Circular orbits under gravity, and satellites


Basic relationships between speed, radius, period, acceleration and force

Here is a demonstration of the relationships between orbital speed, radius, period, centripetal acceleration, and gravitational centripetal force for an object of mass m2 in orbit around an object of mass m1.



Geostationary orbits

notes_page3Polar orbits

A polar orbit is one that goes over the top of each of the north and south poles.  They are often used for surveying: to understand why, it is worth looking at this video of a polar orbit in Kerbal Space Program or this one from EUMETSAT.

Not every circular polar orbit will cover the whole surface of a planet.  If you have a globe, or a good imagination, it is worth trying to work out what happens with polar orbits whose period is some exact multiple of the planet’s day.


11 thoughts on “Circular orbits under gravity, and satellites

  1. Hello,

    Can I please ask a question, thanks for the help with the PAT this year.

    I derived the same equation using centripetal force=gravitational attraction, but didn’t have to neglect the mass of the satelite because it cancelled itself.

    Yours is definitely the general equation and the more robust one; I would just like to ask why did my method not yield the same (m1+m2); yet yours did. We seemed to have used the same first principles..

    My guess is that it has something to do with you taking the accleration of the system to be the sum of the acceleration of the two indidvual bodies; and that plays a significant role if the two bodies are of simlar mass. I can’t quite follow the same logic using forces..?

    Thanks, don’t worry if you dont have time; i apprictate everything you have done for the PAT for everyone!!!


    1. I’m glad this has been helpful and I hope the test went well for you.

      I think your intuition is right, and the difference relates to how you treat the different frames of reference — this is a bit subtle so I’ll try and work out a good way of explaining the details, but it will take a few days before I can put anything up. Watch this space.

  2. Binary Star Systems came up this year. They asked about three stars instead of two equal distance away from each other….

  3. Hi,

    With M2 orbiting M1. Equating forces via gravitational force and circular motion (GM1M2)/(R^2) = (M2v^2)/R leads to the formula v^2 = (GM1)/R. However, you state v^2 = (G(M1+M2))/R.

    These equations give conflicting results and I’m not too sure how to explain that fact? I would appreciate any help.

    Thank you for all of the materials you have produced and all the help you have given me in revising for the PAT.

    Thank you in advance.

    1. It’s a subtle point and I wouldn’t worry too much about it — remember that when M1 is much bigger than M2 then these values are equivalent. When the masses are of similar size, the discrepancy is explained by the fact that really the objects are moving in a circle about the centre of mass of the whole system. For example, imagine the case where you have a binary star — two stars of basically the same mass orbiting each other — and do some calculations with that system and things should make more sense.

      But don’t get too hung up on this — maybe better to just have a rest and make sure you are relaxed for the exam.

  4. Hello,
    I’m just wondering what were you trying to calculate when you did -d^2R/dt^2? Is that the total acceleration of m1 and m2?

    1. Hi Claire, thanks for your question.

      Yes that is absolutely right. Another way of stating this is to think about the centre of gravity of m1 and m2, which lies on the line between them. The masses m1 and m2 are both accelerating towards this centre of gravity, and the value -d^2R/dt^2 is the sum of the accelerations of the two masses.

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