Orbital Mechanics

by Rian Cartmell, January 1998

Table of Contents

Here's the relevant quantities:
These quantities are explained on other pages. 

New Quantities

Sorry! There are no new quantities here.


Here are the relevant formulae that we have so far:
  1. Fg = Gm1m2/r2
  2. Fc = m2v2/r
  3. v/r = 2pi/T
    And now for the only new thing here:
  4. For the orbital condition to be satisfied, Fg = Fc
    Simple substitution then gets us to the most common variant, Gm1m2/r2=m2w2r.  
    This simplifies to v2r = Gm1
    Note that the satellite's mass drops out here.  Substituting  2pi/T for v/r leads  to 
    Kepler's observation that T2/r3 is constant for all planets in our solar system.
Go back to:  Table of Contents

General Problem Solving Strategy:

  1. Read the problem.
  2. Go through the problem and figure out what is given or implied
    Make a list, and identify the quantities you know.
  3. Set up the orbital condition. If necessary, replace v/r with 2pi/T.
  4. 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:
So you set up the orbital condition:  w2r3 = Gm1

Solve it for v, and you get v = (Gm1/r)1/2.  

Toss in the numbers, and you should get  v = (6.67(10-11)Nm2/kg25kg/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(106), and a mass of 7.36(1022)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 sub-trick here is that the period is in hours-it 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. (note-some problems may expect you to know astronomical data. If that is the case, use the real figures.)
So here's what you really know: