Specific angular momentum

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. In the case of two orbiting bodies it is the vector product of their relative position and relative 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" in this context indicates angular momentum per unit mass. The 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 position vector <math> \mathbf{r}</math> and the relative 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 orbital plane, which coincides with the instantaneous perturbed orbit. It is not necessarily perpendicular to the average orbital plane over time.

Proof of constancy in the two body case

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.
Each object can be treated as a spherically symmetrical 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 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:

\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

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 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:

-\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