Orbital Mechanics
by Rian Cartmell, January 1998
Table of Contents
Here's the relevant quantities:
r distance between centers
m_{1} mass of central object
m_{2} mass of satellite
F_{g} force due to gravity
F_{c} centripetal force
G universal gravitational constant = 6.67(10^{11})Nm^{2}/kg^{2}
v orbital velocity
T orbital period
These quantities are explained on other pages.
New Quantities
Sorry! There are no new quantities here.
Formulae
Here are the relevant formulae that we have so far:

F_{g} = Gm_{1}m_{2}/r^{2}

F_{c} = m_{2}v^{2}/r

v/r = 2pi/T
And now for the only new thing here:

For the orbital condition to be satisfied, F_{g} = F_{c}
Simple substitution then gets us to the most common variant, Gm_{1}m_{2}/r^{2}=m_{2}w^{2}r.
This simplifies to v^{2}r = Gm_{1}
Note that the satellite's mass drops out here. Substituting 2pi/T for v/r leads to
Kepler's observation that T^{2}/r^{3} is constant for all planets in our solar system.
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General Problem Solving Strategy:
 Read the problem.
 Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
 Set up the orbital condition. If necessary, replace v/r with 2pi/T.
 Solve it for what you're looking for. Replace known values with their numerical equivalents. For this type of problem, this is all you should ever need to do.
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Example problem 1
A 1 gram fly is in orbit around a 5 kilogram bowling ball.
If it revolves at a distance of 1 m, what is the velocity of the fly?
Here is what you start with:
m_{1} = 5kg
m_{2} = .001kg
r = 1m
F_{g} = ?
F_{c} = ?
T = ?
v = ?
So you set up the orbital condition: w^{2}r^{3} = Gm_{1}
Solve it for v, and you get v = (Gm_{1}/r)^{1/2}.
Toss in the numbers, and you should get v = (6.67(10^{11})Nm^{2}/kg^{2}5kg/1)^{1/2}=.000 018m/s.
This answer is absurdly small because the masses were also absurdly small.
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Example problem 2
A Vogon ship is in lunosynchronous orbit on the far side of the moon so that it can escape detection while waiting for final verification of Earth's termination orders. If the moon has a rotational period of 27.32 days and an equatorial radius of 1.738(10^{6}), and a mass of 7.36(10^{22})kg, how high above the Moon's surface is it orbiting?
WARNING! This problem contains two new tricks! The first is that it's a synchronous orbit. This means that it will always be above the same spot, thus the orbital period will be the same as the rotational period of the planet it's orbiting. (A subtrick here is that the period is in hoursit needs to become seconds, as always.) The other is "Above the Moon's surface," which you have to convert to orbital radius by adding the equatorial radius. (notesome problems may expect you to know astronomical data. If that is the case, use the real figures.)
So here's what you really know: