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Centripetal acceleration of objects in orbit around the Earth

  1. Dec 11, 2014 #1
    Hi all. The answer to this might be trivial. If it is, sorry for posting. If you calculate the acceleration due to gravity of an orbiting satellite, it could be around 8.5-9.5m.s-2, depending. So, it's tangential velocity is such that as it falls towards earth, earth curves away and the satellite never comes closer to earth - here's my problem. If it's is accelerating towards earth at say 9m.s-2, then every second it's downward velocity increases by 9m.s-1. If it continually accelerates in this way, eventually you have massive downward velocity, yet tangential velocity remains constant...the satellite should come crashing down. So, perhaps it's reached terminal downward velocity? But there's scant air resistance.. what gives?
  2. jcsd
  3. Dec 11, 2014 #2


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    "downwards" is not a fixed direction. As the satellite moves along its orbital path, "downwards" now is different from "downwards" a moment ago and from "downwards" a moment from now. After 1/4 of a complete orbit, the new "downwards" acceleration is no longer adding velocity along the original direction at all.
  4. Dec 11, 2014 #3


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    OK You realise it doesn't actually get any closer so there has to be an explanation. Try this:
    If you realise that every time the satellite goes round once, the centripetal force has pointed in all possible radial directions, each instant of acceleration in one direction is balanced out by an instant, radially opposite acceleration, when it gets round there. So the net effect will be zero change in radius (or speed). As jbriggs has already pointed out, "downwards" really means radial.
    The above only applies in the case of circular motion. If the orbit is elliptical (most / all are like this) the effect of g at any point will be to add to or subtract from the tangential speed (except at perigee and apogee, of course).

    If there is any significant quantity of air at the orbital height, energy is continually lost through friction so the orbit will decay, catastrophically.
  5. Dec 12, 2014 #4
    Hmm. Thanks for the quick replies. I like the explanation that the net radial acceleration is zero, as for every radial acceleration, there is an opposite equal in magnitude acceleration. Is there another way of explaining this, perhaps?
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