Posted: 8/19/2005 5:37:22 PM EDT
[Last Edit: 8/19/2005 6:14:31 PM EDT by cmjohnson]
Ok, scientific types, here's a challenge:
Review your orbital mechanics classes in high school? No? Well, how about some basic info you got on orbital mechanics during your physics class, or any other? OK, here's a very brief review, instead: For any given celestial body, earth, for example, there is a minimum theoretical velocity at which an object can be in orbit around it. This is the minimum orbital velocity. At this velocity, the object in orbit is falling toward the planet at the same rate that the planet is falling away from it. On earth, orbit isn't practical unless the orbit is high enough that the atmosphere is not a significant drag factor for the item in orbit. And of course the orbit must clear all obstacles such as mountain ranges, etc. The minimum orbital height is also the point at which the time of orbit is least. The fastest circuits around the planet are at the minimum orbital distance. However, in terms of raw speed, the fastest orbital time is achieved at the SLOWEST possible orbital velocity. If you have surplus orbital velocity, you end up in a higher orbit, and since the distance to travel around the planet increases greatly, being a much larger circumference, you will end up in an orbit that takes longer to complete even though you're actually moving faster. This is why satellites that are in high geosynchronous orbits (they hover over one given spot on earth) have to be boosted to much higher orbital velocities. A geosynchronous orbit of earth occurs at 22,400 miles altitude, assuming you have the velocity to maintain it. So the higher you orbit, the faster your satellite must be moving, and the longer the orbit takes. Now let's complicate things by changing a condition, and that condition is gravity. As you get farther away from earth, the strength of gravity diminishes. As the strength of gravity is much less at a high orbit, the orbital velocity has to be REDUCED to maintain an orbit. The farther away from earth, the slower the orbital velocity must be, BEYOND A CERTAIN DISTANCE. To launch a satellite into a VERY high orbit, it has to be sent out fast to get high enough, and then, at a certain point, a stable orbit beyond it requires the satellite to be slowed DOWN for insertion into the desired high orbit. Otherwise, it'll just fly off into space as gravity won't hold it. My challenge to you is, post formulas or charts that show this. Orbital velocities required to achieve different orbits, from low earth orbit to the very large orbits (like the moon's orbit) where the orbital velocity is LOWER than that of satellites in lower orbits. I've never yet seen any charts or formulas that explain this. Can you come up with it? CJ 

"Now they will know why they are afraid of the dark.
Now they will learn why they fear the night."....Thulsa Doom "Hey, does this rag smell like chloroform to you?" 
Hitting the bottle early, Johnson? I'd rather hear you talk guitar for a while.


Oderint dum metuant  Let them hate, so long as they fear

Calculus makes me vomit.
That's why I abandoned my engineering degree. 

"Computer games don't affect kids. I mean if PacMan affected us as kids, we'd all be running around in darkened rooms, munching magic pills and listening to repetitive electronic music."
Kristian Wilson, Nintendo, Inc, 1989 
http://www.hq.nasa.gov/office/pao/History/conghand/traject.htm
? 


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) 

"Never Knows Best"
GO BUCKS! 
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.... 

It is substantially true that virtue or morality is a necessary spring of popular government.
George Washington 
Marge: Homer, the plant called. They said if you don't show up tomorrow don't bother showing up on Monday.
Homer: Woohoo. Fourday weekend! 
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. 

"Never Knows Best"
GO BUCKS! 
Yes. The two interact strongly beyond a certain point. I have yet to see this rolled together neatly into one formula, graph, or chart. But somebody must know. We couldn't have made it to the moon without it. CJ 

"Now they will know why they are afraid of the dark.
Now they will learn why they fear the night."....Thulsa Doom "Hey, does this rag smell like chloroform to you?" 
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. 

"Never Knows Best"
GO BUCKS! 
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 twobody equations to another once the satellite reaches this point. There are several discrete locations around every twobody 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://wwwspof.gsfc.nasa.gov/stargaze/Slagrang.htm 


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. 


No, I wasn't wrong there, THAT is the whole idea, summed up. For a given body....earth....the required velocity to achieve a circular orbit (always assume circular rather than any other orbit to keep this from being too hairy) increases with distance (altitude) but only to a certain point. Beyond that point, gravitational attraction has reduced the orbital velocity requirement. What I want to know is, what IS the point of maximum orbital velocity and what's that orbital height? And, what's a graph, chart, or forumula that allows you to calculate the CORRECT orbital velocity for any given distance, assuming the body being orbited is always the same (earth) and is massive, and the satellite has negligble mass? (Fewest variables.) Nice to see some brains being put to work here! CJ 

"Now they will know why they are afraid of the dark.
Now they will learn why they fear the night."....Thulsa Doom "Hey, does this rag smell like chloroform to you?" 
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. 

"Never Knows Best"
GO BUCKS! 
I'm tempermental. 5% temper, the rest is mental.
FL, USA

Ahhhh, the dreaded N body problem. Sorry, cant be answered.

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