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Jim Starluck
2004-May-16, 08:09 AM
How many does the Earth/Moon system have, and where are they in relation to the two bodies? I'm looking for distances and co-ordinates, so I could map them in 3D...

Kullat Nunu
2004-May-16, 09:43 AM
How many does the Earth/Moon system have, and where are they in relation to the two bodies? I'm looking for distances and co-ordinates, so I could map them in 3D...

There are five Lagrange points in the Earth-Moon system.

L1 point: between the Earth and the Moon, near the Moon
L2 point: like L1, but opposite side of the Moon
L3 point: on the Moon's orbit, but opposite side of the Earth
L4 point: 60° ahead on the Moon's orbit
L5 point: 60° behind on the Moon's orbit

Only L4 and L5 are stable.

Coordinates (on the Moon's orbital plane):

L1 = (R*(1-(a/3)^(1/3)), 0)
L2 = (R*(1+(a/3)^(1/3)), 0)
L3 = (-R*(1+5*a/12), 0)
L4 = (R/2*((M1-M2)/(M1+M2)),sqrt(3)/2*R])
L5 = (R/2*((M1-M2)/(M1+M2)),-sqrt(3)/2*R])

where

R = secondary's distance from the binary (in this case the Moon's distance from the Earth)
a = M2/(M1+M2) (M1 = the mass of primary, M2 = the mass of secondary; I'm too lazy to calculate a for the Earth-Moon system)

You can find all this from this math intensive PDF file (http://www.physics.montana.edu/faculty/cornish/lagrange.pdf).

milli360
2004-May-16, 11:45 AM
L4 = (R/2*((M1-M2)/(M1+M2)),sqrt(3)/2*R])
L5 = (R/2*((M1-M2)/(M1+M2)),-sqrt(3)/2*R])

where

R = secondary's distance from the binary (in this case the Moon's distance from the Earth)
a = M2/(M1+M2) (M1 = the mass of primary, M2 = the mass of secondary; I'm too lazy to calculate a for the Earth-Moon system)

You can find all this from this math intensive PDF file (http://www.physics.montana.edu/faculty/cornish/lagrange.pdf).
That PDF file may be just the thing that Glom was looking for:

I was trying to derive the LaGrange points and asked for advice (http://www.badastronomy.com/phpBB/viewtopic.php?p=163610#163610) on my equation. kilopi offered a lot of help, but none of the other knowledgeable people came.
However, the coordinates given for L4/L5 don't translate into 60 degrees, so I'm suspicious. I'll have to look at that PDF file more closely.

PS: OK, just before equation 14 on page 5, he says "Demanding that the parallel component of the force vanish leads to the condition that the equilibrium points are at a distance R from each mass." And then apparently he uses that to derive the coordinates for L4/L5--which means the coordinates given in equation 14 are wrong. They must be:
L4 = (R/2, sqrt(3)/2*R])
L5 = (R/2, -sqrt(3)/2*R])

Glom
2004-May-16, 12:06 PM
I now resent LaGrange points.

Kullat Nunu
2004-May-16, 03:03 PM
However, the coordinates given for L4/L5 don't translate into 60 degrees, so I'm suspicious. I'll have to look at that PDF file more closely.

PS: OK, just before equation 14 on page 5, he says "Demanding that the parallel component of the force vanish leads to the condition that the equilibrium points are at a distance R from each mass." And then apparently he uses that to derive the coordinates for L4/L5--which means the coordinates given in equation 14 are wrong. They must be:
L4 = (R/2, sqrt(3)/2*R])
L5 = (R/2, -sqrt(3)/2*R])

Don't forget that M1 must be >> M2 so

(M1-M2)/(M1+M2) ~ 1.

That 60° is good approximation, but not accurate.

2004-May-16, 04:56 PM
How come Earth's L4 and L5 Lagrange Points are empty? :-k

The Supreme Canuck
2004-May-16, 05:22 PM
Are they? There may be some stuff caught there that we just haven't found yet.

