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Old 2007-09-02, 11:07   #1
davieddy
 
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"Lucan"
Dec 2006
England

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Default Lagrange points L4 and L5

We have three different masses located
at the vertices of an equilateral triangle.

Show that their accelerations due to gravity
are directed towards the centre of mass and
inversely proportional to the square of their
distances from it.

If their initial velocities (relative to the c of m)
are at 120 degrees to each other and proportional
to their distance from the c of m, show that their
elliptical orbits have identical periods.

David

Last fiddled with by davieddy on 2007-09-02 at 11:21
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Old 2007-09-02, 19:57   #2
davieddy
 
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"Lucan"
Dec 2006
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A neater way of posing the problem is as follows:
if L is the side of the equilateral triangle and r is
the position of a mass relative to the C of M, show that
acceleration = -k*r/L^3. What is k?

(The similarity of orbits follows from this)

PS where can I find out why [bold] isn't working?

Last fiddled with by davieddy on 2007-09-02 at 20:55
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Old 2007-09-02, 20:34   #3
Wacky
 
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Quote:
Originally Posted by davieddy View Post
PS where can I find out why [bold] isn't working?
Who says that it isn't working? Just because it doesn't do what you suggest it should ...

Now if you were looking to place a part of your response in boldface type, you should use "B" rather than "bold". That is much more likely to have the desired effect.
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Old 2007-09-02, 20:57   #4
davieddy
 
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THX Wacky. I can't find where this sort of info is hiding ATM
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Old 2007-09-03, 05:55   #5
S485122
 
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FAQ
Reading and Posting Messages
Are there any special codes/tags I can use to format my posts?

vB Code List

Last fiddled with by S485122 on 2007-09-03 at 05:57
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Old 2007-09-03, 07:44   #6
mfgoode
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Jan 2004
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Thumbs up



THX S-----. I have not been aware of this and have never used this powerful code before. I must try a post in Tex.

Mally
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Old 2007-09-03, 20:13   #7
davieddy
 
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Quote:
Originally Posted by S485122 View Post
FAQ
"Nobody bothers to read this"

BTW if you want to know what the three masses are called
they ar z1, p2 and j42.
I thought that went without saying

Last fiddled with by davieddy on 2007-09-03 at 20:18
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Old 2007-09-04, 12:50   #8
davieddy
 
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Don't know whether this "puzzle" was too easy/hard/badly posed
but since there have been no takers, here is my solution which IMHO
is quite neat:


Quote:
Originally Posted by davieddy View Post
Label the masses 1,2 and 3 and consider the acceleration
of m1 due to the gravitational attraction of 2 and 3.

This is (m2(r2 -r1)+ m3(r3 -r1))G/L3

We also have m1r1 + m2r2 + m3r3 = 0
by definition of centre of mass.

It follows that the acceleration of m1 is

-G(m1 + m2 + m3)r1/L3

David

Last fiddled with by davieddy on 2007-09-04 at 13:08
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