AGWists are sore losers
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From: Eric Gisse on 1 Aug 2007 20:51 On Aug 1, 4:43 pm, kdth...(a)yahoo.com wrote: > On Aug 1, 2:06 pm, Jonathan Kirwan <jkir...(a)easystreet.com> wrote: > > > > > On Sun, 29 Jul 2007 17:40:35 -0700, kdth...(a)yahoo.com wrote: > > ><snip> > > >Orbits of equal area have equal energy. This is where one begins with > > >proper orbital mechanics. > > > Where in the hell did you find this? > > > Setting aside, (1) your idiocy about circular orbits being impossible > > (due to your confusion, not knowing that circular orbits are a special > > case of elliptical ones) and, (2) your almost religious fundamentalist > > manner in interpreting Kepler's use of the word 'ellipse', so that you > > believe your interpretation of his authority on this matter _must_ > > then somehow exclude circular orbits, ... setting all that aside for > > the moment... > > > The total energy in a captured orbit, elliptical or circular, is not > > in some proportion to the elliptical area. It relates to the semi- > > major axis, only. The actual figure, if the energy at a distance of > > infinity is taken to be zero, is: -G*M*m / (2*a), with (2*a) being > > the semi-major axis and 'e' being the eccentricity. Note that the > > area, which is: > > > PI*a^2*sqrt(1-e^2) > > > isn't present in the energy term. > > > I didn't pull the energy equation out of a hat, by the way. It is > > easily derived and I've done so entirely on my own without the aid of > > some book or web page. > > > In any case, the expression, -G*M*m / (2*a), makes dimensional sense, > > as well. G is in units of (distance^3 * mass^-1 * time^-2), M and m > > are obviously in terms of (mass), and 'a' is obviously (distance). > > Dimmensionally, this works out to (use fixed-space font to read): > > > distance^3 * mass^-1 * time^-2 * mass * mass distance^2 * mass > > -------------------------------------------- = ----------------- > > distance time^2 > > > which just happens to be in the units required for energy, namely in > > terms relatable to the Joule, which is 1 kg*m^2/s^2. This is a "good > > thing", because otherwise we'd be in a peck of trouble. > > > But note that the orbital energy is related to some linear term of the > > ellipse and not to its area. If it had been related to the area, as > > you seem to assert instead, then (1) the energy equation's dimensions > > wouldn't work out right, or else; (2) G would have to have units other > > than those it actually does, or else; (3) you'd need to introduce some > > other linear measure as a factor in the equation, undermining what you > > said earlier but in some new weird way, or else; (4) you would need to > > rework your claim backwards through a great deal of other mechanical, > > physical fundamentals and thus completely redo modern physics -- which > > is rather unlikely of you to achieve. > > > Using the Earth's orbit as a reference, with an eccentricity of about > > 0.016710219 and an assumed semimajor axis arbitrarily set to 1, and > > comparing that to another orbit with the same area but with an > > eccentricity assumed to be 0.5, we can easily compute the required > > semimajor axis of the new ellipse to be 1.196355. This flows from: > > > PI*(a<earth>)^2*sqrt(1-(e<earth>)^2) = PI*a^2*sqrt(1-e^2) > > > where we set a<earth> = 1, e<earth> = 0.16710219, and e = .5 and then > > solve for the required 'a'. > > > These two ellipses have the same area. However, their energies are > > NOT the same, given what I've said above. > > > I didn't add the rather simple derivation of the total energy equation > > for an orbit here. You can easily find the resulting equation on the > > web. But if you need me to do so, I can readily add that derivation > > here. It flows quite simply and naturally from the basic idea that > > acceleration is the second time derivative of r, f=m*acceleration, and > > f=G*M*m/r^2. Nothing magic here. > > > Jon > > Orbits of equal area have equal energy. Orbits of less area have > greater energy. This is how I was taught by a very brilliant and > orthodox astrophysicist. ....and you think you were taught orbital mechanics the way he was taught orbital mechanics? You have no education in math or physics, so he most likely gave you simplified explanations that were "true enough". [...]
