Astrodynamics/Motion Constants

Astrodynamics

Specific Energy and Angular Momentum

Angular momentum, kinetic energy and gravitational potential energy are all proportional to the mass of the orbiting body. However, when on the topic of orbital mechanics, it is more convenient to discuss the specific angular momentum and specific energy which are simply the angular momentum and energy per unit mass. Thus two objects in identical orbits may have very different masses, but will have the same specific angular momentum and the same specific energy (assuming, of course, that the masses of both bodies are negligible compared to the mass of the central body). Throughout this text, unless otherwise noted, the terms "angular momentum" and "energy" will refer to their specific counterparts.

Conservation of Angular Momentum

We already have two fundamental vectors that are used to analyze orbits: the position vector r, and the velocity vector v. We know from basic Newtonian mechanics that the velocity vector is the time-derivative of the position vector:

${\dot {\mathbf {r} }}=\mathbf {v}$

And that:

${\dot {r}}=v$

Starting with our motion equation we derived earlier, we can derive the angular momentum of the body.

${\ddot {\mathbf {r} }}+{\frac {\mu }{r^{3}}}\mathbf {r} =0$

We use the cross-product with the position vector r on both sides of the equation:

$\mathbf {r} \times {\ddot {\mathbf {r} }}+\mathbf {r} \times {\frac {\mu }{r^{3}}}\mathbf {r} =0$

The second term cancels out because we know that r × r = 0.

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

As an aside, consider the following derivative, obtained using the product rule from calculus:

${\frac {d}{dt}}(\mathbf {x} \times {\dot {\mathbf {x} }})={\dot {\mathbf {x} }}\times {\dot {\mathbf {x} }}+\mathbf {x} \times {\ddot {\mathbf {x} }}$

The first term cancels because both vectors in the cross product are the same. We are left with the result:

${\frac {d}{dt}}(\mathbf {x} \times {\dot {\mathbf {x} }})=\mathbf {x} \times {\ddot {\mathbf {x} }}$

We can substitute our above results into our equation to produce:

${\frac {d(\mathbf {r} \times {\dot {\mathbf {r} }})}{dt}}=0$

Because the derivative is zero, we know that the result must be a constant. We can integrate both sides, introducing the constant vector of integration, h:

[Angular Momentum Vector]

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

Because we know that h is a constant, this equation shows that angular momentum is conserved in an orbital system. This constant is known as the angular momentum of the orbit, and is highly important in future calculations.

Meaning of h

h is a vector that is perpendicular to both r and v. r and v are both located in a single plane, known as the orbital plane. The entire orbit is located in this plane. However, the vector h is normal to the orbital plane, which means that it is perpendicular to the orbital plane. The scalar value h is the magnitude of this vector, and is defined as:

$h=|\mathbf {h} |$

Alternate Definitions of h

The magnitude value h can be derived from the magnitude values r and v as such:

$h=rv\cos \phi$

$h=rv\sin \gamma$

Where γ is the angle between the vectors r and v, and φ is the compliment of γ. φ is known as the flight angle of the satellite. γ is known as the zenith angle.

Conservation of Energy

Similar to the angular momentum, we can derive the conservation of energy equation. Instead of performing a cross product on both sides by the position vector r, we can perform the dot product by the velocity vector v:

${\dot {\mathbf {r} }}\cdot {\ddot {\mathbf {r} }}+{\dot {\mathbf {r} }}\cdot {\frac {\mu }{r^{3}}}\mathbf {r} =0$

We know that the second derivative of r is the same as the first derivative of v:

$\mathbf {v} \cdot {\dot {\mathbf {v} }}+{\frac {\mu }{r^{3}}}\mathbf {r} \cdot {\dot {\mathbf {r} }}=0$

In general, a result of the dot product is that:

$\mathbf {x} \cdot {\dot {\mathbf {x} }}=x{\dot {x}}$

We can apply this result to our equation:

$v{\dot {v}}+{\frac {\mu }{r^{3}}}r{\dot {r}}=0$

From the chain rule, we know that:

${\frac {d}{dt}}{\frac {v^{2}}{2}}=v{\dot {v}}$

And:

${\frac {d}{dt}}{\frac {\mu }{r}}=-{\frac {\mu }{r^{2}}}{\dot {r}}$

Plugging these results into our equation gives us:

${\frac {d}{dt}}\left({\frac {v^{2}}{2}}-{\frac {\mu }{r}}\right)=0$

To generalize the equation, an additional constant term c must be added:

${\frac {d}{dt}}\left({\frac {v^{2}}{2}}-{\frac {\mu }{r}}+c\right)=0$

Again, the derivative is equal to zero, so we know that the function has a constant value. We integrate both sides, and we call our new constant of integration the energy of the system:

${\mathcal {E}}={\frac {v^{2}}{2}}-{\frac {\mu }{r}}+c$

Kinetic Energy

From classical mechanics, the first term van be identified as the specific kinetic energy of the moving body:

[Kinetic Energy]

${\mathcal {E}}_{k}={\frac {v^{2}}{2}}$

This represents the amount of work necessary to accelerate the body from a state of rest to its current velocity.

Potential Energy

The second and third term combined represent the specific potential energy of the moving body:

[Potential Energy]

${\mathcal {E}}_{p}=-{\frac {\mu }{r}}+c$

This represents the amount of work necessary to move the body from a given reference distance to its current distance from the center of the central body. The reference distance is defined by the c term in the equation. For instance, to define the surface of the Earth as having zero potential energy, c would be set to equal μ/r₀, where r₀ is the radius of the Earth. In this particular case, bodies below the radius of the earth would have negative potential energy and bodies above would have positive potential energy. It is most convenient, however, to set the reference distance to be infinite, which reduces the expression for the potential energy to:

${\mathcal {E}}_{p}=-{\frac {\mu }{r}}$

This will result in potential energy values that are globally negative. This may seem counter-intuitive at first, but calculations are greatly simplified.