Ut
2004-May-16, 05:35 PM
I had a post doc teaching my E&amp;M class last year who's currently looking for Trojan companions to Earth. Apparently, they have a few candidates.

He's recently been involved in studies of 3753 Cruithne and Asteroid 2002 AA29, too.

Paul Wiegert's Web Page (http://www.astro.uwo.ca/~wiegert/)

Kullat Nunu
2004-May-16, 05:52 PM
I had a post doc teaching my E&amp;M class last year who's currently looking for Trojan companions to Earth. Apparently, they have a few candidates.

He's recently been involved in studies of 3753 Cruithne and Asteroid 2002 AA29, too.

Paul Wiegert's Web Page (http://www.astro.uwo.ca/~wiegert/)

That's interesting.

How come Earth's L4 and L5 Lagrange Points are empty?

It is not that strange that no Earth Trojans have been found since they never come close -- they stay at 1 AU and only 60° from the Sun, naturally.

Jim Starluck
2004-May-16, 06:11 PM
R = secondary's distance from the binary (in this case the Moon's distance from the Earth)
a = M2/(M1+M2) (M1 = the mass of primary, M2 = the mass of secondary; I'm too lazy to calculate a for the Earth-Moon system).

What units would I use? I assume kilometers for distance...metric tons for mass?

Tobin Dax
2004-May-16, 06:28 PM
Jim, in this case, R can be in which ever units you want the coordinates in. Also, since the masses cancel out, you can use any unit you want for that as well.

Glom, I gotta ask why you resent LaGrange Points. :)

Kullat Nunu
2004-May-16, 06:59 PM
Tobin Dax, in "real" physical equations units are always SI units. So in this case mass is in kilograms and r in meters.

Tobin Dax
2004-May-16, 10:48 PM
Blasphemy! Real astronomers use cgs. I know how long an AU is in cm and what the sun's mass is in grams, but I have to think for a second to convert them ito SI/mks. :D

That being said, it sounds like Jim Starluck doesn't have much physcial background at the moment, and so I wan't going to try to force units on him. They're inconsequential anyway. If he finds it in km, he'll likely end up converting to miles (well, *I* do this often enough for smaller numbers). As for the mass, it cancels out of the equation, so he could use slugs or galactic masses and it wouldn't matter. As long as his values are correct, he can use any units he chooses.

kylenano
2004-May-16, 11:32 PM

Ut
2004-May-16, 11:34 PM
Blasphemy! Real astronomers use cgs.

And I hate them for it...

*shakes fist in rage*

Do you have any idea how hard it is to use cgs in ONE out of twelve classes every year, and mks in the remaining eleven?

Tobin Dax
2004-May-17, 01:47 AM
Ut, I got my ** using mks (in physics). I'm still not used to all of the cgs units. What's worse is that profs can tend to use their favorite system, so yes, I can understand your complaint.

Jim Starluck
2004-May-17, 06:53 AM
Blasphemy! Real astronomers use cgs. I know how long an AU is in cm and what the sun's mass is in grams, but I have to think for a second to convert them ito SI/mks. :D

That being said, it sounds like Jim Starluck doesn't have much physcial background at the moment, and so I wan't going to try to force units on him. They're inconsequential anyway. If he finds it in km, he'll likely end up converting to miles (well, *I* do this often enough for smaller numbers). As for the mass, it cancels out of the equation, so he could use slugs or galactic masses and it wouldn't matter. As long as his values are correct, he can use any units he chooses.

Actually, I *was* going to use kilometers. =P

Okay, another question...if I wanted to find the LaGrange points of, say, the Jupiter-Sol system...would I need to factor the mass of all the Jovian moons, or would they not affect the equation?

Edit - Finally went to calculate these...couldn't quite understand the second coordinate of L4/L5

sqrt(3)/2*R])

Is that the square root of three divided by (two times R), or the square root of (three divided by (two times R))?

