Orbital mechanics – Part 1 | Luigi Amedeo Bianchi

How to stay in orbit

This introduction only focuses on some aspects of celestial or orbital mechanics. In particular, we will stick to planar orbits, which we can assume equatorial for simplicity. Moreover, we will always consider two-body approximations. There will be some other simplifications, but we will mention them later when they become relevant.

The questions we want to answer are: how to go in orbit, how to stay in orbit and how to change orbit. We start, in this post, with the second one.

Circular orbits

First of all, we went to consider an object of some kind (a spacecraft, a rock…) orbiting Earth on a circular path. To write the mathematical model, we just need two laws from physics: Newton’s second law, $\vec{F} = m\cdot \vec{a}$ , where $F$ is the force acting on our object, $m$ is its mass and $a$ is the acceleration and (Newton’s) universal law of gravitation

$F=\frac{G\cdot M\cdot m}{r^2},$

where $G$ is the gravitational constant ( $G = 6.67\times 10^{-11}$ ), $M$ is the mass of the Earth ( $M = 5.97\times 10^{24}$ ), and $r$ the radius between the center of mass of the Earth and the center of mass of our object.

Notice that for our purpose right now, both the Earth and the object can be considered points with mass. This can only be acceptable if the radius is big enough to go beyond the Earth’s surface and the Earth’s atmosphere. Notice moreover that there is nothing really special about considering Earth, we could consider a different body, just by changing $M$ (and possibly $r$ ).

Some notation: since the product $MG$ comes up quite often, it makes sense to define $\mu = MG = 4\times 10^{14}$ .

Now, we want our object to stay in orbit at a fixed height of 300 km above the surface (i.e. $r= 6.678 \times 10^{6}$ m). We want the speed (the intensity of the velocity vector) to be constant along our orbit (we are looking for a circular orbit with no other force than gravitational pull acting on our object).

This brings another remark: technically speaking, our object is also pulling the Earth. However, we expect the mass $m$ to be negligible compared to $M$ , so we can safely consider the Earth pulling the object.

Back to our object in orbit. we have the following situation:

where the distance $l$ travelled in a brief amount of time $dt$ can be considered (by means of linear approximation) $l = v dt$ .

Let’s now consider the second triangle. The two sides $v_1$ and $v_2$ have the same length, but they are (as vectors) perpendicular to $r_1$ and $r_2$ respectively, so the angle between them is $\theta$ , as the angle between $r_1$ and $r_2$ . Since the two triangles are both isosceles and have the angle in common, they are similar. In particular

$\begin{equation*} \frac{dv}{v\cdot dt}=\frac{v}{r},\end{equation*}$

that is $a = v^2 \cdot r^{-1}$ .

Now we can plug this into Newton’s second law, to get

$F=m\cdot \frac{v^2}{r}$

and, since the force is the gravitational one,

$\mu \cdot \frac{m}{r^2} = m\cdot \frac{v^2}{r},$

which can be reduced to $v^2 = \mu\cdot r^{-1}$ .

There are a few things to read here:
• the orbital speed does not depend on the mass of the object, but only on the radius of the (circular) orbit
• the higher the orbit (that is the greater $r$ is), the lower $v$ is.

Also, let’s see an example with some numbers: if we are looking at an orbit of radius $6.678\times 10^{6}$ m around the Earth. That is all we need to compute the orbital velocity:

$v = \sqrt{\frac{\mu}{r}}=\sqrt{\frac{4\times 10^{14}}{6.678\times 10^{6}}}=7739\frac{\text{m}}{\text{s}}.$

Elliptical orbits

Kepler’s laws tell us that in general orbits are ellipses, with Earth (in our case) in one focus. In this setting, the orbit is parameterised as follows:

$r(\theta)=\frac{a(1-e^2)}{1+e\cdot \cos (\theta)},$

where $e$ is the eccentricity of the orbit, $r$ the radius, $a$ the semimajor axis (we won’t be needing acceleration anymore), and $\theta$ the true anomaly, that is the angle between the radius at periapsis (or perigee) and the current point. Periapsis is the closest point to Earth and apoapsis (or apogee) is the farthest. The semimajor axis $a$ is the arithmetic mean of the two radii at
perigee and apogee

$a = \frac{r_p+r_a}{2},$

and we can rewrite the radii at apogee ( $r_a$ ) and perigee ( $r_p$ ) in terms of $a$ and $e$ : $r_p=a(1-e)$ , $r_a=a(1+e)$ .

For the moment, we are assuming that there is no other force acting on our object in orbit, so its total energy will be a constant. Let us recall that the total energy is given by

(1) $\begin{equation*} E=E_K-E_P=\frac{1}{2}mv^2-\frac{GMm}{r}\end{equation*}$

that is by the contributions of kinetic and potential energy.

We can also notice that both terms in the right-hand side of (1) have a mass term $m$ . Since mass is a constant (at least for the moment) and we have already seen that the velocity does not depend on $m$ , we can focus on $\mathcal{E}=E/m$ instead, called the specific energy.

We want to compute this specific energy. To do so, we can start by focusing on its value at perigee and apogee, and equal them out, since we want energy to be constant along the orbit:

(2) $\begin{equation*}\frac{1}{2}v_a^2-\frac{\mu}{r_a} = \frac{1}{2}v_p^2-\frac{\mu}{r_p}.\end{equation*}$

We can use the conservation of angular momentum and the fact that $v_p\perp r_p$ and $v_a \perp r_a$ to get $r_a\cdot v_a = r_p\cdot v_p$ , which we can also write $v_p=v_a\cdot r_a\cdot r_p^{-1}$ .

Going back to (2) and plugging in this expression for $V_p$ , we get

$\frac{1}{2}v_a^2\left(1-\frac{r_a^2}{r_p^2}\right) = \mu\left(\frac{1}{r_a}-\frac{1}{r_p}\right),$

from which we can get an explicit expression in $v_a$ :

$\begin{align*}v_a^2 & = \mu \cdot 2 \cdot \frac{r_p^2}{(r_p+r_a)(r_p-r_a)}\cdot\frac{r_p-r_a}{r_a\cdot r_p}\\&=\mu\cdot \frac{1}{a}\cdot \frac{r_p}{r_a}\\&=\mu\left(\frac{2}{r_a}-\frac{1}{a}\right),\end{align*}$

where we have used the definition of semimajor axis and in general the relationship between radii at apoapsis, periapsis and semimajor axis.

We can also (either by following similar computations or by substituting $v_a$ and $r_a$ into the conservation of momentum identity above) get a similar expression for $v_p^2 = \mu(2\cdot r_p^{-1}-a^{-1})$ .

Our focus, however, was the specific energy of the orbit, so we can now plug $v_a$ or $v_p$ in to get

$\mathcal{E} = \frac{1}{2}\cdot v_a^2-\frac{\mu}{r_a} = \mu\left(\frac{1}{r_a}-\frac{1}{2a}-\frac{1}{r_a}\right)=-\frac{\mu}{2a}.$

This is independent of the point used to compute it and is, for $a>0$ , that is for closed orbits, a negative quantity. As it is independent of the point in the orbit, we can use it to obtain the orbital speed point by point:

$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}.$

(Let’s stress that we are talking about speed. We know that the velocity is tangent to the ellipse at the point, but this does not mean anymore that the velocity is tangent to the radius.)

There are a few things left behind here: when is the energy zero or positive? What happens with open orbits?

Next time, we are going to see how we can get from one orbit to another and, more importantly, how can we get to orbit at all.

Orbital mechanics – Part 1

How to stay in orbit

Circular orbits

Elliptical orbits

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