Orbits

2015-08-01

There is a force in the earth which causes the moon to move. – Johannes Kepler, discoverer of the laws of orbits

Before launching a rocket, we need to know where we are going. Up is only part of the answer.

Orbital trajectory

Concept

First, the obligatory quote when explaing how orbiting works:

The knack [of flying] lies in learning how to throw yourself at the ground and miss. – Double Adams, author of The Hitchhiker’s Guide to the Galaxy

The part of throwing oneself at the ground is easy:

On the figure above, notice that, as we go right, the ground gets lower and lower due to the curvature of the planet. Can we go right fast enough so that the ground will lower as fast as we fall? Well, yes:

Central force

The point $S$ is the satellite whose trajectory we want to determine. The celestial objects it orbits is called the primary. Thanks to the shell theorem, we can assimilate the primary with a mass-point $\left(P,M\right)$. Using Newton’s Second Law, the acceleration due to the force of gravitation is:

$\stackrel{\bullet \bullet }{\stackrel{\to }{r}}=-\frac{\stackrel{\mathit{\mu }}{\stackrel{⏞}{GM}}}{{r}^{2}}\stackrel{^}{r}$

where $\stackrel{\to }{r}=\stackrel{\to }{\mathit{PS}}$ and $\stackrel{^}{r}=\frac{\stackrel{\to }{\mathit{PS}}}{\mathit{PS}}$ is the direction (unit vector) from $P$ to $S$.

This already tells us that the acceleration, and hence the trajectory, does not depend on the mass of the satellite.

r = ^2 (r r)
= ( r r + r )
= ( r - r ^2) r + (2 r + r )

Hence the equation can be written as::

$\left\{\begin{array}{cc}\hfill \stackrel{\bullet \bullet }{r}-r{\stackrel{\bullet }{\mathit{\theta }}}^{2}& =-\frac{\mathit{\mu }}{{r}^{2}}\hfill \\ \hfill 2\stackrel{\bullet }{r}\stackrel{\bullet }{\mathit{\theta }}+r\stackrel{\bullet \bullet }{\mathit{\theta }}& =0\hfill \end{array}$

Kepler’s second law

The second line can be rewritten as:

0 = 2 r + r
= 1 (2 r r + r^2 )
= 1 ( r^2 )

This means that $h={r}^{2}\stackrel{\bullet }{\mathit{\theta }}$ is constant.

Variable substitution

Now, the trick is to introduce a new variable $u=\frac{1}{r}$ and to consider its relation to $\mathit{\theta }$. This means wil will replace $r$, $\stackrel{\bullet }{r}$ and $\stackrel{\bullet \bullet }{r}$ with expressions involving $u\left(\mathit{\theta }\right)$. For short, we will use the following notation:

First, we have:

$r=\frac{1}{u}$

Then, we look at the expression of $u\text{'}$:

$u\text{'}=\frac{d}{d\mathit{\theta }}\frac{1}{r}=\frac{dt}{d\mathit{\theta }}\frac{d}{dt}\frac{1}{r}=\frac{1}{\stackrel{\bullet }{\mathit{\theta }}}×\left(-\frac{\stackrel{\bullet }{r}}{{r}^{2}}\right)=-\frac{\stackrel{\bullet }{r}}{h}$

From this, we get that:

$\stackrel{\bullet }{r}=-hu\text{'}$

Similarly, we develop the expression of $u"$:

u" = (- r)
= - 1 t r
= - 1 1 r
= - r  because  h = r^2
= - r

This let us know that:

$\stackrel{\bullet \bullet }{r}=-{h}^{2}{u}^{2}u"$

Solving the equation

We can now rewrite the equation by using $u$, $u\text{'}$ and $u"$:

r - r ^2 = -
- h^2 u^2 u" - 1 (h u^2)^2 = - u^2
- h^2 u^2 u" - h^2 u^3 = - u^2
u" + u =

It can be shown that the solutions for this equation are of the form:

$u=\frac{\mathit{\mu }}{{h}^{2}}-A\mathrm{cos}\left(\mathit{\theta }-{\mathit{\theta }}_{0}\right)$

for some values of $A$ and ${\mathit{\theta }}_{0}$. Thus:

$r=\frac{1}{\frac{\mathit{\mu }}{{h}^{2}}-A\mathrm{cos}\left(\mathit{\theta }-{\mathit{\theta }}_{0}\right)}$

Now, if we assume ${\mathit{\theta }}_{0}=0$ and set $l=\frac{{h}^{2}}{\mathit{\mu }}$ and $e=\frac{A}{p}$, then:

$r=\frac{l}{1-e\mathit{cos}\left(\mathit{\theta }\right)}$

$e$ is called the eccentricity and $l$ the semi-latus rectum

Shape

From this expression of $r$ depending on $\mathit{\theta }$, we can vizualize the possibles shapes for the trajectories of an objects only subjugated to gravitation.

First, we notice that the semi-latus rectum, $l$, only affects the scale: the larger $l$, the larger $r$, in all directions (all values of $\mathit{\theta }$). It remains to look for the influence of the eccentricity, $e$.

First, notice that $\mathrm{cos}\left(\mathit{\theta }\right)$ is an even function. It means that $\mathrm{cos}\left(-\mathit{\theta }\right)=\mathrm{cos}\left(\mathit{\theta }\right)$, for all values of $\mathit{\theta }$. It implies that $r\left(\mathit{\theta }\right)$ is also even: the trajectory is symmetric along the axis where $\mathit{\theta }=0$ (also called the major axis).

Second, notice that $-\mathrm{cos}\left(\mathit{\theta }\right)=\mathrm{cos}\left(\mathit{\theta }+\mathit{\pi }\right)$. This means that reverting the value of $e$ is the same as doing a $180\mathrm{deg}$ rotation. For this reason, we can restrict ourselves to non-negative values of $e$ ($e\ge 0$).

Now, let us look at different cases:

If we set $e=0$, we get rid of most of the expression and end up with $r\left(\mathit{\theta }\right)=l$. This means that the distance to the origin does not depend on the direction; in other words, the trajectory is a circle.

Because of the fraction, we need to avoid its denominator, $1-e\mathrm{cos}\left(\mathit{\theta }\right)$ to be zero. Since $-1\le \mathrm{cos}\left(\mathit{\theta }\right)\le 1$, choosing $0 will ensure that $1-e\mathrm{cos}\left(\mathit{\theta }\right)\ne 0$. That way, $r\left(\mathit{\theta }\right)$ is well defined in all directions; it turns out the shape is that of an ellipse.

On the other hand, if we choose $e\ge 1$, we will encouter a discontinuity: since $|\mathrm{cos}\left(\mathit{\theta }\right)|$ ranges from $0$ to $1$, there will be a values of $\mathit{\theta }$ where $1-e\mathrm{cos}\left(\mathit{\theta }\right)\le 0$, which means an infinite or negative value for $r\left(\mathit{\theta }\right)$. In practice this means that the object will never be in that direction: this is a hyperbola (or parabola, in the edge case where $e=0$).

An object in free fall is always following an orbital trajectory around its primary. It is actually said to be in orbit when that trajectory does not intersect the surface (nor goes to deep into the atmosphere).

When the orbit is closed (circular or elliptical), the oject is said to be orbiting its primary. When the orbit is open (hyperbolic or parabolic), the trajectory is said to be:

• a capture orbit (or trajectory) when moving towards the periapsis
• an escape orbit (or trajectory) when moving away from it

Characterisitics

Apses

We easily find the extremal value of $r$: the extremal values of $\mathrm{cos}\left(\mathit{theta}\right)$ are $-1$ and $1$, which means that $r$ varies between

• ${r}_{\text{per}}=\frac{l}{1+e}$ and ${r}_{\text{ap}}=\frac{l}{1-e}$ when $e<1$
• ${r}_{\text{per}}=\frac{l}{1+e}$ and $+\mathit{\infty }$ when $e\ge 1$

The point where $r={r}_{\text{per}}$ is called the periapsis (peri- is Greek for close), while the point where $r={r}_{\text{ap}}$ is called the apoapsis (apo- is Greek for far).

While the periapsis is always defined, the apoapsis may not, since an infinite (or negative) altitude will never be reached.

