The unit binary separation and unit mean motion imply through Kepler third law that
The centre of mass conditions put at and at . Therefore
In a frame with unit angular velocity, transforming the acceleration introduces the Coriolis acceleration and centrifugal acceleration. Radiation reduces the attraction of by the factor , so
With the effective potential
its Cartesian components are exactly
This is the photogravitational restricted three-body problem.
Multiply the three equations by and add. The Coriolis acceleration does no work because its two terms cancel, leaving
Equivalently, the radiation-modified Jacobi constant is
Define
At an equilibrium point, the velocity and acceleration vanish. Direct differentiation gives
and
Thus the equilibrium conditions are
At a Triangular Lagrange point, , so . The equation then gives
Substitution into and yields
Intersecting these two circles gives
As rises from zero to one, falls from one to zero. The two points move along the unit circle about , from the classical equilateral positions toward , where they coalesce when the attraction of is fully cancelled.
Put . Repeating part (v) with radiation from both bodies gives
The two triangular points are intersections of circles with these radii and centre separation one. If
their coordinates are
For ordinary outward radiation pressure, , the non-collinear points exist precisely when the strict triangle inequality
holds. Equality merges the two points on the line of centres.
The same absorption and re-emission that produce radial radiation pressure also produce Poynting–Robertson drag. This velocity-dependent force removes specific orbital energy and angular momentum, destroys the exact Jacobi constant, and turns the formal equilibrium points into slowly drifting configurations. Stellar-wind drag can provide a similar correction.

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