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SOLUTION TO EXERCISE 34
When an object is in a stable orbit around another, the gravitational acceleration it feels must equal the centripetal acceleration. In other words, GM v^2 ---- = ----- r^2 r Meanwhile, we know that the velocity of a nearly circular orbit can be expressed as v = 2 * pi * r / p, which is the circumference of the orbit divided by the orbital period. Now plug the second equation into the first to get GM (2*pi*r/p)^2 ----- = ------------ which can be "cross-multiplied and divided" to get r^2 r GMr = r^2 * 4 * pi^2 * r^2 / p^2 which can be simplified to p^2 = (4 * pi^2 / GM) * r^3 which is Kepler's third law. If we compare this equation to Kepler's formulation of the third law: P^2 = kr^3 (orbital period squared = a constant * orbital radius cubed) We see that k = (4 ?2)/(G M) where M is the mass of the object being orbited. This formula for k is correct when the orbiting object has a much smaller mass than its primary (the object being orbited). This is almost always the case for planets, asteroids and comets in a solar system, but not always the case for binary star systems, star clusters, or galaxies. In those cases, The mass m of the orbiting object must be taken into account, and the formula becomes k = (4 ?2)/(G (M+m)) The slightly more rigorous derivation of this more general formula can be found, for example, at http://www-astro.physics.uiowa.edu/~lam/teaching/general/10.html. |