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Deriving an equation for orbital period

  1. Apr 11, 2012 #1
    Hi guys, I derived an equation for determining orbital period, given an altitude, speed, and mass of the primary and the object orbiting it. I think it makes sense, but I'd welcome anyone who is willing to check it for conceptual error or nonsensical math.

    Here is the equation:

    [tex]P = \frac{2 \pi ~ \mu }{ \Big( \frac{2 \mu}{r} - v^2 \Big) ^{ \frac{3}{2}} }[/tex]
    Where
    [itex]P[/itex] is the orbital period in seconds,
    [itex]\mu[/itex] is the standard gravitational parameter [itex]G(m_1 + m_2)[/itex]
    ([itex]G[/itex] is the gravitational constant in [itex]m^3~kg^{-1}~s^{-2}[/itex],
    [itex]m_1[/itex] and [itex]m_2[/itex] are the masses of the orbiting bodies),
    [itex]r[/itex] is the distance between them in meters, and
    [itex]v[/itex] is their relative speed in meters per second.

    Concept
    The idea is that, if an orbiting body has a certain speed at a certain instant, its orbital period is not affected by the direction in which it is headed at that moment in time. It could be taking a circular orbit or a heavily elliptical one, and its period would be the same because the orbit's specific energy would be the same.
    So, what this equation does is:
    1) Calculate the specific orbital energy using the givens (altitude, speed, m1, m2);
    2) From this specific energy, use the properties of the circular orbit to find [itex]v_c[/itex] and [itex]r_c[/itex], the speed and altitude/distance values for the equivalent circular orbit;
    3) Calculate the orbital period of this equivalent circular orbit.

    Derivation - the pieces
    So, we start with the equation to calculate specific orbital energy:
    [tex]\epsilon = \frac{1}{2}v^2 - \frac{\mu}{r}[/tex]
    And then we will use the properties of a circular orbit to find [itex]v_c[/itex] and [itex]r_c[/itex]:
    [tex]\frac{1}{2}v_c^2 = -\epsilon[/tex][tex]\frac{-\mu}{r_c} = 2\epsilon[/tex]which simplify to:
    [tex]v_c = \sqrt{-2\epsilon}[/tex][tex]r_c = \frac{-\mu}{2\epsilon}[/tex]
    And finally, we'll use [itex]v_c[/itex] and [itex]r_c[/itex] to get the orbital period:
    [tex]P = \frac{2\pi r_c}{v_c}[/tex]

    Derivation - putting them together
    With the circular orbit equations all plugged together, it looks like this:
    [tex]P = \frac{-2\pi \frac{\mu}{2\epsilon}}{\sqrt{-2\epsilon}}[/tex]which simplifies to:
    [tex]P = \frac{2\pi \mu}{(-2\epsilon)^{\frac{3}{2}}}[/tex]
    And then, plugging in the specific orbital energy equation, we get:
    [tex]P = \frac{2\pi \mu}{\Big( -2(\frac{1}{2}v^2 - \frac{\mu}{r}) \Big)^{\frac{3}{2}}}[/tex]which simplifies to the final equation at the head of this post.

    Behaviour
    If the orbiting object is traveling at a speed greater than its escape velocity, the kinetic energy of the orbit is greater than the potential energy. The term inside the (3/2) exponent works out to be negative, and the result is undefined--not a closed orbit, so no period.
    If the orbiting object is traveling at exactly escape velocity, the kinetic and potential energy cancel out, leaving a denominator of zero--undefined again.
    If the potential energy just barely outweighs the kinetic energy, we have a tiny positive denominator, and so we get a huge orbital period. This seems sane--if we consider an object orbiting at almost its escape velocity, it would probably have a large orbital period.



    So yeah...this seems to be a pretty simple derivation, but I'm very noob at this stuff, so please point out if I botched something. And if you look through it and it checks out, please indicate your approval, because I'll probably be wanting to use this in further questions of mine soon. Many thanks!
     
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 11, 2012 #2
    One simple way to check out your equation, Cephron;

    It must replicate Kepler's 3rd law.
    IOW's you must be able to transform your equation back to the Kepler eqn. namely; the period must transform back to ...

    P = sq.rt.[(4(pi)^2) (R^3)/ GM]

    Give it a try ,
    ... for simplicity assume small orbital mass so you can use u = GM instead of G(M+m), and also substitute v = sq.rt.(GM/R), and then solve for P.

    It appears that you may recover Kepler....although I haven't gone through the algebra ...if not then your eqn. is not acceptable classically.

    Creator
     
    Last edited: Apr 11, 2012
  4. Apr 11, 2012 #3
    Given the substitutions you offered, it appears to check out! Thanks!

    I'm not sure how you knew to substitute v for sqrt(GM/r), though...what assumptions are needed for that substitution to work? Would the equation still hold for elliptical orbits if it uses that substitution to get back to Kepler?
     
