Editing Specific angular momentum

{{Short description|Vector quantity in celestial mechanics}}
In [[celestial mechanics]], the '''specific relative angular momentum''' (often denoted <math>\vec{h}</math> or <math>\mathbf{h}</math>) of a body is the [[angular momentum]] of that body divided by its mass.<ref name="Vallado">{{cite book |last1=Vallado |first1=David A. |title=Fundamentals of astrodynamics and applications |date=2001 |publisher=Kluwer Academic Publishers |location=Dordrecht |isbn=0-7923-6903-3 |pages=20–30 |edition=2nd}}</ref> In the case of two [[orbiting body|orbiting bodies]] it is the [[vector product]] of their relative position and relative [[Momentum|linear momentum]], divided by the mass of the body in question.

Specific relative angular momentum plays a pivotal role in the analysis of the [[two-body problem]], as it remains constant for a given orbit under ideal conditions. "[[Specific (disambiguation)#Physical natural sciences, including physiology, and engineering|Specific]]" in this context indicates angular momentum per unit mass. The [[SI units|SI unit]] for specific relative angular momentum is square meter per second.

== Definition ==
The specific relative angular momentum is defined as the [[cross product]] of the relative [[orbital position vector|position vector]] <math> \mathbf{r}</math> and the relative [[orbital velocity vector|velocity vector]] <math> \mathbf{v} </math>.
<math display="block"> \mathbf{h} = \mathbf{r}\times \mathbf{v} = \frac{\mathbf{L}}{m} </math>

where <math>\mathbf{L}</math> is the angular momentum vector, defined as <math> \mathbf{r} \times m \mathbf{v}</math>.

The <math> \mathbf{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 perpendicular to the average orbital plane over time.

== Proof of constancy in the two body case ==
[[File:FlightPathAngle.svg|thumb|Distance vector <math> \mathbf{r} </math>, velocity vector <math> \mathbf{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>).]]

Under certain conditions, it can be proven that the specific angular momentum is constant. The conditions for this proof include:
* 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 proof starts with the [[Two-body problem|two body equation of motion]], derived from [[Newton's law of universal gravitation]]:

<math display="block"> \ddot{\mathbf{r}} + \frac{G m_1}{r^2}\frac{\mathbf{r}}{r} = 0</math>

where:
* <math>\mathbf{r}</math> is the position vector from <math>m_1</math> to <math>m_2</math> with scalar magnitude <math>r</math>.
* <math>\ddot{\mathbf{r}}</math> is the second time derivative of <math>\mathbf{r}</math>. (the [[acceleration]])
* <math>G</math> is the [[Gravitational constant]].

The cross product of the position vector with the equation of motion is:

<math display="block"> \mathbf{r} \times \ddot{\mathbf{r}} + \mathbf{r} \times \frac{G m_1}{r^2}\frac{\mathbf{r}}{r} = 0</math>

Because <math>\mathbf{r} \times \mathbf{r} = 0</math> the second term vanishes:

<math display="block"> \mathbf{r} \times \ddot{\mathbf{r}} = 0 </math>

It can also be derived that:
<math display="block">
\frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbf{r}\times\dot{\mathbf{r}}\right) =
\dot{\mathbf{r}} \times \dot{\mathbf{r}} + \mathbf{r} \times \ddot{\mathbf{r}} =
\mathbf{r} \times \ddot{\mathbf{r}}
</math>

Combining these two equations gives:
<math display="block">\frac{\mathrm{d}}{\mathrm{d}t} \left(\mathbf{r}\times\dot{\mathbf{r}}\right) = 0</math>

Since the time derivative is equal to zero, the quantity <math>\mathbf{r} \times \dot{\mathbf{r}}</math> is constant. Using the velocity vector <math>\mathbf{v}</math> in place of the rate of change of position, and <math>\mathbf{h}</math> for the specific angular momentum:
<math display="block"> \mathbf{h} = \mathbf{r}\times\mathbf{v}</math> is constant.

This is different from the normal construction of momentum, <math>\mathbf{r} \times \mathbf{p}</math>, because it does not include the mass of the object in question.

== Kepler's laws of planetary motion ==
{{Main|Kepler's laws of planetary motion}}
Kepler's laws of planetary motion can be proved almost directly with the above relationships.

