Langrange points

Let the two masses M and m orbit about the origin such that M is at (R, π) and m is at (r, 0).

The co-ordinate system rotates in synchronism with the masses with angular velocity ω.

The gravitational force on each mass is:
           G M m
( R + r ) 2
This will be exactly balanced by the centripetal force when the orbit is stable.
           G M m
( R + r ) 2 		  =   M R ω 2   =   m r ω 2

Lagrange points not on the line joining M and m.

The centripetal force acts outwards from the common centre of mass which we have taken as the origin of our rotating co-ordinate system. If we resolve forces at right angles to this line we get just the balance of the gravitational forces from the two masses. If the gravitational field is to cancel out, we need:
           G M
S 2 		sin θ   =   G m
s 2 		sin φ

           M R
S 2 		  sin θ
R 		  =   m r
s 2 		  sin φ
r

By the sine rule:
           M R
S 2 		  sin β
S 		  =   m r
s 2 		  sin α
s

As   M R = m r   and   sin β = sin α , we conclude that   S = s , i.e. the triangle is isosceles.

We need another equation in order to relate S and s to R and r. We can resolve about any line not perpendicular to OP, e.g. the line joining M and m.
The centripetal force must balance the two gravitational forces if the gravitational field is to vanish at this point. We can exploit our knowledge of the isosceles triangle to get:
           G M
s 2 		sin γ   +   G m
s 2 		sin γ   =   ρ ω 2 sin α   =   s ω 2 sin γ

Remembering that:
           G M m
( R + r ) 2 		  =   M R ω 2   =   m r ω 2

we have:
      G M   =   r ω 2 ( R + r ) 2
and:
      G m   =   R ω 2 ( R + r ) 2
leading to:
           ( G M + G m )
s 2 		sin γ   =   s ω 2 sin γ

      ( R + r ) 3 ω 2 sin γ   =   s 3 ω 2 sin γ
and hence:
      R + r   =   s   =   S
           but only so long as   sin γ   ≠   0 , i.e. not on the line joining the two masses.

Lagrange points on the line joining M and m.

Lagrange points along the axis (ρ, 0) will feel a “gravitational” acceleration composed of the gravitational attraction of the two masses and the centripetal force ρ ω 2, which will all cancel to zero.

If   0 < ρ < r,   M will attract against the centripetal force and m with it:
           ρ ω 2   +   G m
( r – ρ ) 2 		  =   G M
( R + ρ ) 2

The formula is actually valid for   – R < ρ < r, because it is linear in ρ.
      ( r – ρ ) 2   ( R + ρ ) 2   ρ ω 2   –   ( r – ρ ) 2   G M   +   ( R + ρ ) 2   G m   =   0
      ( r – ρ ) 2   ( R + ρ ) 2   ρ ω 2   –   ( r – ρ ) 2   r ω 2 ( R + r ) 2   +   ( R + ρ ) 2   R ω 2 ( R + r ) 2   =   0
      ( r – ρ ) 2   ( R + ρ ) 2   ρ   –   ( r – ρ ) 2   r ( R + r ) 2   +   ( R + ρ ) 2   R ( R + r ) 2   =   0

Following Arthur, put   r – ρ = X   and   R + ρ = Y.
      X 2 Y 2 ρ   –   X 2 r ( R + r ) 2   +   Y 2 R ( R + r ) 2   =   0
and then put   x ( R + r )   =   X   and   y ( R + r )   =   Y:
      x 2 y 2 ρ   –   x 2 r   +   y 2 R   =   0

Still following Arthur, put
           a   =   R
( R + r ) 		  =   m
( M + m )

and
           b   =   r
( R + r ) 		  =   M
( M + m ) 		  =   1   –   a

Also
           y   =   1   –   x     and     ρ
( R + r ) 		  =   b   –   x

giving us a quintic in x :
      x 2 ( 1 – x ) 2 ( b – x )   –   x 2 b   +   ( 1 – x ) 2 a   =   0

Let M = m, \ R = r ,   and   a = b = ½.
By symmetry we would expect that x = y = ½

Our quintic equation for x now becomes:
      x 2 ( 1 – x ) 2 ( 1 – 2 x )   –   x 2   +   ( 1 – x ) 2   =   0
      x 2 ( 1 – x ) 2 ( 1 – 2 x )   –   x 2   +   1 – 2 x + x 2   =   0
      x 2 ( 1 – x ) 2 ( 1 – 2 x )   +   1 – 2 x   =   0
      [ x 2 ( 1 – x ) 2   +   1 ] ( 1 – 2 x )   =   0
      \ x   =   ½   is a root   Q.E.D.
The term in square brackets is positive definite, so there are no other real roots.

Lagrange points in line with M and m beyond m.

The equations are as for the previous case except that ρ > r so that the gravitaional attractions both act against the centripetal force.
           ρ ω 2   =   G m
( r – ρ ) 2 		  +   G M
( R + ρ ) 2

leading to the quintic (in which x is negative):
      x 2 ( 1 – x ) 2 ( b – x )   –   x 2 b   –   ( 1 – x ) 2 a   =   0
      x 2 ( 1 – x ) 2 ( b – x )   –   ( 1 – 2 x + x 2 ) a   –   x 2 b   =   0
      x 2 ( 1 – x ) 2 ( b – x )   –   ( 1 – 2 x ) a   –   x 2   =   0
Let M = m, \ R = r ,   and   a = b = ½.

Our quintic equation for x now becomes:
      x 2 ( 1 – x ) 2 ( 1 – 2 x )   –   ( 1 – 2 x )   –   2 x 2   =   0
      [ x 2 ( 1 – x ) 2   –   1 ] ( 1 – 2 x )   –   2 x 2   =   0

Given that x < 0 , ( 1 – 2 x ) > 0 , so that the square bracket term must be positive.
      \ x ( 1 – x )   >   1   or   x ( 1 – x )   <   – 1
The first inequality can never be satisified for negative x.
The second inequality becomes:
      x – x 2   <   – 1
      0   <   x 2 – x – 1
      x   <   ½ ( 1 – √5 )

\ ρ   >   ½ √5 ( R + r )