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Orbital Mechanics 101

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posted on May, 5 2009 @ 04:50 PM
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Thanks for your kind words!

Yea the space elevator idea would be neat, but that is another thread.

What is your major?




posted on May, 5 2009 @ 07:05 PM
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Ive completed with a degree in geophysics. Wondering what i should do for post-graduate studies at the moment, as there are very little jobs going round for graduates these days.....



posted on May, 5 2009 @ 10:39 PM
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reply to post by nzldude
 


Not much happening with petroleum exploration these days? That seems a bit surprising.

Sorry about going OT but it will help keep the thread alive.



posted on May, 6 2009 @ 03:59 AM
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reply to post by Phage
 



Yea, thanks for trying to keep the thread alive, although there isn't much to talk about, lol it is kinda a 'read only' thread, you know what I mean?



posted on May, 6 2009 @ 05:37 PM
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Here is a nice site where you can track in real time spacecraft and satellites.

www.n2yo.com...



posted on May, 9 2009 @ 07:13 PM
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Bumping to get the thread excited again....

A fellow ATSer in another thread had questions about HOW Apollo got to the Moon without a disastrous fly-by and loss of the vehicle, based on the "lack of computing technolgy" at the time (late 1960s).

Rather than deflecting that thread, I'm putting this video here (it tends to graphically depict what the OP put into his beginning info).

I'm still hunting for a graphical representation of the mechanics of TLI and subsequent insertion into Lunar orbit once arriving.

The member's erroneous impression that the Apollo spacecraft system was too massive to be maneuvered is at the crux of his confusion. (Of course, the Shuttle is heavier, as we know...)




posted on May, 9 2009 @ 07:16 PM
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reply to post by weedwhacker
 


Excellent my friend... Oh and sorry about that u2u, I meant to send it to someone else, but you are always welcome too!lol.......

Maybe you should refer that member here.....I am glad to see some people aren't board to death with these more educational and scientific topics!


Thanks to you and Phage for bumping the thread!



posted on Jun, 18 2009 @ 07:32 PM
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Little bump and addition..Here is a nice example of perigee and apogee...




Courtesy:en.wikipedia.org...



posted on Jun, 18 2009 @ 08:33 PM
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Could somebody please explain to me the physics behind an observation of mine that I made a while ago when browsing through a table of planetary data ... I'm sure there's a perfectly valid scientific explanation but darned if i know what it is !

If you perform a simple mathematical process on the orbital velocity of any body (planet or smaller body) that is in orbit around the sun and it's orbital radius, then you ALWAYS get a value of approx 885 give or take a very small percentage ... which begins to start looking like some kind of constant if you perhaps were to use extremely accurate data.

k = r[adius] x v[elocity](squared)

here's some examples:

Mercury
- mean distance from sun = 0.39AU
- mean orb vel = 47.89 km/s
k = 894

Venus
- mean distance from sun = 0.72AU
- mean orb vel = 35.03 km/s
k = 883

Earth
- mean distance from sun = 1AU
- mean orb vel = 29.79 km/s
k = 887

Mars
- mean distance from sun = 1.52AU
- mean orb vel = 24.13 km/s
k = 885

Jupiter
- mean distance from sun = 5.2AU
- mean orb vel = 13.06 km/s
k = 886

Saturn
- mean distance from sun = 9.54AU
- mean orb vel = 9.64 km/s
k = 886

Uranus
- mean distance from sun = 19.18AU
- mean orb vel = 6.81 km/s
k = 889

Neptune
- mean distance from sun = 30.06AU
- mean orb vel = 5.43 km/s
k = 886

And this also seems to apply to smaller objects:

Ceres
- mean distance from sun = 2.7659 AU
- mean orb vel = 17.882 km/s
k = 884

Pluto
- mean distance from sun = 39.44 AU
- mean orb vel = 4.74 km/s
k = 886

Haumea
- mean distance from sun = 43.335 AU
- mean orb vel = 4.484 km/s
k = 871


So this "k" value looks to be in the vicinity of an average of 885

Am I completely of base by suggesting that it looks like you can get a very close estimate of a bodies orbital distance (radius) simply by dividing this k value by it's velocity (squared)

OR

a very close estimate of a bodies orbital velocity by dividing this k value by it's orbital radius ?