From: kdthrge on 1 Aug 2007 21:13 On Aug 1, 7:51 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > On Aug 1, 4:43 pm, kdth...(a)yahoo.com wrote: > > > > > > > On Aug 1, 2:06 pm, Jonathan Kirwan <jkir...(a)easystreet.com> wrote: > > > > On Sun, 29 Jul 2007 17:40:35 -0700, kdth...(a)yahoo.com wrote: > > > ><snip> > > > >Orbits of equal area have equal energy. This is where one begins with > > > >proper orbital mechanics. > > > > Where in the hell did you find this? > > > > Setting aside, (1) your idiocy about circular orbits being impossible > > > (due to your confusion, not knowing that circular orbits are a special > > > case of elliptical ones) and, (2) your almost religious fundamentalist > > > manner in interpreting Kepler's use of the word 'ellipse', so that you > > > believe your interpretation of his authority on this matter _must_ > > > then somehow exclude circular orbits, ... setting all that aside for > > > the moment... > > > > The total energy in a captured orbit, elliptical or circular, is not > > > in some proportion to the elliptical area. It relates to the semi- > > > major axis, only. The actual figure, if the energy at a distance of > > > infinity is taken to be zero, is: -G*M*m / (2*a), with (2*a) being > > > the semi-major axis and 'e' being the eccentricity. Note that the > > > area, which is: > > > > PI*a^2*sqrt(1-e^2) > > > > isn't present in the energy term. > > > > I didn't pull the energy equation out of a hat, by the way. It is > > > easily derived and I've done so entirely on my own without the aid of > > > some book or web page. > > > > In any case, the expression, -G*M*m / (2*a), makes dimensional sense, > > > as well. G is in units of (distance^3 * mass^-1 * time^-2), M and m > > > are obviously in terms of (mass), and 'a' is obviously (distance). > > > Dimmensionally, this works out to (use fixed-space font to read): > > > > distance^3 * mass^-1 * time^-2 * mass * mass distance^2 * mass > > > -------------------------------------------- = ----------------- > > > distance time^2 > > > > which just happens to be in the units required for energy, namely in > > > terms relatable to the Joule, which is 1 kg*m^2/s^2. This is a "good > > > thing", because otherwise we'd be in a peck of trouble. > > > > But note that the orbital energy is related to some linear term of the > > > ellipse and not to its area. If it had been related to the area, as > > > you seem to assert instead, then (1) the energy equation's dimensions > > > wouldn't work out right, or else; (2) G would have to have units other > > > than those it actually does, or else; (3) you'd need to introduce some > > > other linear measure as a factor in the equation, undermining what you > > > said earlier but in some new weird way, or else; (4) you would need to > > > rework your claim backwards through a great deal of other mechanical, > > > physical fundamentals and thus completely redo modern physics -- which > > > is rather unlikely of you to achieve. > > > > Using the Earth's orbit as a reference, with an eccentricity of about > > > 0.016710219 and an assumed semimajor axis arbitrarily set to 1, and > > > comparing that to another orbit with the same area but with an > > > eccentricity assumed to be 0.5, we can easily compute the required > > > semimajor axis of the new ellipse to be 1.196355. This flows from: > > > > PI*(a<earth>)^2*sqrt(1-(e<earth>)^2) = PI*a^2*sqrt(1-e^2) > > > > where we set a<earth> = 1, e<earth> = 0.16710219, and e = .5 and then > > > solve for the required 'a'. > > > > These two ellipses have the same area. However, their energies are > > > NOT the same, given what I've said above. > > > > I didn't add the rather simple derivation of the total energy equation > > > for an orbit here. You can easily find the resulting equation on the > > > web. But if you need me to do so, I can readily add that derivation > > > here. It flows quite simply and naturally from the basic idea that > > > acceleration is the second time derivative of r, f=m*acceleration, and > > > f=G*M*m/r^2. Nothing magic here. > > > > Jon > > > Orbits of equal area have equal energy. Orbits of less area have > > greater energy. This is how I was taught by a very brilliant and > > orthodox astrophysicist. > > ...and you think you were taught orbital mechanics the way he was > taught orbital mechanics? You have no education in math or physics, so > he most likely gave you simplified explanations that were "true > enough". For a dishonest twit like you, everything is in what you say and what you get people to believe. The fact is in the reality. If the false mathematics and orbital mechanics to which you ascribe are used in real life, there will be catastrophe. If the proper orbital mechanics are used such as I was taught, actual space exploration can occur. The point then is to keep you and your false theoretics out of any actual consideration with the use of public funds. In the meantime, you can beat on your little drum all you want. Just leave us the hell alone. KDeatherage
From: Eric Gisse on 1 Aug 2007 21:36 On Aug 1, 5:13 pm, kdth...(a)yahoo.com wrote: > On Aug 1, 7:51 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > > > > > On Aug 1, 4:43 pm, kdth...(a)yahoo.com wrote: > > > > On Aug 1, 2:06 pm, Jonathan Kirwan <jkir...(a)easystreet.com> wrote: > > > > > On Sun, 29 Jul 2007 17:40:35 -0700, kdth...