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Deriving an equation for orbital period 
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#1
Apr1112, 06:33 AM

P: 125

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 undefinednot 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 zeroundefined 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 saneif 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! 


#2
Apr1112, 01:20 PM

P: 547

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 


#3
Apr1112, 03:39 PM

P: 125

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? 


#4
Apr1112, 10:14 PM

P: 547

Deriving an equation for orbital period
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). Kepler's equation allows you to substitute the semimajor 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 semimajor axis. Good questions, BTW. Creator 


#5
Apr1212, 02:44 AM

P: 125

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: 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 believeplease correct me if I'm wrong!that there is a onetoone 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 onetoone 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. 


#6
Apr1212, 09:29 AM

Mentor
P: 11,574

The orbital period of an object can be expressed as [tex]T=2\pi\sqrt{\frac{a^3}{\mu}}[/tex]
where a is the semimajor 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. 


#7
Apr1212, 01:02 PM

Sci Advisor
HW Helper
P: 2,278

Yes, your formula works and it does eliminate the need to calculate the semimajor 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. 


#8
Apr1312, 01:00 AM

P: 547

e = u / 2a ....where a is the semimajor axis. And since by Kepler's law the Period squared must be proportional to semimajor axis cubed, then the period must also be proprotional to 'e'. http://en.wikipedia.org/wiki/Visviva_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.. ... 


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