Here's
an example. Consider the system shown to the right of two satellites
orbiting a planet. Let's compare forces, accelerations, speeds, and
periods. Define these symbols:
*m*_{o},* m*_{i} represent the masses of the blue and red satellites. (*o* =
outer, *i* = inner)
*M*_{p} represents the mass of
the planet.
*r*_{o},* r*_{i} represent the radii of the satellites' orbits.
Suppose we're given that *r*_{o}/r_{i} = 2 and *m*_{o}/m_{i} = 3.
How do the accelerations of the satellites
compare? The equation that applies is Eq. 2 above. In addition to G, *M* will be a constant, since *M* represents the mass of
the planet, which is the same for both satellites. Therefore, we set
up a proportion between acceleration and orbital radius.
Note that the radii are squared and are inverted
compared to the accelerations. This, of course, is the nature of an
inverse-squared relationship. Substituting the inverse ratio of the
radii and squaring it, *a*_{o}/*a*_{i} = 1/4. It
makes sense that the satellite which is further away experiences
lesser acceleration.
Let's suppose now that you want to compare
gravitational forces between the satellites and the planet. Now we use Eq. 1. We have the same constants *G* and *M* as before.
Our proportion is:
Note that the force depends directly on the mass.
Since *m*_{o}/*m*_{i} = 3, then *F*_{o}/*F*_{i} = 3/4.
Now compare orbital speeds using Eq. 3. Once
again, the constants are *G* and *M*. Our proportion is:
We substitute 1/2 for the inverse of the given ratio of the orbital
radii and take the square root to obtain *v*_{o}/*v*_{i} = sqrt(1/2). The outer satellite moves slower as expected.
Finally, we compare periods. We have the same
constants as before. The proportion is:
We substitute *r*_{o}/*r*_{i} =
2, cube that, and take the square root. The result is *T*_{o}/*T*_{i} = sqrt(8) or 2sqrt(2). The outer planet has the greater period. |