Total Mechanical Energy

The expression for the total specific mechanical energy of the system thus becomes:

[Total Mechanical Energy]

${\mathcal {E}}={\frac {v^{2}}{2}}-{\frac {\mu }{r}}$

This quantity is important in determining the shape of the orbit and will be discussed in detail in the next chapter.

Derivation of Kepler's Laws from Newton's Laws

Now that the constants of motion have been defined, it is possible to derive Kepler's laws from Newton's laws of motion and gravitation.

First Law

As stated previously, the equation of motion for an object in the restricted two body problem is:

${\ddot {\mathbf {r} }}=-{\frac {\mu }{r^{3}}}\mathbf {r}$

Taking the cross product of both sides with the angular momentume vector yields:

${\ddot {\mathbf {r} }}\times \mathbf {h} =-{\frac {\mu }{r^{3}}}\mathbf {r} \times \mathbf {h}$

Because the h vector does not change with time, the left side is equivalent to:

${\ddot {\mathbf {r} }}\times \mathbf {h} ={\ddot {\mathbf {r} }}\times \mathbf {h} +{\dot {\mathbf {r} }}\times {\dot {\mathbf {h} }}={\frac {d}{dt}}({\dot {\mathbf {r} }}\times \mathbf {h} )$

Meanwhile the left side becomes:

$-{\frac {\mu }{r^{3}}}\mathbf {r} \times (\mathbf {r} \times {\dot {\mathbf {r} }})=-{\frac {\mu }{r^{3}}}\left(\mathbf {r} ({\dot {\mathbf {r} }}\cdot \mathbf {r} )-{\dot {\mathbf {r} }}(\mathbf {r} \cdot \mathbf {r} )\right)$

$={\frac {\mu }{r^{3}}}\left({\dot {\mathbf {r} }}r^{2}-\mathbf {r} r{\dot {r}}\right)$

$=\mu \left({\frac {\dot {\mathbf {r} }}{r}}-{\frac {\mathbf {r} {\dot {r}}}{r^{2}}}\right)$

$=\mu {\frac {d}{dt}}\left({\frac {\mathbf {r} }{r}}+\mathbf {B} \right)$

Where B is a constant of integration. Thus the equation becomes:

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

Taking the dot product of both sides with r:

$\mathbf {r} \cdot ({\dot {\mathbf {r} }}\times \mathbf {h} )=\mu r+\mathbf {B} \cdot \mathbf {r}$

Recalling properties of triple vector products:

$\mathbf {r} \cdot ({\dot {\mathbf {r} }}\times \mathbf {h} )=\mathbf {h} \cdot (\mathbf {r} \times {\dot {\mathbf {r} }})=\mathbf {h} \cdot \mathbf {h} =h^{2}$

The equation becomes:

$h^{2}=\mu r+Br\cos \nu$

Rearranging the equation:

[Trajectory Equation]

$r(\nu )={\frac {h^{2}/\mu }{1+B/\mu \cos \nu }}$

This equation is identical to the polar equation of a conic section centered at one focus:

$r(\theta )={\frac {p}{1+e\cos \theta }}$

Where e is the eccentricity of the conic section and p is the semi-latus rectum of the conic section. For an ellipse:

$p=a(1-e^{2})$

Where a is the semi-major axis.

This derivation shows that in addition to elliptical orbits, a body is also capable of parabolic or hyperbolic trajectories (both of which will be discussed shortly). Thus, stated generally, Kepler's first law becomes:

The trajectory of an object in a gravitational field is a conic section with the central body at one focus

Second Law

The angular momentum vector:

$||\mathbf {h} ||=||\mathbf {r} \times \mathbf {v} ||=rv\cos \phi$

Where φ is the flight path angle and vcosφ is the component of the velocity vector which is normal to the position vector. From polar calculus, this can be re-expressed as:

$v\cos \phi =r{\dot {\nu }}$

Thus the equation for angular momentum becomes:

$h=r^{2}{\frac {d\nu }{dt}}$

Recalling from calculus, the differential element of area in polar form is:

$dA={\frac {r^{2}d\nu }{2}}$

Combining these equations gives:

[Kepler's Second Law]

${\frac {dA}{dt}}={\frac {h}{2}}$

The angular momentum is already known to be constant in the restricted two body problem, so this is the mathematical expression of Kepler's second law. This equation holds true for parabolic and hyperbolic as well as elliptical trajectories.

Third Law

Because the third law only deals with closed orbits, it is only valid for elliptical orbits. The result from the previous section can be rewritten as:

$dA={\frac {h}{2}}dt$

Integrating both sides over one complete orbit yields:

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

Where πab is the area enclosed by the entire ellipse and T is the orbital period. The semi-minor axis can be rewritten in terms of semi-latus rectum and semi-major axis:

$b={\sqrt {a^{2}(1-e^{2})}}={\sqrt {ap}}$

Recall that angular momentum is related to the semi-latus rectum via:

$p={\frac {h^{2}}{\mu }}$

Combining these equations and reducing gives the expression for orbital period:

[Period of an Elliptical Orbit]

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

Which can be reduced to Kepler's third law:

[Kepler's Third Law]

$T^{2}\propto a^{3}$