Home Table of Contents About the Authors Glossary Buy This Book Joseph Henry Press


SOLUTION TO EXERCISE 34

back to exercise

motion: TOC for Knowledge Concepts, Exercises, and Solutions



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.