Kullat Nunu
2004-May-17, 08:32 AM
Edit - Finally went to calculate these...couldn't quite understand the second coordinate of L4/L5

sqrt(3)/2*R])

Is that the square root of three divided by (two times R), or the square root of (three divided by (two times R))?

It's sqrt(3) divided by 2 and then the whole thing multipled by R (sorry for the extra bracket).

R*sqrt(3)/2 is safer to use.

ngc3314
2004-May-17, 10:56 AM
Blasphemy! Real astronomers use cgs.

And I hate them for it...

*shakes fist in rage*

Do you have any idea how hard it is to use cgs in ONE out of twelve classes every year, and mks in the remaining eleven?

Well, way back when _this_ professor was in school, physics courses were as likely to use cgs as mks (and I'm still trying to repress all that stuff involving the two different sets of EM units). And once you've imprinted one system, you have a much better intuitive feel for the rightness of many quantities in that system. To this day I'm uncomfortable even having to solve Newton's version of Kepler's third law in mks, because I sort of know sans calculator where a solar mass falls in cgs...

The journals have tried, but made almost no progress in eliminating Angstroms for wavelength in the optical and UV. They're just too useful and give exactly the right precision for too many things. However, get past one micron and almost everyone goes to that SI-acceptable form
for IR wavelengths - your choice whether to list in observed or emitted frames..

Ut
2004-May-17, 11:01 AM
Well, I'll be honest... The cgs/mks systems interchange more easily than the SI/Imperial systems. Is all chemistry done using Imperial? Or were my first year chem professors just out of their minds?

milli360
2004-May-17, 05:21 PM
However, the coordinates given for L4/L5 don't translate into 60 degrees, so I'm suspicious. I'll have to look at that PDF file more closely.

PS: OK, just before equation 14 on page 5, he says "Demanding that the parallel component of the force vanish leads to the condition that the equilibrium points are at a distance R from each mass." And then apparently he uses that to derive the coordinates for L4/L5--which means the coordinates given in equation 14 are wrong. They must be:
L4 = (R/2, sqrt(3)/2*R])
L5 = (R/2, -sqrt(3)/2*R])

Don't forget that M1 must be >> M2 so

(M1-M2)/(M1+M2) ~ 1.

That 60° is good approximation, but not accurate.
Are you sure? In that paper, and in the analysis I've done in the past, it's not. Notice, he says that the points are at a distance R from both objects--that makes the whole system an equilateral triangle.

I don't see how he got the answers he got, from the analysis he presents. Either way, there's an inconsistency in the paper.

Kullat Nunu
2004-May-17, 07:53 PM
Are you sure? In that paper, and in the analysis I've done in the past, it's not. Notice, he says that the points are at a distance R from both objects--that makes the whole system an equilateral triangle.

I don't see how he got the answers he got, from the analysis he presents. Either way, there's an inconsistency in the paper.

Well, I won't argue since I'm definitely not an expert and you seem to know better :).

Donnie B.
2004-May-17, 09:41 PM
How come Earth's L4 and L5 Lagrange Points are empty? :-k
Maybe Brad Pitt's career is parked there.

(Nudge, nudge. Get it? Trojan points? Har har.)

Tobin Dax
2004-May-17, 11:43 PM
Actually, I *was* going to use kilometers. =P

Yeah, if I wasn't too busy defending myself, Ii could have scrooled up and realized that. Oh, well. :-?

milli360
2004-May-18, 01:25 AM
Are you sure? In that paper, and in the analysis I've done in the past, it's not. Notice, he says that the points are at a distance R from both objects--that makes the whole system an equilateral triangle.

I don't see how he got the answers he got, from the analysis he presents. Either way, there's an inconsistency in the paper.

Well, I won't argue since I'm definitely not an expert and you seem to know better :).
To tell you the truth, it's not much. Found it (http://www.badastronomy.com/phpBB/viewtopic.php?p=197796&amp;highlight=lagrange#197796), we were talking about it in Jan. in Bad TV.