### Orbital Equations

• The "centripetal" force required to keep a particle with a mass, m, travelling on a circular path of radius, R is:

where v is the velocity of the particle.

• This equation can be partially derived intuitively by noting that if you swing a ball around your head on a string...

• If you increase the mass of the ball you will have to pull harder on the string.

• If you swing the ball around your head at a faster speed you will have to pull (much) harder on the string.

• If you lengthen the string while keeping the speed of the ball constant your tug on the string decreases.

• Equating this force and gravitational force we get:

In the equation above M is the mass of the object being orbited (e.g. the Sun) and m is the mass of the object in orbit. R is the separation between the two objects (i.e. the length of the string). Note that m cancels out on both sides of the equation.

• Rearranging this equation you can get:

• These relations show that you can:

• find the circular orbital velocity needed to launch a satellite into Earth orbit.

• determine the mass of a planet (or star or galaxy...) which has a satellite with a known velocity and separation from the planet.

• Applying Newton's Laws permits us to determine the mass of just about any structure in the Universe.

• For those who are interested... the equations above are actually an alternative experession of Kepler's Third Law.... (Orbital Period(years))2=(Orbital Radius(A.U.))3

• We can relate the period of the orbit (P) to the velocity (v) above by noting that the planet completes a circular orbit in each time interval, P.

• Re-arranging this equation we get

which is Kepler's Third Law. The constants appear because this equation is arranged for using meters and seconds (rather than A.U. and Years, in which case the constants all work out to 1 and thus disappear).

Updated October 1, 2007