When we want to be explicit about the object being orbited, we can use more specific suffixes that -apsis. For instance, the perigee is the periapsis of an object orbiting the Earth.

Suffixes for various bodies
Body Suffix
Galactic center -galacticon
Star -astron
Sun -helion
Mercury -hermion
Venus -cytherion
Earth -gee
Moon -lune, -cynthion, -selene
Mars -areion
Jupiter -zene, -jove
Saturn -krone, -saturnium
Uranus -uranion
Neptune -poseidon
Various names come from Latin an Greek. For instance, Poseidon is the Greek god of the ocean, while Neptune is the Roman equivalent. The suffix -gee is related to Gaia, the Greek goddess of the Earth.

Angles

Astronomers use several ways to measure what are essentially angles.

The true anomaly is simply the angle $\mathit{\theta }$ we have been using in this chapter; it is also conventionally named $f$. In the case of an object in a non-circular orbit, this angle does not change uniformly. There are two reasons:

1. similar changes in position will result in smaller changes in true anomaly the further $S$ is from $P$
2. the object is moving slower when it is far from $P$

The first issue is solved by measuring the angle from the center $O$, and projecting the trajectory on a regular circle. This new angle is named the eccentric anomaly, since it is measured from (ex in Latin) the center. It is usually notated $E$.

As for the second issue, Kepler’s equation defines a new quantity $M$, the mean anomaly:

$M=E-e\mathrm{sin}\left(\angle E\right)$

It can be shown that $M$ grows linearly with time. It means that we can easily predict the mean anomaly of an object at time $t$ as:

$M\left(t\right)=M\left(0\right)+n×t$

where $n$ is a constant value called the mean motion. By knowing the mean anomaly at epoch $M\left(0\right)$ of an object as well as its mean motion, we can retrieve its mean anomaly; we can then solve Kepler’s equation to retrieve $E$, and then $\mathit{\theta }$.

Orbital elements

The state of a mass-point is its position and velocity. In space, we need three values for each the position and the velocity, for a total of six parameters. The orbital elements are six such parameters that make it easy to also know the orbit of an object.

First, we have seen that the eccentricity, $e$, determines the shape of the orbit. Second, the semi-latus rectum determines the shape, but the periapsis ${r}_{\text{per}}$ conveys the same information (it is always defined), and is more intuitive.

To tell where the object is on its orbit, we need an angle. The true anomaly is intuitive, but does not change regularly through time; for this reason, we will usually prefer the mean anomaly $M$.

We now need to rotate the orbit in its orbital plane. Since the periapsis is not necessarily on the $x$ axis, we need an additional angle: the argument of periapsis $\mathit{\omega }$.

Our orbit is now fully determined in two dimension; we just need to orient the orbital plane in space. For this, we will need to know the inclnation $i$, as well as where the inclination occurs.

The axis around which the orbit is inclined is named the line of nodes, and the two points its intersection with the the orbit are called the ascending node and the descending node. Their names corresponds to wether the object is going up (along the $z$ axis) or down, when going through it. The line of nodes is simply determined by the longitude of ascending node, $\mathit{\Omega }$.

To sum up, the six orbital elements are:

• periapsis, ${r}_{\text{per}}$
• eccentricity $e$
• argument of periapsis $\mathit{\omega }$
• inclination $i$
• longitude of ascending node $\mathit{\Omega }$
• mean anomaly $M$

Vis viva equation

The vis viva equation is a simple relation between speed and distance, which uses only one orbital characteristic.

The principle of energy conservation tells us that the total energy (kinetic and potential energies) of an isolated system is constant:

Let us consider this at the two most notable points of a closed orbit: periapsis and apoapsis. At these points, the distance from center $r$ is extremal (and so is the altitude, or distance from surface); thus, $\stackrel{\bullet }{r}=0$ in both situations. It comes that:

$v=|\stackrel{\bullet }{\stackrel{\to }{r}}|=|\stackrel{\bullet }{r}\stackrel{^}{r}+r\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{\mathit{\theta }}|=|r\stackrel{\bullet }{\mathit{\theta }}\stackrel{^}{\mathit{\theta }}|=r\stackrel{\bullet }{\mathit{\theta }}$

