Orbital mechanics – Part 2

For part one, follow this link: there we answer the question “how to stay in orbit”. In this post, we want to figure out how to change the orbit we are in and, in particular, how to get into orbit.

How to change our orbit

Let us assume we are in an elliptical orbit, and that we change our velocity with an impulse (instantaneous change of velocity) along the same velocity vector we had previously when we are at perigee (this is just for simplicity, to fix a special point). What happens?

We will still be on an orbit, and this orbit is once again going to be a conic. Moreover, this new curve has to go through the point we are in when we perform the burn (that is the impulsive manoeuvre). Also, we know how to compute the specific energy of this new orbit:

    \[\widetilde{\mathcal{E}}=\frac{1}{2}\widetilde{v}_p^2 - \frac{\mu}{r_p} = -\frac{\mu}{2\widetilde{a}},\]

where \widetilde{v}_p is the new velocity at perigee, r_p is the same radius as before the manoeuver, and \widetilde{a} is the new semimajor axis of the orbit, which means that r_a has changed and has been replaced with something new. We can actually compute this new quantity (that is, the radius of the point in the orbit opposite to where we are currently, which might not be the apogee anymore), and in particular, we can see that:

  • if the new speed is higher than the previous one, then the specific energy of the new orbit is higher, that is also -\mu/(2\widetilde{a}) is higher;
  • consequently, \mu/(2\widetilde{a}) is smaller;
  • and 2\widetilde{a}=r_p+\widetilde{r}_a is bigger;
  • so the new radius \widetilde{r}_a is bigger than the previous one (and in particular remains the apogee);
  • if the new speed is lower than the previous one, \widetilde{r}_a is going to be smaller than r_a, and possibly also smaller than r_p, thus replacing it as perigee.

It could also be that the new apogee is +\infty, if we are moving to a parabolic orbit, or negative, if we are moving to a hyperbolic orbit. In those cases we are now on an escape trajectory and we won’t be coming back, unless we perform another change of trajectory.

We can now state and prove a result on the best (in a sense that we will discuss later on) way to change orbit.

Theorem (Hohmann): Given two circular orbits with the same centre and lying on the same plane (concentric and complanar), of radii r_i and r_f>r_i respectively, the two-impulse manoeuvre from the first orbit to the second one that minimises the total velocity increment \Delta v_{\text{TOT}} is the transfer along the elliptical orbit of periapsis r_i and apoapsis r_f, the same orbital plane as the two given orbits, and the same direction as the starting orbit.

Before moving on to the proof, let us discuss a couple of points raised in the statement above. The fact that we are raising our orbit is not really relevant: we can of course lower our orbit and we will have a very similar result, with an elliptic transfer orbit, only with periapsis and apoapsis swapped. On the other hand, one can show that similar results hold also when we start from or get to an elliptical orbit. However, this is highly non-symmetrical, as one has to consider also the angle between the two orbits (actually, their major axes).

Proof: The orbital velocity of the internal circular orbit is \vec{v}_i, with v_{i}=\sqrt{\mu\cdot r_i^{-1}}, and similarly the velocity of the external circular orbit is \vec{v}_f, with v_f=\sqrt{\mu\cdot r_f^{-1}}. We also have some other quantities of interest: \vec{v}_1, the velocity vector immediately after the first impulse, and \vec{v}_2, the velocity vector immediately before the second (and final) impulse, as well as \phi, the angle between \vec{v}_{i} and \vec{v}_1 and \psi, the one between \vec{v}_f and \vec{v}_2.

