The unit binary separation and unit mean motion imply through Kepler third law thatThe 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 , soWith the effective potentialits Cartesian components are exactlyThis 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, leavingEquivalently, the radiation-modified Jacobi constant is
DefineAt an equilibrium point, the velocity and acceleration vanish. Direct differentiation givesandThus the equilibrium conditions are
At a Triangular Lagrange point, , so . The equation then givesSubstitution into and yieldsIntersecting these two circles givesAs 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 givesThe two triangular points are intersections of circles with these radii and centre separation one. Iftheir coordinates areFor ordinary outward radiation pressure, , the non-collinear points exist precisely when the strict triangle inequalityholds. 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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