Astrodynamics/Motion Constants

Specific Angular Momentum

We already have two funamental 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:

${𝐫}^{\prime }=𝐯$

And that:

$r^{\prime }=v$

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

${𝐫}^{″}+\frac{\mu }{r^3}𝐫=0$

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

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

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

$𝐫\times {𝐫}^{″}=0$

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

$\frac{d(𝐫\times {𝐫}^{\prime })}{dt}=0$

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:

$𝐫\times 𝐯=𝐡$

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:

$h=|𝐡|$

Alternate Definitions of h

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

$h=rv\sin \varphi$

$h=rv\cos \gamma$

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:

${𝐫}^{\prime }\cdot {𝐫}^{″}+{𝐫}^{\prime }\cdot \frac{\mu }{r^3}𝐫=0$

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

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

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

$𝐱\cdot {𝐱}^{\prime }=x{x}^{\prime }$

We can apply this result to our equation:

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

From the chain rule, we know that:

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

And:

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

Plugging these results into our equation gives us:

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

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:

$ℰ=\frac{v^2}{2}-\frac{\mu }{r}+c$

The term E is the total energy of the system, the first term is the kinetic energy, and the second and third terms represent the potential energy. We can see from the derivation that the energy in this system must remain constant.

Kinetic Energy

We can see from the above equation for total energy that there are basically two terms. The first term is defined as the kinetic energy of the moving body:

${ℰ}_k=\frac{v^2}{2}$

This is the amount of energy necessary for the body to remain in motion.

Potential Energy

If the first term in the equation is the kinetic energy, then the second part of the equation must be the potential energy. The potential energy is the amount of stored energy in the object under a constant force, such as gravity. In elementary mechanics, the potential energy of an object is defined typically in terms of it's height above the earth's surface. This is because an object on the surface of the earth has the potential to travel from it's position to the surface of the earth unless it is held up by an equal but opposite force. Because we are looking at large-scale astronomical bodies, the term h, which previously had been the height above the surface of the earth is replaced by the term r, which is the distance from the object to the center of the prime body.

$ℰ=\frac{\mu }{r}+c$

The constant term c is a constant of integration, and can be arbitrarily set to provide a suitable reference frame. Notice that if c is a distance above the surface of the earth, then an object on the surface will have negative potential energy. To equate our new equation to the elementary version, we can say that:

$h=r-r_e$

Where c = r_e is the radius of the earth. Notice again that if our position r is smaller then the radius of the earth (inside the earth, below the surface, or in a cave), then the potential energy in this scheme will be negative.

To reduce the occurrence of negative energies in our calculations, we will typically set c = 0, so that c corresponds to the center of the prime body (the center of the earth, locally). This has the added benefit of simplifying our calculations because we don't need to explicitly account for a c term.