# Specific angular momentum

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In celestial mechanics, the specific relative angular momentum (often denoted ${\displaystyle {\vec {h}}}$ or ${\displaystyle \mathbf {h} }$) of a body is the angular momentum of that body divided by its mass. [1] In the case of two orbiting bodies it is the vector product of their relative position and relative velocity, divided by the mass of the body in question.

## Contents

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 ${\displaystyle \mathbf {r} }$ and the relative velocity vector ${\displaystyle \mathbf {v} }$.

${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} ={\frac {\mathbf {L} }{m}}}$

where ${\displaystyle \mathbf {L} }$ is the angular momentum vector, defined as ${\displaystyle \mathbf {r} \times m\mathbf {v} }$.

The ${\displaystyle \mathbf {h} }$ vector is always perpendicular to the instantaneous osculating orbital plane, which coincides with the instantaneous perturbed orbit. It is not necessarily be perpendicular to the average orbital plane over time.

## Proof of constancy in the two body case

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. (${\displaystyle m_{1}\gg m_{2}}$)
• 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:

${\displaystyle {\ddot {\mathbf {r} }}+{\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0}$

where:

• ${\displaystyle \mathbf {r} }$ is the position vector from ${\displaystyle m_{1}}$ to ${\displaystyle m_{2}}$ with scalar magnitude ${\displaystyle r}$.
• ${\displaystyle {\ddot {\mathbf {r} }}}$ is the second time derivative of ${\displaystyle \mathbf {r} }$. (the acceleration)
• ${\displaystyle G}$ is the Gravitational constant.

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

${\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}+\mathbf {r} \times {\frac {Gm_{1}}{r^{2}}}{\frac {\mathbf {r} }{r}}=0}$

Because ${\displaystyle \mathbf {r} \times \mathbf {r} =0}$ the second term vanishes:

${\displaystyle \mathbf {r} \times {\ddot {\mathbf {r} }}=0}$

It can also be derived that:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)={\dot {\mathbf {r} }}\times {\dot {\mathbf {r} }}+\mathbf {r} \times {\ddot {\mathbf {r} }}=\mathbf {r} \times {\ddot {\mathbf {r} }}}$

Combining these two equations gives:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathbf {r} \times {\dot {\mathbf {r} }}\right)=0}$

Since the time derivative is equal to zero, the quantity ${\displaystyle \mathbf {r} \times {\dot {\mathbf {r} }}}$ is constant. Using the velocity vector ${\displaystyle \mathbf {v} }$ in place of the rate of change of position, and ${\displaystyle \mathbf {h} }$ for the specific angular momentum:

${\displaystyle \mathbf {h} =\mathbf {r} \times \mathbf {v} }$

is constant.

This is different from the normal construction of momentum, ${\displaystyle \mathbf {r} \times \mathbf {p} }$, 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 one multiplies it (cross product) with the specific relative angular momentum

${\displaystyle {\ddot {\mathbf {r} }}\times \mathbf {h} =-{\frac {\mu }{r^{2}}}{\frac {\mathbf {r} }{r}}\times \mathbf {h} }$

The left hand side is equal to the derivative ${\textstyle {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\dot {\mathbf {r} }}\times \mathbf {h} \right)}$ because the angular momentum is constant.

After some steps the right hand side becomes:

${\displaystyle -{\frac {\mu }{r^{3}}}\left(\mathbf {r} \times \mathbf {h} \right)=-{\frac {\mu }{r^{3}}}\left(\left(\mathbf {r} \cdot \mathbf {v} \right)\mathbf {r} -r^{2}\mathbf {v} \right)=-\left({\frac {\mu }{r^{2}}}{\dot {r}}\mathbf {r} -{\frac {\mu }{r}}\mathbf {v} \right)=\mu {\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {\mathbf {r} }{r}}\right)}$

Setting these two expression equal and integrating over time leads to (with the constant of integration ${\displaystyle \mathbf {C} }$)

${\displaystyle {\dot {\mathbf {r} }}\times \mathbf {h} =\mu {\frac {\mathbf {r} }{r}}+\mathbf {C} }$

Now this equation is multiplied (dot product) with ${\displaystyle \mathbf {r} }$ and rearranged

{\displaystyle {\begin{aligned}\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+rC\cos \theta \\\Rightarrow h^{2}&=\mu r+rC\cos \theta \end{aligned}}}

Finally one gets the orbit equation [1]

${\displaystyle r={\frac {\frac {h^{2}}{\mu }}{1+{\frac {C}{\mu }}\cos \theta }}}$

which is the equation of a conic section in polar coordinates with semi-latus rectum ${\textstyle p={\frac {h^{2}}{\mu }}}$ and eccentricity ${\textstyle e={\frac {C}{\mu }}}$.

### 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. [1]

If one connects this form of the equation ${\textstyle \mathrm {d} t={\frac {r^{2}}{h}}\,\mathrm {d} \theta }$ with the relationship ${\textstyle \mathrm {d} A={\frac {r^{2}}{2}}\,\mathrm {d} \theta }$ for the area of a sector with an infinitesimal small angle ${\displaystyle \mathrm {d} \theta }$ (triangle with one very small side), the equation

${\displaystyle \mathrm {d} t={\frac {2}{h}}\,\mathrm {d} A}$

### Third law

Kepler's third is a direct consequence of the second law. Integrating over one revolution gives the orbital period [1]

${\displaystyle T={\frac {2\pi ab}{h}}}$

for the area ${\displaystyle \pi ab}$ of an ellipse. Replacing the semi-minor axis with ${\displaystyle b={\sqrt {ap}}}$ and the specific relative angular momentum with ${\displaystyle h={\sqrt {\mu p}}}$ one gets

${\displaystyle T=2\pi {\sqrt {\frac {a^{3}}{\mu }}}}$

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.

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## References

1. Vallado, David A. (2001). Fundamentals of astrodynamics and applications (2nd ed.). Dordrecht: Kluwer Academic Publishers. pp. 20–30. ISBN   0-7923-6903-3.