Talk:Hyperbolic trajectory

Untitled
Formula on the page is wrong. — Preceding unsigned comment added by 130.207.84.116 (talk • contribs) 2006-02-22T05:33:32


 * Thank you for letting us know; I corrected the formula. 62.245.80.251 11:47, 20 March 2006 (UTC)

Untitled
The first link in the External links section (http://www.cix.co.uk/~sjbradshaw/msc/traject.html) appears to be broken.Ugo 11:50, 6 November 2006 (UTC)

In English please?
Can someone explain what a hyperbolic trajectory is in slightly more layman's terms?

Ans: It is not closed like an ellipse but goes out to infinity, becoming rather straight when far from the central force. The orbiter gradually slows to a terminal velocity as it goes far away. It looks a little like a rope loosely hung between two supports, but that curve is a catenary - not the same. SlowestLoris (talk) 04:09, 2 June 2009 (UTC)

It is in English... Ice.Queen (talk) 04:35, 6 February 2017 (UTC)

error
The formula for velocity is WRONG with negative "a" because if r->infinity you would get imaginary velocity. You need to use the absolute value of a or make a positive. The same error occurs in Wapedia. SlowestLoris (talk) 04:09, 2 June 2009 (UTC)


 * I fixed it. Either "a" is positive and we have "+" in the formula, or negative and "-". I chose the first, corresponding to Semi-major_axis.--Patrick (talk) 09:14, 13 December 2009 (UTC)

Further Apparent Error
The formula for distance of closest approach implies that this distance approaches zero when the eccentricity approaches 1. However, a parabolic orbit doesn't pass through the center of the primary, as the illustration supplied with the article shows. Quadibloc (talk) 20:04, 13 March 2016 (UTC)
 * No error. The periapsis will get closer and closer, as the eccentricity is increased, if you keep the semi major axis constant (and the hyperbolic excess velocity). Considering the impact parameter may help (which will decease to 0 as eccentricity approaches 1). Care is needed when treating a parabola as a limit of a hyperbola - it depends what you keep constant as the eccentricity approaches 1. The semi major axis is not defined for a parabola so the formula a(e-1) is not defined at e=1 - in the situation you end up with a radial hyperbolic trajectory. Keep the semi-latus rectum (p) constant (or equivalently angular momentum) and periapsis is p/(e+1), and there is no problem (but energy will vary). Marqaz (talk) 14:50, 26 February 2017 (UTC)

Error in the formula for the relationship between true anomaly and eccentric anomaly
The formula includes sqrt((1 + e) / (1 - e)), which impossible for trajectories with e > 1 (that is, hyperbolic ones.) Some googling around suggests that the correct version uses sqrt((e + 1) / (e - 1)), which yields real results.77.234.112.153 (talk) 23:46, 2 May 2017 (UTC)

External links modified
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The hyperoblic anomaly formula is wrong
I thing all signs are wrong in this formula. Someone with a proper book may control this. — Preceding unsigned comment added by 87.14.121.169 (talk) 15:24, 30 January 2018 (UTC)

Definition and formula for mean anomaly seems intuititvely incorrect.
" ... The mean anomaly is proportional to time ..." . That suggests that the mean anomaly is periodic. Given that the calculation for eccentric and true anomaly proceeds from this, obviously those two will also be periodic, which obviously for a hyperbolic orbit is not true.

Seeing as many other texts lists this, I'm guessing that the formular is correct, and I'm merely not understanding it. So perhaps a bit of explanation could be in order here? — Preceding unsigned comment added by 2001:8003:E448:D401:E05C:9A1:67AA:1913 (talk) 05:27, 6 November 2019 (UTC)

Spelling mistake
Uhhh, the past participle of "sweep" is "swept", not "sweeped". Who wrote this? 2001:8003:E490:7D01:FD60:F722:A97E:5A50 (talk) 12:23, 26 February 2023 (UTC)


 * Done 71.168.116.87 (talk) 19:37, 13 July 2023 (UTC)

Problem with the distance formula
In a hyperbolic trajectory the true anomaly is linked to the distance between the orbiting bodies by the orbit equation (and I cant paste that in).

The problem with that formula is that the radius does not change correctly as a function of time. In fact it does the opposite. As the secondary object moves further away, it travels faster and faster. So it is slowest at perihelion and fastest at well, infinity.

2001:8003:E490:7D01:FD60:F722:A97E:5A50 (talk) 12:37, 26 February 2023 (UTC)