        SOLUTION TO EXERCISE 34

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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.

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