AGWists are sore losers
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From: Eric Gisse on 31 Jul 2007 21:52 On Jul 31, 4:50 pm, kdth...(a)yahoo.com wrote: > On Jul 30, 11:18 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > > > > > We have had a pretty good record in the last 20 years, especially if > > you discount the craft that never made it _to_ Mars. > > The mission that failed just a few years ago, failed according to the > press release, because two of the laboratories doing the calculations > did not do their calculations in the same units. A few hundred million > dollars wasted. If by "just a few" you mean "eight", then yes. > > With the false orbtial mechanics that irritated eric has learned, > they'll need the balloons to land any mission. They certainly cannot > calculate either the rate of descent or the direction of the > application of force to reduce the orbital momentum. I'm interested to know how you think spacecraft are put near/on other planets if orbital mechanics is wrong. > > KDeatherage
From: kdthrge on 1 Aug 2007 01:45 On Jul 31, 9:07 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > > This is very simple. This is how mean orbital velocity is determined. > > This basic analyses when done by people with mechanical apptitude also > > proves that orbits cannot be circular. These continued fundmentals > > also prove that orbits of lesser area have greater energy. > > Really, orbits cannot be circular? > > Why is that, death-rage? Please use mathematics to prove your > assertion. Try to keep in mind that a circle is simply a special case > of an ellipse. Also keep in mind that a proof that uses only words is > unacceptable. This is the proof that you are inept. Despite your obsession with semantics of physics and your self accredited intelligence. You have no interest or capability in analysing this problem. Only your self infatuation at your ability to quote ideas which you think make you look intelligent and which you presume nobody can understand since you don't understand them yourself. The intial momentum is linear. The force of gravity causes the curve. If the momentum matches the gravitational force to the extent that after 1 second, the object will still be at the same radius, which would describe a circle, it will have acquired kinetic energy from the force of gravity to make the turn to this point from it's original vector of momentum. Gravity is an acceleration. Therefore, the distance that is traveled laterally, will result in a velocity added to the initial velocity that is 2 times greater than the average velocity needed to cover this lateral distance in the time of 1 second. Therefore, in the second second, the object will not have the correct velocity to do this again, and it will not have the perfect 90deg angle to the source of the gravity. The final vector of the momentum may be tangential to the two imposed vectors, but the motion to this point is not tangential, duffus twit. It will travel in an ellipse, not a circle. Therefore, in it's orbit it will always be increasing or decreasing velocity as it increases or decreases it's orbital radius. IN a circlular orbit, the object must be at EXACTLY the same velocity for the entire orbit, and it must EXACTLY travel at 90deg to the gravity. Completely impossible. Even the fact of perturbations of other orbital bodies will always throw the object into a changing orbital radius. According to your application of r x p, Pluto by far would have the greatest anglular momentum of any planet. If this is what you want to believe, you have the right to this fantasy just as you do for your fantasies of your massive intelligence. KDeatherage
From: Eric Gisse on 1 Aug 2007 02:05 On Jul 31, 9:54 pm, kdth...(a)yahoo.com wrote: > On Jul 31, 8:52 pm, Eric Gisse <jowr...(a)gmail.com> wrote: > > > > > I'm interested to know how you think spacecraft are put near/on other > > planets if orbital mechanics is wrong. > > Look, I was taught proper orbital mechanics. Good for you, but that isn't what I asked. [...]
From: Jonathan Kirwan on 1 Aug 2007 15:06 On Sun, 29 Jul 2007 17:40:35 -0700, kdthrge(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
From: kdthrge on 1 Aug 2007 20:43
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. To know the period of an orbit, one needs only the mean orbital radius. No need at all to know the eccentricity. However you may wish to define energy of the orbit, it must be defined by what happens when energy is added to the orbit. If a satellite uses rockets and adds energy to it's momentum or velocity in any direction, this will affect the orbit by affecting mean orbital radius. It will have the effect of diminishing the area of the orbit. Orbits of less area have greater energy. So you must be of the school of thought that if a satellite is in near circular orbit, and uses thrust at 90deg to the force of gravity in the direction of it's motion, that it will move out to an orbit of greater radius?? If so, you should study your basics closer, and get over your invalid presumptions. This most primitive idea of orbits and energy of orbits is invalid, although it is very commonly believed and taught within the schools of theoretical physics. This idea only means the death of any mission in which this assumption is employed. Momentum is a single vector at any given moment. Orbits of less orbital radius have greater orbital velocity. This is the only energy that exists in orbits. The single vector of the momentum of the mean orbital velocity. Where else is there any energy to be refered?? KDeatherage |