Binary Star—Equal Masses. Two identical stars with mass M orbit around their center of mass. Each orbit is circular and has radius R, so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) How much energy would be required to separate the two stars to infinity?
Solution 71P The gravitational force of attraction between two bodies with masses M and M and separated 1 2 by a distance R is given by F =GM M1 2 …..(1) R2 Since the two stars are always on the opposite sides of the circle of radius R, hence the distance between them is = 2R Each star has a mass of M. (a) Substituting these values in equation (1), we get 2 F = GM 2 (2R) GM 2 F = 4R2 This is the gravitational force of one star on the other. (b) Orbital speed The gravitational force calculated in step (a) will be equal to the centripetal force of orbital motion of a star. GM 2 Mv 2 Therefore, 4R2 = R , here v is the orbital speed. v = GM …..(2) 4R v = 1 GM 2 R This is the required orbital speed of each star. Period T = 2R/v T = 2R GR 3/2 T = 4R / GM This is the required period of the orbit. (c) To solve this part, we shall have to first calculate the kinetic energy of each star. The kinetic energy of one star = Mv 2= M × GM = GM 2 2 2 4R 8R GM 2 2 Total kinetic energy of both the stars = 2 × 8R = GM /4R Since the stars are orbiting, hence they possess kinetic energy and in order to separate them to GM 2 infinity, 4R amount of energy would be required.