From the above calculations, it appears that this is a reasonable statement to make (unless I'm missing something really obvious) ... so that begs the question, why isn't the effect of gravity required to be taken into consideration (Newtons gravity equation) as all that seems to be needed is the orbiting bodies velocity or distance and what seems to be (almost) a constant value ?

Comments ?



posted on Jun, 18 2009 @ 09:59 PM
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Planetary Orbits
Easy to prove from Newtonian forces to find the Kepler relations.
But not so easy for elliptical orbits which takes calculus.
From the Gamow "Gravity" book:
Centripetal Acceleration equals the force of gravitational attraction.

m v^2 : R = G M m : R^2
: indicates divide, R radius of orbit being a circle
Length of circular orbit = 2 pi R giving period of one revolution
T = 2 pi R : v thus solve for v = 2 pi R : T and substitute above.

m 4 pi^2 R^2 : T^2 R = G M m : R^2
Which when cleared up gives Kepler's cubes of R proportional to
the squares of T, the third law of Kepler.
To satisfy the elliptical orbits, the periods of revolution is used with
the requirement of mean distance from the Sun.



posted on Jun, 18 2009 @ 11:03 PM
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reply to post by zetabeam
 


Thanks for dropping by!


Someone by the name of Johannes Kepler (lol) discovered this relationship too, about 400 years ago...


And third, Kepler found a universal relationship between the orbital properties of all the planets orbiting the sun. For each planet, the cube of the planet's distance from the sun, measured in astronomical units (AU), is equal to the square of the planet's orbital period, measured in Earth years. Jupiter, for example, is approximately 5.2 AU from the sun and its orbital period is 11.86 Earth years. So 5.2 cubed equals 11.86 squared, as predicted.

en.wikipedia.org...

And people say wikipedia isn't helpful?

Anyways hoped that answered your question.



posted on Jun, 18 2009 @ 11:37 PM
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Thanks for the above, TeslaandLyne and jkrog08 ... but I'm still not absolutely clear why (in my 1st post) that gravity seems not to be required to work out either the objects distance from the body it's orbiting or to work out it's orbital velocity.

Surely the effect of gravity constraining the body into a circular/elliptical path MUST be taken into account ... after all, one would assume a weaker or stronger gravitational attraction would determine the bodies orbital radius ?

In my round about way of trying to get to the point, I think what I'm saying is that if hypothetically we knew NOTHING about gravity or of it's existance ... that we could still get a very close approximation of an orbiting bodies velocity (knowing it's distance and that constant I used) OR it's distance (knowing it's velocity and that constant i used).
Just seems strange to me that those 2 properties of the orbiting body appear to independent of knowing anything about gravity.



posted on Jun, 18 2009 @ 11:48 PM
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reply to post by zetabeam
 


G · m · t² = 4 · π² · r³



Kepler's third law, which is often called the harmonic law, is a mathematical relationship between the time it takes the planet to orbit the Sun and the distance between the planet and the Sun. The time it takes for a planet to orbit the Sun is its orbital period, which is often simply called its period. For the average distance between the planet and the Sun, Kepler used what we call the semi-major axis of the ellipse. The semi-major axis is half the major axis, which is the longest distance across the ellipse. Think of it as the longest radius of the ellipse.

Kepler's third law states that the square of the period, P, is proportional to the cube of the semi-major axis, a. In equation form Kepler expressed the third law as: P^2=ka^3. k is the proportionality constant. To Kepler it was just a number that he determined from the data. Kepler did not know why this law worked. He found it by playing with the numbers.

Newton's Form of Kepler's Third Law
Using Newton's laws it is possible to show why Kepler's third law works. For circular orbits, the centripetal force required to keep the planet moving in a circular path equals the gravitational force between the Sun and planet. For elliptical orbits, the idea is similar but a little more complex. Because the gravitational force depends on the mass it turns out that the proportionality constant in Kepler's third law involves the mass of the Sun or other object being orbited. See the figure for the equation for Newton's form of Kepler's third law. In the case of a planet and a star, the mass of the planet is negligible and can be dropped from the equation. In the case of two stars or other orbiting objects of similar mass, both masses must be included.



