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cephron
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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!
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!
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