Kepler's third law for two point masses in a Newtonian gravitational field isHere is the semimajor axis of their relative orbit, the orbital period and the gravitational constant. Equivalently , to the accuracy with which these astronomical units normalize the solar mass.
Assume a circular orbit and an orbital inclination , as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,Thus , and Kepler's third law givesThe orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give ; the numerical answer requires the circular assumption.
Adopt the approximate mass-luminosity relation and equal available fuel fractions. The main-sequence lifetime then scales as , givingUsing an exponent instead gives about and years; these are model-dependent estimates. At the Newtonian gravitational potential energy released from infinity is . Half remains as kinetic energy in a circular orbit, so the radiatable binding energy is :Using the whole potential energy loss would give , which does not subtract the final kinetic energy. A fully relativistic calculation for a nonrotating black hole gives . When the surviving star exhausts hydrogen burning, it expands off the main sequence and can fill its Roche lobe. Binary mass transfer to the black hole can form an accretion disk and an X-ray binary; unstable binary mass transfer may instead produce a common envelope. This is conditional on the binary star remaining bound after the first remnant forms. The supplied black hole endpoint is an assumption, not a universal outcome for an isolated star.
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