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Posted: 8/19/2005 5:37:22 PM EDT
Link Posted: 8/19/2005 5:43:56 PM EDT
[#1]
Hitting the bottle early, Johnson? I'd rather hear you talk guitar for a while.
Link Posted: 8/19/2005 5:47:45 PM EDT
[#2]
Calculus makes me vomit.

That's why I abandoned my engineering degree.
Link Posted: 8/19/2005 5:57:49 PM EDT
[#3]
http://www.hq.nasa.gov/office/pao/History/conghand/traject.htm

?
Link Posted: 8/19/2005 6:34:48 PM EDT
[#4]
Maybe I am misunderstanding your question,
but the equation for calculating the orbital
velocity of a body around the earth is:

SQRT(398600.441* / r)

This will always give you a lessor velocity
for larger values of r.

* geocentric grav. const. (in km^3/s^2)
Link Posted: 8/19/2005 6:36:35 PM EDT
[#5]
Whenever I try orbital mechanicing, my sockets keep floating away on me. Plus it is darn near impossible to get the carburetor to flow gas in the vaccum of space. That and the whole not being able to breathe thing....

I gave it up years ago....
Link Posted: 8/19/2005 6:37:56 PM EDT
[#6]
The answer is three.
Link Posted: 8/19/2005 6:40:23 PM EDT
[#7]
I see where you are coming from. I am
using the gravitational constant whereas
you are assuming the reduction in gravity
as the radius increases.
Link Posted: 8/19/2005 6:49:37 PM EDT
[#8]
Link Posted: 8/19/2005 6:57:28 PM EDT
[#9]

Quoted:
I have yet to see this rolled together neatly into one formula, graph, or chart.



This is easy to calculate for geostationary
orbits since the orbital period is constant
(one day) and the mass of the orbital body
is inconsequential.

Once you start considering all orbit types,
the number of variables increase quite a
lot.
Link Posted: 8/19/2005 7:20:31 PM EDT
[#10]
You forgot one little detail: orbit eccentricity.

A satellite's orbital velocity depends not only on it's altitude, but also the eccentricity of the orbit. The Russkies perfected a highly eccentric orbit for their communication satallites since geostationary orbits are over the equator whereas most of Russia is at higher lattitudes. This orbit, the Molniya orbit, has a very high apogee at high northern latitudes and a low perigee at low southern latitudes. As a result, it "hangs" over Russia for most of it's orbit, then falls and zips around quickly to "hang" again. A couple of properly timed satellites and one is always overhead.

As for gravity "cancelling" out at larger distances, technically it never does. In theory, for a two body system both objects are orbiting each other no matter how far apart they are. In practice, other objects (planets, moons etc.) start to affect the satellite's course more than the home planet at a certain point. The satellite is then said to be "out of orbit" of the original planet at that point.

Note: no one has ever solved the equations for a three body system. Orbital mechanicists just switch from one set of two-body equations to another once the satellite reaches this point.

There are several discrete locations around every two-body system where both bodies have equal gravitational attraction (note that I did not say "cancels out"). These are the Lagrange Points, named after the mathematician who solved the equations. Here is the derivation:

http://www-spof.gsfc.nasa.gov/stargaze/Slagrang.htm
Link Posted: 8/19/2005 7:52:17 PM EDT
[#11]


So the higher you orbit, the faster your satellite must be moving, and the longer the orbit takes.



You have one thing wrong, and I admit it does seem a bit confusing.  Objects at higher orbits actually orbit at slower speeds in addition to having a longer path to orbit.  They do not travel faster.  The farther away you get, gravity has less of an effect so you don't have to travel as fast.



To launch a satellite into a higher orbit you need a higher initial speed, which is lost as the satellite ascends to a higher orbit.  After it reaches the highest point of its very eliptical orbit, extra velocity must be added to turn the orbit into a (mostly) circular one at the desired altitude - otherwise the object would fall back down very close to the body it is orbiting, pick up speed, and repeat the whole process in an eliptical orbit.

If I misunderstood what you were saying, I apologize.  And if I'm wrong, sorry
Link Posted: 8/19/2005 8:05:43 PM EDT
[#12]

Quoted:


So the higher you orbit, the faster your satellite must be moving, and the longer the orbit takes.



You have one thing wrong, and I admit it does seem a bit confusing.  Objects at higher orbits actually orbit at slower speeds in addition to having a longer path to orbit.  They do not travel faster.  The farther away you get, gravity has less of an effect so you don't have to travel as fast.

users.ihost.biz/greg/images/planetvelocity.gif

To launch a satellite into a higher orbit you need a higher initial speed, which is lost as the satellite ascends to a higher orbit.  After it reaches the highest point of its very eliptical orbit, extra velocity must be added to turn the orbit into a (mostly) circular one at the desired altitude - otherwise the object would fall back down very close to the body it is orbiting, pick up speed, and repeat the whole process in an eliptical orbit.

If I misunderstood what you were saying, I apologize.  And if I'm wrong, sorry



You did not misunderstand him, he definitely has that wrong.  

He has several other things wrong as well.  One major micsconception is that you can have too much velocity to be in orbit.  Your velocity and your orbit are determoined simultaneously, they are one and the same.  If you "have too much velocity"" you still have an orbit unless your velocity is so high you have escape velocity and are not in orbit at all but just apssing through.  It may not be the orbit you want, but you still have one.  There are other misconceptions, too.
Link Posted: 8/20/2005 5:44:16 AM EDT
[#13]
Link Posted: 8/20/2005 11:45:27 AM EDT
[#14]
v = SQRT( u / r )

u = (4 pi^2 r^3) / T^2

v = SQRT ( (4 pi^2 r^2)/T^2)

Where:

v = orbital velocity (in km/sec)
r = orbital radius (in km)
u = standard gravitational parameter (solved for)
T = orbital period (in seconds)


Solve for all values of T and r.
Link Posted: 8/20/2005 11:49:59 AM EDT
[#15]
Ahhhh, the dreaded N body problem. Sorry, cant be answered.
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