(a)yahoo.com wrote: > > > > ><snip> > > > > >Orbits of equal area have equal energy. This is where one begins with > > > > >proper orbital mechanics. > > > > > Where in the hell did you find this? > > > > > Setting aside, (1) your idiocy about circular orbits being impossible > > > > (due to your confusion, not knowing that circular orbits are a special > > > > case of elliptical ones) and, (2) your almost religious fundamentalist > > > > manner in interpreting Kepler's use of the word 'ellipse', so that you > > > > believe your interpretation of his authority on this matter _must_ > > > > then somehow exclude circular orbits, ... setting all that aside for > > > > the moment... > > > > > The total energy in a captured orbit, elliptical or circular, is not > > > > in some proportion to the elliptical area. It relates to the semi- > > > > major axis, only. The actual figure, if the energy at a distance of > > > > infinity is taken to be zero, is: -G*M*m / (2*a), with (2*a) being > > > > the semi-major axis and 'e' being the eccentricity. Note that the > > > > area, which is: > > > > > PI*a^2*sqrt(1-e^2) > > > > > isn't present in the energy term. > > > > > I didn't pull the energy equation out of a hat, by the way. It is > > > > easily derived and I've done so entirely on my own without the aid of > > > > some book or web page. > > > > > In any case, the expression, -G*M*m / (2*a), makes dimensional sense, > > > > as well. G is in units of (distance^3 * mass^-1 * time^-2), M and m > > > > are obviously in terms of (mass), and 'a' is obviously (distance). > > > > Dimmensionally, this works out to (use fixed-space font to read): > > > > > distance^3 * mass^-1 * time^-2 * mass * mass distance^2 * mass > > > > -------------------------------------------- = ----------------- > > > > distance time^2 > > > > > which just happens to be in the units required for energy, namely in > > > > terms relatable to the Joule, which is 1 kg*m^2/s^2. This is a "good > > > > thing", because otherwise we'd be in a peck of trouble. > > > > > But note that the orbital energy is related to some linear term of the > > > > ellipse and not to its area. If it had been related to the area, as > > > > you seem to assert instead, then (1) the energy equation's dimensions > > > > wouldn't work out right, or else; (2) G would have to have units other > > > > than those it actually does, or else; (3) you'd need to introduce some > > > > other linear measure as a factor in the equation, undermining what you > > > > said earlier but in some new weird way, or else; (4) you would need to > > > > rework your claim backwards through a great deal of other mechanical, > > > > physical fundamentals and thus completely redo modern physics -- which > > > > is rather unlikely of you to achieve. > > > > > Using the Earth's orbit as a reference, with an eccentricity of about > > > > 0.016710219 and an assumed semimajor axis arbitrarily set to 1, and > > > > comparing that to another orbit with the same area but with an > > > > eccentricity assumed to be 0.5, we can easily compute the required > > > > semimajor axis of the new ellipse to be 1.196355. This flows from: > > > > > PI*(a<earth>)^2*sqrt(1-(e<earth>)^2) = PI*a^2*sqrt(1-e^2) > > > > > where we set a<earth> = 1, e<earth> = 0.16710219, and e = .5 and then > > > > solve for the required 'a'. > > > > > These two ellipses have the same area. However, their energies are > > > > NOT the same, given what I've said above. > > > > > I didn't add the rather simple derivation of the total energy equation > > > > for an orbit here. You can easily find the resulting equation on the > > > > web. But if you need me to do so, I can readily add that derivation > > > > here. It flows quite simply and naturally from the basic idea that > > > > acceleration is the second time derivative of r, f=m*acceleration, and > > > > f=G*M*m/r^2. Nothing magic here. > > > > > Jon > > > > Orbits of equal area have equal energy. Orbits of less area have > > > greater energy. This is how I was taught by a very brilliant and > > > orthodox astrophysicist. > > > ...and you think you were taught orbital mechanics the way he was > > taught orbital mechanics? You have no education in math or physics, so > > he most likely gave you simplified explanations that were "true > > enough". > > For a dishonest twit like you, everything is in what you say and what > you get people to believe. The fact is in the reality. If the false > mathematics and orbital mechanics to which you ascribe are used in > real life, there will be catastrophe. If the proper orbital mechanics > are used such as I was taught, actual space exploration can occur. Which brings me back to my original question: If what I was taught is wrong, how do you think men and machinery are being sent into space? On backs of hoping and praying? While you are contemplating that, why don't you sit down and think about Lagrange points. Where do they fit into your overall framework? > > The point then is to keep you and your false theoretics out of any > actual consideration with the use of public funds. In the meantime, > you can beat on your little drum all you want. Just leave us the hell > alone. Who is this "us", death rage? > > KDeatherage
From: Jonathan Kirwan on 2 Aug 2007 04:03 On Wed, 01 Aug 2007 20:04:23 -0700, I wrote: ><snip> >Then I will quite politely show you the math involved so that you >can learn a little. ><snip> After reading Eric's earlier post, a few days back, I no longer imagine any exposition will help Kent. He needs to develop some skills in math and physics and seems unrepentant about his ignorance. Jon
From: Eric Gisse on 2 Aug 2007 21:01
On Aug 2, 4:38 pm, kdth...(a)yahoo.com wrote: > On Aug 2, 3:07 am, Jonathan Kirwan : > > > > > >: --> Kent wrote: > > >: Orbits of lesser area have greater energy. This is where your > > >: anlalyses is defunct and where the other idiots of theoretical science > > >: like yourself, will not listen to reason, just as yourself, and in > > >: their idiocy, dominate the academics of theoretical science. > > > Listening to reason is listening to rigorous theory and a quantitative > > deduction to specifics via math. Eric tried that path without effect. > > It's very hard to reason with remorseless ignorance. > > So all of this in order for you to answer the question of what happens > if an object in orbit is accelerated in the direction of it's > motion??? > > Your answer is simple. From a simple mind without any valid mechanics > of motion. Taken from the fools of theoretical physics. From theory > derived from classical physics which does not even respect Newtonian > physics. It doesn't obey Newtonian physics? Mind supporting your latest spew with some actual evidence? [...] |