Since we already know that ${r}^{2}\stackrel{\bullet }{\mathit{\theta }}$ is a constant value, it comes that:

${r}_{a}{v}_{a}={r}_{p}{v}_{p}$

Now, we apply the equation of energy (1) at both apses and get:

1 m v_a^2 - G M m = 1 m v_p^2 - G M m
v_a^2 - v_r^2 = 2 G M ( 1 - 1 )
(1 - r_a^2) v_a^2 = 2 G M ( 1 - 1 )
v_a^2 = 2 G M ( 1 - 1 ) r_p^2
= 2 G M r_a - r_p r_p^2
= 2 G M 1 r_p
= G M 2 a - r_a
= G M ( 2 - 1 )

We now have an expression of ${v}_{a}$ that depends only on ${r}_{a}$ and $a$. If we use the equation of energy (1) again on the with an arbitrary point with distance $r$ and speed $v$:

1 m v_a^2 - G M m = 1 m v^2 - G M m
v^2 = G M ( 2 - 2 ) + v_a^2
v^2 = G M ( 2 - 1 )

From this, we can easily determine the speed of an object from its distance to the primary.

State prediction

Finding the position from the orbital parameters if relatively easy. First, we can find the true anomaly $\mathit{\theta }$ from the mean anomaly $M$. Then we can retrieve the distance from the trajectory equation:

$r=\frac{l}{1-e\mathrm{cos}\left(\angle \mathit{\theta }\right)}$

For this, we need $l$, that we can easily find back using the fact that ${r}_{\text{per}}=\frac{l}{1+e}$. With this, we have the polar coordinates $\left(r,\mathit{\theta }\right)$ in the oriented orbital plane.

Then, it is only a matter of rotating around the proper axes ($\mathit{\omega }$ around $\left(z\right)$, $i$ around $\left(x\right)$ and $\mathit{\Omega }$ around $\left(z\right)$) to find the coordinates in the general frame of reference

Finding the velocity is very similar. We simply use the vis viva equation to determine the speed.

Inferring an orbit

We consider the inverse problem, where we want to determine the orbital parameters from our current position $\stackrel{\to }{r}$ and velocity $\stackrel{\to }{v}$.

First, let us define the specific relative angular momentum $\stackrel{\to }{h}=\stackrel{\to }{r}×\stackrel{\to }{v}$. Since both $\stackrel{\to }{r}$ and $\stackrel{\to }{v}$ are contained in the orbital plane and $\stackrel{\to }{h}$ is orthogonal to both these vectors, it means that $\stackrel{\to }{h}$ is a normal vector of the orbital plane. The angle between $\stackrel{\to }{h}$ and $\left(z\right)$ is the inclination $i$.

If we now compute $\stackrel{\to }{n}=\stackrel{\to }{{u}_{z}}×\stackrel{\to }{h}$, we get a vector that is both in the $\left(\mathit{xOy}\right)$ plane and in the orbital plane: this is the line of nodes. The angle between $\stackrel{\to }{n}$ and $\left(x\right)$ is thus the longitude of ascending node, $\mathit{\Omega }$.

The eccentricity vector is defined as:

$\stackrel{\to }{e}=\frac{\stackrel{\to }{v}×\stackrel{\to }{h}}{\mathit{\mu }}-\frac{\stackrel{\to }{r}}{r}$

It can be shown that its magnitude is the eccentricity and that it points towards the periapsis. We can thus compute $e=|\stackrel{\to }{e}|$, set $\mathit{\omega }$ as the angle between $\stackrel{\to }{n}$ and $\stackrel{\to }{e}$, and $\mathit{\theta }$ as the angle between $\stackrel{\to }{e}$ and $\stackrel{\to }{r}$.

It can be shown that $l=\frac{{h}^{2}}{\mathit{\mu }}$. Since we know the semi-latus rectum $l$ and the eccentricity $e$, we know the periapsis ${r}_{\text{per}}=\frac{l}{1+e}$.

1. Orbit angles,
2010, SVG diagram, User:Theoprakt,
https://en.wikipedia.org/wiki/File:Orbit1.svg