Read more: mechanical-physics.suite101.com...&C

mechanical-physics.suite101.com...

After the discovery of gravity Newton modified the law to the equation shown above. That proved a direct relationship between gravity effects and distance from source of the focus.



posted on Jun, 19 2009 @ 01:16 AM
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reply to post by zetabeam
 


Sounds like you are making us think like Kepler.

The Period of a planet rotation, a year or two and ten years
is the known quantities from Tycho Brahe, Danish astronomer,
had all the data.

Kepler also use the area of swept out in a small period of time.
He might have assumed the faster planets were closer to the
Sun.
With the period information, T in my example, you find R the
distance from the Sun.
Seemingly the mass of the planet means nothing.
Or does it. The mass is in the proportionality.

Einstein might not like mass being ineffective.
4 pi^2 R^3 = G M T^2
or
G · m · t² = 4 · π² · r³
above.
M is mass of the Sun, so the mass of the orbiting planet means nothing.

ED: You recorded the time. Found the R. Velocity is the 2piR distance
divided by the time.


[edit on 6/19/2009 by TeslaandLyne]



posted on Jun, 19 2009 @ 02:07 AM
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Ok, guys ... I can see what you're both explaining and I thank you for it.

I'm going to simply assume that in that approximate "constant" (I know, I know :-) ) that I used in my 1st post, that the gravity term and a mass term must be concealed within it.

Just nice to know that if there was a hypothetical planet orbiting the sun somewhere inbetween the orbits of say, Venus & Earth, that without knowing anything about the Gravitational constant or the masses of the sun or hypothetical planet, that we could still quite accurately predict it's orbital velocity using it's orbital radius and the "constant" ... or it's orbital radius using it's orbital velocity and the "constant".

The universe works in mysterious ways ! :-)

Thanks again, guys ....



posted on Jun, 19 2009 @ 03:03 PM
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I have this book

Way too complicated.
Not for me and had it for years.

The information I used came from this book.

Easier to understand.



posted on Jun, 22 2009 @ 11:25 AM
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I've always found the "Newton Cannonball" thought experiment to be the simplest way to describe an orbit:

Newton's Cannonball

Some people seem to think that an orbit has something to do with the lack of gravity. However, in reality Gravity is required for an object to be in orbit. Objects in Earth's orbit (such at the space station) are actually being affected by 99% of the Earth's gravity. The space station is actually "falling" due to earth's gravity (but it falls sideways, thus missing the Earth due to the Earth's spherical shape.)

Theoretically, if the Earth was a perfectly smooth sphere (no hills or bumps), and object could orbit at an altitude of one centimeter from the surface (but it would need to move very, very fast.)



posted on Jun, 22 2009 @ 11:30 AM
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reply to post by Soylent Green Is People
 


Good explanation, that is a common misconception of orbits. That is an interesting thought about orbiting one cm from the surface of a perfect sphere. Thanks for the addition.



PS: May I ask what is up with your screen name? lol, I have ben wondering that seeing you on the boards.



posted on Jun, 22 2009 @ 11:56 AM
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Originally posted by jkrog08
reply to post by Soylent Green Is People
 


Good explanation, that is a common misconception of orbits. That is an interesting thought about orbiting one cm from the surface of a perfect sphere. Thanks for the addition.



PS: May I ask what is up with your screen name? lol, I have ben wondering that seeing you on the boards.


Off topic (sorry):

"Soylent Green" is a famous sci-fi movie from the early 1970s, and "Soylent Green is People" is a famous line from the end of that film. My screen name is actually a "spoiler" for the end of the film.

Wikipedia Article: Soylent Green

My screen name is a bit of a "tongue-in-cheek" name that seems fitting for a conspiracy forum such as ATS.

So if you didn't know the plot of the film but were planning to see it someday, I'm sorry if my screen name spoiled the ending for you.

ps...My signature is also one that adds a little "humor" -- it's from the movie "Caddyshack" and is my reference to the spirituality and mysticism that I often see on ATS.



Back on topic:

I said it could "theoretically" orbit at one centimeter -- but that theoretical idea would also need to neglect friction and other forces.


[edit on 6/22/2009 by Soylent Green Is People]




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