What is the separation at which an orbiting object has its average speed, relative to the primary object?

Find a general expression for the separation between two gravitationally bound masses in a elliptical orbits around a common center of mass have their average speed relative to each other. (Hint: it isn't the length of the semimajor axis.)

2 Answers

Relevance
  • 1 year ago
    Favourite answer

    I'll consider a case where a small body is orbiting a large one, so I won't deal with the motion of the large body.  My "ANSWER" is noted below as "ANSWER".

    The orbital period is 2*pi*a^(3/2)*(GM)^(-1/2), where "a" is the semi-major axis of the orbit.  The speed at any particular separation "r" is sqrt[GM(2/r - 1/a)].  These facts are given in various websites, including Wikipedia.

    A decent approximation for the arc-length of an ellipse is

    L = pi*{3(a + b) - sqrt[(3a + b)(a + 3b)]}.  This approximation is due to Ramanujan.  So the average speed of the orbiting body would be

    {3(a+b) - sqrt[(3a+b) (a+3b)]} * (GM)^(1/2) / [2a^(3/2)].

    So we're looking for "r" such that

    sqrt(2/r - 1/a) = {3(a+b) - sqrt[(3a+b)(a+3b)] / [2a^(3/2)].

    From here on, one could solve for "r" in terms of a and b, by methods of high-school algebra!  Whether Ramanujan's approximation is robust enough to support such an answer, I cannot say.  

    Let's see if I can do it without frying my brain:

    (2/r - 1/a)*(4a^3) 

    = 9(a+b)^2 + (3a^2+10ab+3b^2) - 6(a+b)*sqrt(3a^2+10ab+3b^2) =>

    8a^3/r - 4a^2 = 9a^2 + 18ab + 9b^2 + 3a^2 + 10ab + 3b^2 - 6(a+b)*sqrt(3a^2 + 10ab + 3b^2) =>

    8a^3/r = 16a^2+28ab+12b^2 - 6(a+b)*sqrt(3a^2+10ab+3b^2)  =>

    4a^3/r = 8a^2+14ab+6b^2 - 3(a+b)*sqrt(3a^2+10ab+3b^2)

    =>

    ANSWER:   r = (4a^3) /

    [8a^2+14ab+6b^2 - 3(a+b)*sqrt(3a^2+10ab+3b^2)].  

    As a check on reasonableness, let's see what happens when we have a circular orbit (a=b).  You'd get

    r = 4a^3/(28a^2 - 6a*sqrt(16a^2))

    = 4a^3/(28a^2 - 24a^2) = a,

    which of course is the right answer for a circular orbit.

    Well, at least THAT case tends to support my analysis !!!

  • 1 year ago

    The bunch of words is unclear. Do you want to know where in the orbit, the object's instantaneous speed equals its average orbital speed?

Still have questions? Get answers by asking now.