1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Keppler's third law and the speed of cellestial bodies.

  1. Aug 9, 2011 #1
    Hi. Keppler's third law implicates that the velocity of an object orbiting around our sun is directly dependant on the object's average distance to the sun... Afterall, if the velocity of be more irregular, then no constant would exist.

    I realize this probably has something to do with the average centripetal force provided by the gravity of the sun...

    I was thinking that this is because the longer away an object is the bigger its orbit and thus the smaller the average centripetal force and thus one could assume that it is possible that the speed must be smaller as well? mass*(v^2)/r=centripetal force. But then again the radius r is also bigger if the orbit is bigger and that allows for a larger speed..

    bah.

    And:

    The next sub-chapter of my physics book it is about Newton's work with gravity and using that jazz to prove keppler's laws.. So maybe I will find out eventually tho still I'd appreciate it if you people could make some simple explanations.
     
    Last edited: Aug 9, 2011
  2. jcsd
  3. Aug 9, 2011 #2

    Drakkith

    User Avatar

    Staff: Mentor

    The larger the radius, the lower the velocity must be in order to sustain an orbit. In an elliptical orbit the orbiting body moves faster as the distance between it and the larger body decreases, with a max velocity at its closest approach, perigee. After that it slows down until it reaches apogee, or farthest approach.
     
  4. Aug 9, 2011 #3

    Janus

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In a simple case, consider a circular orbit where the centripetal force needed to maintain the circular path is equal to the force of gravity or:

    [tex]\frac{GMm}{r^2}= \frac{mv^2}{r}[/tex]

    This reduces to

    [tex]\frac{GM}{r} = v^2[/tex]

    Given that the period(P) is the time it would take for the object to travel a distance of [itex]2\pi r[/itex]

    A little algebra will get you

    [tex]P^2 = 4 \pi^2 \frac{r^3}{GM}[/tex]

    Which demonstrates Kepler's third law.
     
  5. Aug 10, 2011 #4
    thanks=) everyone

    janus: I'll be learning about that formula you posted (force of gravity) in the next sub-chapter chapter
     
  6. Aug 10, 2011 #5
    After reading the next chapter, I understand everything, Janus. A very elegant explanation imo^^
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Keppler's third law and the speed of cellestial bodies.
  1. Newton's Third Law (Replies: 3)

Loading...