  5. Apr 11, 2012 #4
    The same assumption applies as we used for u = GM, namely, that the orbiting mass is almost negligible compared to the central mass, but in addition, there is the more important assumption of very little eccentricity (which was the initial constraint with your derivation using circular orbit.)

    Why could I use that substitution? Because for circular orbits the kinetic orbital energy always equals -1/2 the potential energy.
    IOW;
    K.E. =-(1/2)U
    1/2mv^2 = - (1/2)GMm/r

    and solving for orbital vel.(ignoring the sign convention)...
    v = sqrt(GM/r) ....for CIRCULAR orbits. (v is actually the mean orbital speed).

    Not sure, but your equation would probably not be applicable because mean orbital speed actually decreases with increased eccentricity and that would alter the substitution.

    Kepler's equation allows you to substitute the semi-major axis of elliptical orbits for the radius (in the eqn.) and you can still use his same equation.
    But in your case you derived with the initial assumption of circular orbit, and unfortunately the mean orbital speed v changes with eccentricity.

    Fundamentally, the problem arises since you are using a velocity dependent term whereas Kepler uses only period and radius which sweeps out equal areas in equal time periods in spite of a changing radius and thus circumvents the eccentricity problem. Now you know the genius of Kepler.

    Furthermore, the ratio of R^3/P^2 is the same value for ALL planets in the solar system, namely GM/(2pi)^2...(where M is solar mass)....in spite of their different orbital eccentricities as long as R is used as the semi-major axis.

    Good questions, BTW.

    Creator
     
    Last edited: Apr 11, 2012
  6. Apr 12, 2012 #5
    http://burtleburtle.net/bob/physics/orbit101.html

    I'm not certain how reliable this website is, but I'm guessing from his apparently successful gravity simulations that he knows something about all this.

    Anyways, the key thing I noticed on this webpage was where he says:
    and had the little java applet showing the simulation of a bunch of objects departing a point in different directions, but arriving back there at the same time. This is essentially what prompted me to build the above equation: if period is a function of position and speed (and of course mass), then there should be an equation which gives us that.

    My idea for how to calculate the period of any elliptical orbit was to calculate the period of a circular orbit with the same specific orbital energy. It looks to me like specific orbital energy is dependant on the exact same things, no more, no less, as orbital period: distance, absolute velocity, and mass of the bodies. This makes me believe--please correct me if I'm wrong!--that there is a one-to-one mapping between orbital period and specific orbital energy. Thus, if you've calculated the orbital period for one orbit of energy [itex]\epsilon[/itex], you've calculated them all, right? This is the assumption upon which my equation is based...if the one-to-one mapping between specific orbital energy and orbital period is true, then I think my equation would work for calculating the period of any elliptical orbit.
     
  7. Apr 12, 2012 #6

    mfb

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    Staff: Mentor

    The orbital period of an object can be expressed as [tex]T=2\pi\sqrt{\frac{a^3}{\mu}}[/tex]
    where a is the semi-major axis (see Wikipedia for example). This can be related to the orbital energy via [itex]a=\frac{-\mu}{2\varepsilon}[/itex] with the energy [itex]\varepsilon=\frac{v^2}{2}-\frac{\mu}{r}[/itex]. Putting everything together, I think you get your formula.
     
  8. Apr 12, 2012 #7

    BobG

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    Yes, your formula works and it does eliminate the need to calculate the semi-major axis.

    I would note one thing. The geocentric gravitational constant (mu) is just G * M, with M being the mass of the Earth and G being the universal gravitational constant.

    You're interested in the motion of orbiting objects, and you only need the acceleration due to gravity to figure that out (and since F=ma, all objects accelerate at the same rate regardless of their mass).

    That's also why specific energy (actually, specific energy per unit of mass) is used instead of your actual energy level (which does require mass).

    Your mu works for a satellite, since the mass of the satellite is miniscule compared to the mass of the Earth, but the mass of the satellite really shouldn't be there.
     
  9. Apr 13, 2012 #8
    I see what you are trying to do Cephron, but there is no need to derive a new equation. The correspondance between orbital period of an elliptical orbit and specific orbital energy is already established through this other eqn. for total specific orbital energy, e .....

    e = -u / 2a ....where a is the semi-major axis.

    And since by Kepler's law the Period squared must be proportional to semi-major axis cubed, then the period must also be proprotional to 'e'.

    Again , I don't think so. The reason is because in your derivation you constrained the velocity to a circular orbit, using v = sqrt(-2Gm/r), . However, for elliptical orbits the velocity goes as the more general vis-viva equation which is different and must include the semi-major axis....see here...(first equation at top of page)..
    http://en.wikipedia.org/wiki/Vis-viva_equation

    However, I could be wrong; I haven't done the substitution to see how it comes out.
    Sorry it took so long to respond..

    ...
     
    Last edited: Apr 13, 2012
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