost one) from local library. Has voice CD and some explanatory text by someone. There is a Java script on the Internet to demo some of this approach .... but I found it incomplete (for my understanding). Hi Don Could you post the link to this please, I'd like to take a look. Never mind, found it. http://www.lostlecture.host.sk/JFeynmanEn.htm Well ... I do stand corrected. It appears that he can produce both elliptic and hyperbolic orbits on a purely geometric basis. the problem appears to be with parabolic orbits. The parabolic orbits are only approximations and only approach a parabola in the limiting case. It also looks strange because the tangential velocity appears to be at a minimum at periapsis (this also appears to be the case for the hyperbolic orbits). Perhaps this was the problem originally alluded to. |
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