Talk:Circular orbit

message: I don't know if I'm just retarded and I'm missing something complete obvious, but it seems to me that the speed is given by v=\sqrt{2\mu\over{r}} and not v=\sqrt{\mu\over{r}}. I mean, the kinetic energy is mv^2\over{2} and the potential energy is :-GMm\over{r}. On the orbit, total energy is minimized (E=0). 
Therefore :v=\sqrt{2\mu\over{r}} .
This error carries on to the period. T=\pi\sqrt{2r^3\over{\mu}} not T=2\pi\sqrt{r^3\over{\mu}}
And also, if we're talking about circular orbits, might as well give the real orbital energy conservation equation {v^2\over{2}}-{\mu\over{r}}=0 . I have no clue why it was equal to something other than 0 before.
It's as if someone went through this and purposely screwed things up. Anyway I cleaned things up.
Headbomb 03:22, 17 May 2007 (UTC)
replies: text: The energy is not zero. The formulas were correct.--Patrick 12:53, 22 June 2007 (UTC)
thread_title:  WTF 

message: The equation of motion became redundant, and I really don't see what mentionning the Virial theorem or the delta V brought to the topic.
Headbomb 03:33, 17 May 2007 (UTC)
replies: text: I restored it.--Patrick 12:58, 22 June 2007 (UTC)
thread_title:  Equation of Motion, delta V and Virial theorem 

message: Is the orbital velocity of circular orbits in general relativity exactly the same as in the Newtonian case or are there any subtle differences? Agge1000 12:57, 11 November 2007 (UTC)
replies: text: I'm aware this is an old question, but it looks bad unanswered. Yes, there are differences and I attempted to describe how to calculate the speed in GR. Ebvalaim (talk) 23:34, 21 February 2012 (UTC)

text: I think you mean v = r d\phi/dt? The problem here is that dt doesn't actually measure time relative to the local observer. It's just an arbitrary coordinate. The actual time difference is \sqrt{g_{tt}}dt in Schwarzschild metric. If you take this into account, \frac{1}{\sqrt{g_{tt}}} r \frac{d\phi}{dt} is actually the same result. Ebvalaim (talk) 12:41, 13 November 2020 (UTC)

text: How do you calculate the angular momentum? The conserved quantity is r^2 \dot{\phi}, which isn't equivalent to rv (because of the issue with time mentioned above in my answer to Dundanox). Ebvalaim (talk) 12:41, 13 November 2020 (UTC)
thread_title: Orbital velocity in general relativity

message: This is the first time I'm looking at these equations, can someone clarify what the two different R's are in the equation: \mathbf{a} = - \frac{v^2}{r} \frac{\mathbf{r}}{r} = - \omega^2 \mathbf{r}  — Preceding unsigned comment added by Nhilton (talk • contribs) 20:07, 27 September 2012 (UTC)
replies: text: As far as I can tell, \mathbf{r} in all cases is radius. I do not understand why it is presented as it is, and further, the article Circular_motion#Uniform_circular_motion presents the formula as  a\, = \frac {v^2} {r} \, = {\omega^2} {r}  , and for the sake of clarity, I am going to replace the former with the later. T.Randall.Scales (talk) 19:43, 2 January 2015 (UTC)

text: Late to the party, but the bold r is the position relative to the center of the massive body as a vector. The regular r is the magnitude of this vector, which is also the distance from the center, or the radius. Ebvalaim (talk) 13:54, 13 November 2020 (UTC)
thread_title:  More details please 

message: The definition of the standard gravitational parameter from here:
 http://en.wikipedia.org/wiki/Standard_gravitational_parameter   
does not agree with the text from here:
 http://en.wikipedia.org/wiki/Circular_orbit#Velocity  — Preceding unsigned comment added by 98.125.78.211 (talk) 09:05, 19 February 2015 (UTC)
thread_title:   standard gravitational parameter does not agree with link.