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Archive 1 |
Astrodynamics#Historical_approaches: needs to be re-written. Just throwing in names of famous scientists in history is useless. -- PFHLai 23:04, 2004 Oct 22 (UTC)
Well, folks, this is the article that taught me I'm no good at writing large articles from scratch. I'll be happy to contribute any way I can, but three years after I had first intended to revamp this article, I haven't done much. If someone else wants to give it ago, they'll have my full support. -- Doradus 19:02, 19 April 2007 (UTC)
Not every one is a physicist, please write out your equations in full, like this, force = mass*distance/time squared, Newtons. Force of gravitational attraction = G*M*m/r^2, Newtons. Energy = force*distance, hence [G*M*m/r^2]*r = G*M*m/r joules. and your energy/mass = G*M/r, joules per kg. velocity v = r/t =[2*G*M/r]^1/2 = {2*[r^3/m*t^2]*M}/r, m/s, since f^2=1/t^2, then velocity, r/t = frequency*wave length = f*{r*[2*M/m]^1/2}, m/s. Since hf = mc^2, joules, then mass, m = h*f/c^2, kg, substituting for mass gives us, v=f*r[2*[h*f(M)/c^2]*[1/h*f/C^2]]^1/2, m/s. which contracts to, v=r*[2*f*f(M)]^1/2, m/s, where f(M) is the frequency of the wave between mass M and the test mass, see www.QuantumMatter.com.-- 79.68.156.33 23:57, 3 November 2007 (UTC)
This is something about orbits I read a little while back, that might be appropriate in this article:
In order to slow down, you speed up; to speed up, you slow down.
Slowing down puts you into a lower orbit, which is faster in relation to a point on the surface of the body you are orbiting around. Speeding up puts you into a higher orbit, which is slower in relation to the aforementioned point.
Can anyone think of a place to put this?
Phædrus 12:01, 20 March 2007 (UTC)
Phraedrus: You are incorrect, your speed is dependent on your semi major axis and speeding up puts you in a larger orbit which means you are going slower, simple derivation from Kepler's equations.
Is the "Application" subsection of Kepler's Laws partly redundant with the "Position as a function of time" section of Kepler's laws of planetary motion? Not sure where this material belongs or in what form. -- Beland ( talk) 19:22, 6 August 2008 (UTC)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
Astrodynamics#Historical_approaches: needs to be re-written. Just throwing in names of famous scientists in history is useless. -- PFHLai 23:04, 2004 Oct 22 (UTC)
Well, folks, this is the article that taught me I'm no good at writing large articles from scratch. I'll be happy to contribute any way I can, but three years after I had first intended to revamp this article, I haven't done much. If someone else wants to give it ago, they'll have my full support. -- Doradus 19:02, 19 April 2007 (UTC)
Not every one is a physicist, please write out your equations in full, like this, force = mass*distance/time squared, Newtons. Force of gravitational attraction = G*M*m/r^2, Newtons. Energy = force*distance, hence [G*M*m/r^2]*r = G*M*m/r joules. and your energy/mass = G*M/r, joules per kg. velocity v = r/t =[2*G*M/r]^1/2 = {2*[r^3/m*t^2]*M}/r, m/s, since f^2=1/t^2, then velocity, r/t = frequency*wave length = f*{r*[2*M/m]^1/2}, m/s. Since hf = mc^2, joules, then mass, m = h*f/c^2, kg, substituting for mass gives us, v=f*r[2*[h*f(M)/c^2]*[1/h*f/C^2]]^1/2, m/s. which contracts to, v=r*[2*f*f(M)]^1/2, m/s, where f(M) is the frequency of the wave between mass M and the test mass, see www.QuantumMatter.com.-- 79.68.156.33 23:57, 3 November 2007 (UTC)
This is something about orbits I read a little while back, that might be appropriate in this article:
In order to slow down, you speed up; to speed up, you slow down.
Slowing down puts you into a lower orbit, which is faster in relation to a point on the surface of the body you are orbiting around. Speeding up puts you into a higher orbit, which is slower in relation to the aforementioned point.
Can anyone think of a place to put this?
Phædrus 12:01, 20 March 2007 (UTC)
Phraedrus: You are incorrect, your speed is dependent on your semi major axis and speeding up puts you in a larger orbit which means you are going slower, simple derivation from Kepler's equations.
Is the "Application" subsection of Kepler's Laws partly redundant with the "Position as a function of time" section of Kepler's laws of planetary motion? Not sure where this material belongs or in what form. -- Beland ( talk) 19:22, 6 August 2008 (UTC)