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→Proof that the specific relative angular momentum is constant under ideal conditions: - rewrote section for clarity, removed extraneous bits at the end.
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The <math> \vec{h}</math> vector is always perpendicular to the instantaneous [[Osculating orbit|osculating]] [[orbital plane (astronomy)|orbital plane]], which coincides with the instantaneous [[Perturbation (astronomy)|perturbed orbit]]. It is not necessarily be perpendicular to the average orbital plane over time.
 
== Proof of constancy in the two body case ==
== Proof that the specific relative angular momentum is constant under ideal conditions ==▼
[[File:FlightPathAngle.svg|thumb|Distance vector <math> \vec{r} </math>, velocity vector <math> \vec{v} </math>, [[true anomaly]] <math> \theta </math> and flight path angle <math> \phi </math> of <math> m_2 </math> in orbit around <math> m_1 </math>. The most important measures of the [[ellipse]] are also depicted (among which, note that the [[true anomaly]] <math>\theta</math> is labeled as <math>\nu</math>).]]▼
=== Prerequisites ===
The following is only valid under the simplifications also applied to [[Newton's law of universal gravitation]].
 
▲==Under Proofcertain conditions, it can be proven that the specific relative angular momentum is constant. under idealThe conditions ==for this proof include:
One looks at two point masses <math> m_1 </math> and <math> m_2 </math>, at the distance <math> r </math> from one another and with the gravitational force <math> \vec{F} = G\frac{m_1 m_2}{r^2}\frac{\vec{r}}{r} </math> acting between them. This force acts instantly, over any distance and is the only force present. The coordinate system is inertial.
* The mass of one object is much greater than the mass of the other one. (<math> m_1 \gg m_2 </math>)
* The coordinate system is [[Inertial frame of reference|inertial]].
* Each object can be treated as a spherically symmetrical [[point particle|point mass]].
* No other forces act on the system other than the gravitational force that connects the two bodies.
 
=== Proof ===▼
The further simplification <math> m_1 \gg m_2 </math> is assumed in the following. Thus <math> m_1 </math> is the [[central body]] in the origin of the coordinate system and <math> m_2 </math> is the [[satellite]] orbiting around it. Now the reduced mass is also equal to <math> m_2 </math> and the equation of the two-body problem is
:<math> \ddot{\vec{r}} = -\frac{\mu}{r^2}\frac{\vec{r}}{r} </math>▼
 
The proof starts with the [[Two-body problem|two body equation of motion]], derived from [[Newton's law of universal gravitation]]:
with the [[standard gravitational parameter]] <math> \mu = Gm_1 </math> and the distance vector <math> \vec{r} </math> (absolute value <math> r </math>) that points from the origin (central body) to the satellite, because of its negligible mass.<ref group="Notes">The derivation of the specific angular momentum works also if one does not make this assumption. Then the gravitational parameter is <math> \mu = G\left(m_1 + m_2\right) </math>.</ref>
 
▲:<math> \ddot{\vec{r}} =+ -\frac{\muG m_1}{r^2}\frac{\vec{r}}{r} = 0</math>
It is important not to confound the gravitational parameter <math> \mu </math> with the reduced mass, which is sometimes also denoted by the same letter <math> \mu </math>.
 
where:
▲=== Proof ===
* <math>\vec{r}</math> is the position vector from <math>m_1</math> to <math>m_2</math> with scalar magnitude <math>r</math>.
▲[[File:FlightPathAngle.svg|thumb|Distance vector <math> \vec{r} </math>, velocity vector <math> \vec{v} </math>, [[true anomaly]] <math> \theta </math> and flight path angle <math> \phi </math> of <math> m_2 </math> in orbit around <math> m_1 </math>. The most important measures of the [[ellipse]] are also depicted (among which, note that the [[true anomaly]] <math>\theta</math> is labeled as <math>\nu</math>).]]
* <math>\ddot{\vec{r}}</math> is the second time derivative of <math>\vec{r}</math>. (the [[acceleration]])
* <math>G</math> is the [[Gravitational constant]].
 
OneThe obtains the specific relative angular momentum by multiplying (cross product) the equation of the two-bodyposition problemvector with the distanceequation vector <math>of \vec{r}motion </math>is:
 
:<math> \vec{r} \times \ddot{\vec{r}} = -+ \vec{r} \times \frac{\muG m_1}{r^2}\frac{\vec{r}}{r} = - \frac{\mu}{r^3} \left( \vec{r} \times \vec{r} \right) 0</math>
 
Because <math>\vec{r} \times \vec{r} = 0</math> the second term vanishes:
 
:<math> \vec{hr} = \vec{r}\times \dotddot{\vec{r}} \text{= is const.}0 </math>▼
 
It can also be derived that:
 
The cross product of a vector with itself (right hand side) is 0. The left hand side simplifies to
:<math>
\frac{\mathrm{d} \left(\vec{r} \times \ddotdot{\vec{r}}\right) }{\mathrm{d}t} =
\dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}} =
\frac{\mathrm{d} \left(\vec{r} \times \dotddot{\vec{r}}\right) }{\mathrm{d}t} = 0
</math>
 
Combining these two equations gives:
according to the product rule of differentiation.
:<math>\frac{\mathrm{d} \left(\vec{r}\times\dot{\vec{r}}\right) }{\mathrm{d}t} = 0</math>
 
This means that <math> \vec{r}\times\dot{\vec{r}} </math> is constant (i.e., a [[conserved quantity]]). And this is exactly the angular momentum per mass of the satellite:<ref group="References">{{Cite book | last = Vallado | first = David Anthony | title = Fundamentals of Astrodynamics and Applications | page = 24 | publisher = Springer | year = 2001 | isbn = 0-7923-6903-3 | url = https://books.google.com/books?id=PJLlWzMBKjkC}}</ref>
▲:<math> \vec{h} = \vec{r}\times\dot{\vec{r}} \text{ is const.} </math>
 
Since the time derivative is equal to zero, the quantity <math>\vec{r} \times \dot{\vec{r}}</math> is constant. Using the velocity vector <math>\vec{v}</math> in place of the rate of change of position, and <math>\vec{h}</math> for the specific angular momentum:
This vector is perpendicular to the orbit plane, the orbit remains in this plane because the angular momentum is constant.
 
:<math> \vec{h} = \vec{r}\times\vec{v}</math> is constant.
One can obtain further insight into the two-body problem with the definitions of the flight path angle <math> \phi </math> and the transversal and radial component of the velocity vector (see illustration on the right). The next three formulas are all equivalent possibilities to calculate the absolute value of the specific relative angular momentum vector
* <math> h = rv\cos\phi </math>
* <math> h = r^2\dot{\theta} </math>
* <math> h = \sqrt{\mu p} </math>
 
This is different from the normal construction of momentum, <math>\vec{r} \times \vec{p}</math>, because it does not include the mass of the object in question.
Where <math>p</math> is called the [[conic section#Conic parameters|semi-latus rectum]] of the curve.
 
== Kepler's laws of planetary motion ==
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