=== First law ===
The proof starts again with the equation of the two-body problem. This time the cross product is multiplied with the specific relative angular momentum
<math display="block"> \ddot{\mathbf{r}} \times \mathbf{h} = - \frac{\mu}{r^2}\frac{\mathbf{r}}{r} \times \mathbf{h} </math>

The left hand side is equal to the derivative <math display="inline"> \frac{\mathrm{d}}{\mathrm{d}t} \left(\dot{\mathbf{r}}\times\mathbf{h}\right)</math> because the angular momentum is constant.

After some steps (which includes using the [[Triple product#Vector triple product|vector triple product]] and defining the scalar <math>\dot{r}</math> to be the <em>radial velocity</em>, as opposed to the norm of the vector <math>\dot{\mathbf{r}}</math>) the right hand side becomes:
<math display="block">
  -\frac{\mu}{r^3}\left(\mathbf{r} \times \mathbf{h}\right) =
  -\frac{\mu}{r^3} \left(\left(\mathbf{r}\cdot\mathbf{v}\right)\mathbf{r} - r^2\mathbf{v}\right) =
  -\left(\frac{\mu}{r^2}\dot{r}\mathbf{r} - \frac{\mu}{r}\mathbf{v}\right) =
   \mu \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathbf{r}}{r}\right)
</math>

Setting these two expression equal and integrating over time leads to (with the constant of integration <math> \mathbf{C} </math>)
<math display="block"> \dot{\mathbf{r}}\times\mathbf{h} = \mu\frac{\mathbf{r}}{r} + \mathbf{C} </math>

Now this equation is multiplied ([[dot product]]) with <math> \mathbf{r} </math> and rearranged
<math display="block">\begin{align}
            \mathbf{r} \cdot \left(\dot{\mathbf{r}}\times\mathbf{h}\right) &= \mathbf{r} \cdot \left(\mu\frac{\mathbf{r}}{r} + \mathbf{C}\right) \\
\Rightarrow \left(\mathbf{r}\times\dot{\mathbf{r}}\right) \cdot \mathbf{h} &= \mu r + r C\cos\theta \\
\Rightarrow                                                            h^2 &= \mu r + r C\cos\theta
\end{align}</math>

Finally one gets the [[orbit equation]]<ref name="Vallado" />
<math display="block"> r = \frac{\frac{h^2}{\mu}}{1 + \frac{C}{\mu}\cos\theta} </math>

which is the [[Conic Sections in Polar Coordinates|equation of a conic section in polar coordinates]] with [[semi-latus rectum]] <math display="inline"> p = \frac{h^2}{\mu} </math> and [[Eccentricity (mathematics)|eccentricity]] <math display="inline"> e = \frac{C}{\mu} </math>.

=== Second law ===
The second law follows instantly from the second of the three equations to calculate the absolute value of the specific relative angular momentum.<ref name="Vallado" />

If one connects this form of the equation <math display="inline"> \mathrm{d}t = \frac{r^2}{h} \, \mathrm{d}\theta </math> with the relationship <math display="inline"> \mathrm{d}A = \frac{r^2}{2} \, \mathrm{d}\theta </math> for the area of a sector with an infinitesimal small angle <math> \mathrm{d}\theta </math> (triangle with one very small side), the equation
<math display="block"> \mathrm{d}t = \frac{2}{h} \, \mathrm{d}A </math>

=== Third law ===
Kepler's third is a direct consequence of the second law. Integrating over one revolution gives the [[orbital period]]<ref name="Vallado" />
<math display="block"> T = \frac{2\pi ab}{h} </math>

for the area <math> \pi ab </math> of an ellipse. Replacing the semi-minor axis with <math> b=\sqrt{ap} </math> and the specific relative angular momentum with <math> h = \sqrt{\mu p} </math> one gets
<math display="block"> T = 2\pi \sqrt{\frac{a^3}{\mu}} </math>

There is thus a relationship between the semi-major axis and the orbital period of a satellite that can be reduced to a constant of the central body.

== See also ==
* [[Specific orbital energy]], another conserved quantity in the two-body problem.
* {{slink|Classical central-force problem#Specific angular momentum}}

==References==
{{reflist|group="References"}}

{{orbits}}
[[Category:Angular momentum]]
[[Category:Astrodynamics]]
[[Category:Orbits]]