Author Topic: Please Explain the Velocity Numbers  (Read 47811 times)

Offline smartcooky

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Re: Please Explain the Velocity Numbers
« Reply #60 on: May 25, 2013, 06:21:14 AM »
It seems to me that an elliptical orbit is necessary

No kidding, really? Kepler's First Law, published over 400 years ago, specified that planets orbit in ellipses, with the Sun at one focus of the ellipse.

In fact we know today that in all cases, objects orbit each other in ellipses. While not technically correct, for all practical purposes, even a circular orbit is an ellipse that just happens to have both foci in the same place.

So your statement "an elliptical orbit is necessary" is somewhat redundant!
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Offline ka9q

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Re: Please Explain the Velocity Numbers
« Reply #61 on: May 25, 2013, 06:43:49 AM »
It's more correct to say that a conic section is necessary. Ellipses and circles are conic sections, but so is the parabola, the hyperbola and even the straight line. These last three correspond to open trajectories, i.e., the object is on an escape trajectory.

circle: eccentricity=0
ellipse: 0 < e < 1
parabola: e = 1
hyperbola: e > 1
straight line: e = infinity


Offline smartcooky

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Re: Please Explain the Velocity Numbers
« Reply #62 on: May 25, 2013, 08:06:56 AM »
It's more correct to say that a conic section is necessary. Ellipses and circles are conic sections, but so is the parabola, the hyperbola and even the straight line. These last three correspond to open trajectories, i.e., the object is on an escape trajectory.

circle: eccentricity=0
ellipse: 0 < e < 1
parabola: e = 1
hyperbola: e > 1
straight line: e = infinity

With respect, parabolas and hyperbolas cannot, by definition, be "orbits" 
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Offline Glom

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Re: Please Explain the Velocity Numbers
« Reply #63 on: May 25, 2013, 08:12:14 AM »
It's more correct to say that a conic section is necessary. Ellipses and circles are conic sections, but so is the parabola, the hyperbola and even the straight line. These last three correspond to open trajectories, i.e., the object is on an escape trajectory.

circle: eccentricity=0
ellipse: 0 < e < 1
parabola: e = 1
hyperbola: e > 1
straight line: e = infinity

With respect, parabolas and hyperbolas cannot, by definition, be "orbits" 

That's a terminology thing.  In a strict astrodynamic sense, they are "open orbits" because the satellite will only come round once.  However, I understand those who want to keep the term "orbit" limited to closed orbits, using it in a more practical context.

Offline gillianren

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Re: Please Explain the Velocity Numbers
« Reply #64 on: May 25, 2013, 10:52:50 AM »
No matter your terminology, I'm not sure how a hyperbola can be an orbit, open or otherwise.
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Offline Echnaton

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Re: Please Explain the Velocity Numbers
« Reply #65 on: May 25, 2013, 02:25:46 PM »
http://en.wikipedia.org/wiki/Kepler_orbit

Quote
In celestial mechanics, a Kepler orbit (or Keplerian orbit) describes the motion of an orbiting body as an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space.

That is the scientific, not common, use of the term orbit.

Here is an image of examples of orbits that KA9Q referred too.
« Last Edit: May 25, 2013, 02:28:08 PM by Echnaton »
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Offline Noldi400

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Re: Please Explain the Velocity Numbers
« Reply #66 on: May 25, 2013, 02:40:04 PM »
No matter your terminology, I'm not sure how a hyperbola can be an orbit, open or otherwise.
Technically true, I think...  the terminology I usually see is  a parabolic or hyperbolic trajectory.

But either way, it still has a point of closest approach (periapsis), and if it passes close enough to a planet with an atmosphere, friction takes over.

BTW, what "logbook" are we referencing here?  I've been unable to follow the exact discussion - I've searched every reference I can think of and keep coming up empty.

Also, please forgive my lay ignorance, but isn't the moon's orbital velocity allowed for by exactly where in the orbit TEI is initiated, since that determines the apoapsis of the resultant orbit?

Last I heard, velocity consists of a speed and direction. And, as has been said repeatedly, the speed of the return velocity is not nearly as relevant as the direction. So long as the CM hit atmosphere at an angle that allowed speed to be bled off as heat, reentry was possible.
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