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From: kdthrge on 4 Aug 2007 20:37 On Aug 4, 5:34 pm, Jonathan Kirwan > >So answer the question. > > Goody. Love questions. > > >If a body is in a very near circular orbit, > > (Your refusal to accept the idea of a truly circular orbit, even in > concept, noted.) > > >and thrust is applied which directly adds to it's direction of > >velocity, > > By this, I take it that you mean that we have a rocket in orbit and we > point the rocket so that its centerline is oriented in line with its > instantaneous velocity vector.... > > >will the resultant orbit have more or less area than the > >near circular orbit it was in???? > > I'll assume a short, non-zero burst from a rocket, here, so that the > acceleration vector can be reasonably taken pointing the same, for the > entire time, as the velocity vector. > l> The orbit will now possess more energy (which means the semimajor axis l> will increase) and because the rocket's propulsion cannot possibly l> avoid contributing to the angular momentum, as well, the area of the l> orbit will also increase. The period of the orbit would also l> increase, since the energy increased. You say the area would increase and the period would increase. This is wrong. The thrust takes the object out from it's near circular orbit. The distance it must travel to get out of this quadrant of the orbit is greater. Even if enough velocity is added to account for the greater distance, it will have too high a velocity for the greater orbital radius. Thus it will not achieve an orbit of greater area, but the added energy in this direction will cause an ellipse which has less area than the intitial near circle. This is because the gravity will quickly reduce it's velocity also with the greater angle at which the momentum is directed away from the gravity. The area of the orbit it will achieve is according to a^3 = p^3. Mars has a mean orbital velocity that is .80 that of earth. This lower orbital velocity and the greater distance of it's circumference of orbit, defines it's period. This is because it is in gravitational field of less force. A greater radius orbit, which has greater area, has a lower mean orbital velocity according to, a / a^3/p^2. You also have invalid numbers for the velocities induced by the force of gravitation. This means that any accounting you have for the effects of decreasing the forward velocity are invalid also, since in this case, remaining in the same quadrant causes a great increase in velocity which will result in a greater radius, area and period. It is clear that with your math and concepts, there is no possiblitly of success in space missions. There is no point arguing with you and your personal attachment to your dogma which means all your rational is towards supporting the false conclusions that you hold, and you cannot objectively analyses these problems. Sorry it irritates you so to hear valid mechanics which are not in line with your dogmatic beliefs which you cannot support by valid math or science. KDeatherage
From: Jonathan Kirwan on 4 Aug 2007 21:14 On Sat, 04 Aug 2007 17:37:26 -0700, kdthrge(a)yahoo.com wrote: ><snip> >Sorry it irritates you so to hear valid mechanics which are not in >line with your dogmatic beliefs which you cannot support by valid math >or science. Actually, I think I'm going to ignore you pretty much in the future. Not always, but mostly. I hadn't realized before (because I didn't bother reading you much) quite the level of your incapacity. But you don't have a clue and can't acquire one. You have no skills here. Too bad. Jon
From: Jonathan Kirwan on 4 Aug 2007 21:17 On Sun, 05 Aug 2007 01:10:51 GMT, I wrote: ><snip> >The way to look at it in the case of orbits is that kinetic energy is >exchanged for potential energy during the total period of the orbit. ><snip> More precisely, SOME kinetic energy is exchanged. Sorry, Jon
From: kdthrge on 5 Aug 2007 07:19 On Aug 4, 8:10 pm, Jonathan Kirwan > >> The potential energy varies with distance, per -GMm/r. The specific > >> potential energy is usually good enough for many purposes, though, as just > >> -GM/r. Note that "mean orbital radius" isn't present. > The actual proper use of the idea of potential energy is that it is set at an arbitrary point. One can define the potential energy from the floor, from the desk, or from the top of a shelf. Potential energy is not defined as distance from the center of the gravitational attracting body. So it means nothing unlless you set it at a point. It is useful to set potential energy at mean orbital radius. Therefore when doing proper evaluation of the potential energy of an orbit, it is proper to determine the potential energy at the mean orbital radius. This potential energy is GMm / r^2 x R. In centimeters this becomes, GMmR. Any point on the ellipse has a different value of R. Subtracted or added to the intital determination of R gives the applied force and the change in kinetic energy and thus the velocity. Since KE = f x d x cosine of theta, the changing cosine affects the rate that velocity is affected by the gravitiation. But the actual potential energy is not described by GMmR. Because the gravitation increases and decreases as an inverse square to distance. So it is invalid to describe potential energy as you do as always a function of r. > > In celestial mechanics, negative values of energy represent captured > orbits, a zero energy value represents the highly improbable exact > parabolic trajectory, and positive energy values represent hyperbolic > trajectories. This shows your basic invalid appication of physics which you have acquired from your training in the invalid schools of theoretical physics. Energy is a quantity and can never be negative. The force of gravity is not a quantity. It can be refered mathematically as negative, -GMm/r^2, so that when multiplied by the negative cosine, it gives a postive value for the energy. The important point is that the acting force of gravity and the vectors it imposes on a falling object are simply added to the vectors of the momentum. This occurs when cosine is zero or less. When the momentum is in contradiction to the force of gravity, the rate that velocity is subtracted is determined by cosine of theta. > > Let's develop some of this, since you asked. (I love it when folks > ask for something I enjoy talking about.) Mental masturbation is fun. HAHAHAHhahahahahHAHAH You are dead wrong. You have no valid concept of the energy of an orbit. Your prediciton that to add energy to the velocity of an orbital body will increase the area of the orbit is false. But you have the right to live in your world of radial momentum which has nothing to do with reality. > > Start with: > > F = G M m / r^2 (gravitational force) > > Also, > > F = m a (force required for an acceleration) > > In the general case of a falling object, we know that the magnitudes > of these two forces are equal... in the sense that the force of > gravitation results in an acceleration: > > | m a | = | G M m / r^2 | > > (I used | | to indicate 'absolute value of.') > > However, because the acceleration (a= d^2r/dt^2) acts to reduce 'r', > it follows that the sign is opposite. So the equation becomes: > > m a = -G M m / r^2 > > This can also be expressed as: > > m (dv/dt) = - G M m / r^2 > > or even better as, > > m (d^2r/dt^2) = - G M m / r^2 > Kinetic energy is defined as,,, m a * d * cosine of theta, when cosine is 1, this simplifies to ( m * final velocity - intitial velocity ) / time,,,x (average velocity * time). The time cancels and when intitial velocity is zero leaves, final velocity x average velocity or v * 1/2v * mass, or 1/2mv^2. What other energy are you talking about??? Energy is a quantity and can never be negative. Sorry, but you have no valid physics or orbital mechanics since you have no valid understanding of this quantity. Your orbital mechanics define nothing and allow no viable application in space exploration. But have fun and enjoy, HAHAHAHhahahahaHAHAHAHhahahahHAHAHA KDeatherage
From: Jonathan Kirwan on 5 Aug 2007 11:05
On Sun, 05 Aug 2007 04:19:44 -0700, kdthrge(a)yahoo.com wrote: ><snip> >What other energy are you talking about??? I developed the energy equation from basic principles. You need only read it. Of course, you can't. In any case, it's also easy to see the final result -- that the energy of an orbit is determined by (or determines) the major axis. That conclusion is inescapable. Of course, I already know you can't follow the math. Oh, well. Jon |