Physics PAT 2017 question 20 https://www2.physics.ox.ac.uk/sites/default/files/page/2011/02/14/pat-2017-paper-42578.pdf

(Unfortunately the question did not fit in the box, but I have provided a link and can add it to the interview)We begin by considering the binary system. As the two stars are identical (i.e. they have equal mass), there orbits must have an equal radius and period, and they must orbit their centre of mass (i.e. half way between them). Labelling the distance between them d, then each star experiences and acceleration due to gravitational attraction of mG/d^2 and for circular orbit must have a centripetal acceleration mv^2/R = 2mv_2^2/d. Equating these, we find v_2^2 = mG/2d = mG/4RIn the three body case, we again note that all three stars must have the same speed to maintain them at equal angular distances. We must now calculate the sum of two forces on each star. Using a force diagram, we see that the forces acting on star A are equal but reflected in the radius of A, so we can ignore the tangential components of the two forces and focus on the radial components. Using geometry, we can draw a triangle of distances to show that the separation of the stars is sqrt(3)R, so the net force from each is m^2G/3R^2. We can then use trigonometry to determine that the radial component of this is sqrt(3)m^2G/6R^2, for a total radial force m^2*G/sqrt(3)*R^2 on each star. Inserting this into the centripetal force equation F=mv_3^2/R, we find v_3^2 = mG/sqrt(3)R and thus v_3 = 2/3^(1/4) * v_2