We want to minimise the total delta-v, that is

    \[\Delta v_{\text{TOT}} = |\vec{v}_1-\vec{v}_i|+|\vec{v}_f-\vec{v}_2|=\Delta v_1 +\Delta v_2.\]

We can simplify the minimisation problem by using the conserved quantities that we already know:

    \[\mathcal{E}=\frac{v_1^2}{2}-\frac{\mu}{r_i}=\frac{v_2^2}{2}-\frac{\mu}{r_f} \qquad \Longrightarrow\qquad v_2^2=v_1^2-\frac{2\mu(r_f-r_i)}{r_i\cdot r_f},\]

and also, because of the conservation of angular momentum,

    \[c=r_i\cdot v_1\cdot \cos\phi=r_f\cdot v_2\cdot \cos\psi \quad \Longrightarrow\quad \cos\psi = \frac{r_i\cdot v_1}{r_f\cdot v_2}\cdot \cos \phi.\]

With them, we can write both \Delta v_1 and \Delta v_2 in terms of v_1 and \phi:

    \begin{align*}\Delta v_1 & = |\vec{v}_1-\vec{v}_i| = \sqrt{v_1^2 + v_i^2 -2v_1v_i\cos\phi}\\\Delta v_2 & = |\vec{v}_f-\vec{v}_2| = \sqrt{v_f^2 + v_2^2 -2v_fv_2\cos\psi}\\&=\sqrt{v_f^2 + v_1^2 -\frac{2\mu(r_f-r_i)}{r_i\cdot r_f}-2v_fv_1\frac{r_i}{r_f}\cos\phi}.\end{align*}

We can now rescale everything by the speed in the internal circular orbit v_i, by setting x=v_1/v_i, so that:

    \begin{align*}\frac{\Delta v_{\text{TOT}}}{v_i} &= \sqrt{x^2-2x\cos\phi+1}+\sqrt{x^2-2\left(\frac{r_i}{r_f}\right)^{3/2}x\cos\phi-\frac{r_i-2(r_f-r_i)}{r_f}}\\& =: D(x,\phi),\end{align*}

where we have put in evidence in the new quantity D that we want to minimise, that the two parameters we can play around with are x and \phi. Let us start by checking how D varies in terms of \phi:

    \[\frac{\partial D}{\partial \phi}=\frac{v_1v_i}{\Delta v_1}\sin \phi + \frac{v_1v_fr_1}{\Delta v_2 r_f}\sin\phi,\]

which attains minimum for phi=0, independently of the speed and consequently of x. Now we can differentiate D(x)=D(x,0):

    \[\frac{dD}{dx}=v_i\frac{x-1}{\Delta v_1}+v_i\frac{x-(r_i/r_f)^{3/2}}{\Delta v_2}.\]

This quantity is always positive since x>1 (as we want to reach the new orbit, so we need to raise the apoapsis of the first orbit), so the solution we are looking for is the minimal v_1 such that we actually reach the required new radius r_f. This happens when the elliptic orbit has apoapsis at height r_f, that is when v_f is orthogonal to r_f and

    \[\frac{v_1^2}{2}-\frac{\mu}{r_i}=-\frac{\mu}{r_i+r_f},\qquad v_1^2 = v_i^2\left(2-\frac{2}{1+r_i/r_f}\right).\]

Notice that in this case the transfer orbit (called Hohmann transfer lies on the same plane as the two circular orbits. This should be no surprise since changing plane (even if we have not seen it yet) comes at an additional cost in terms of delta-v. Also, we can read the time needed for this transfer directly from what we have written: we need to travel half the orbit, and the time needed for that is T=\pi\sqrt{a^3/\mu}, with a denoting the semimajor axis.

Finally, we could also write the total delta-v required in terms of the ratio of the radii:

    \begin{eqnarray*}   \frac{\Delta v_{\operatorname{TOT}}}{v_{i}} & = &       \frac{\Delta   v_1}{v_{i}} + \frac{\Delta v_f}{v_{f}} = \frac{v_1 -       v_{i}}{v_{i}} +   \frac{v_{f} - v_2}{v_{i}}\\   & = &       \sqrt{\frac{2 \cdot {r_f} /{r_i}}{1 + {r_f} /{r_i}}} - 1 +         \sqrt{\frac{1}{r_f} /{r_i}}} - \sqrt{\frac{2}{{r_f} /{r_i}        \left(   1 +{r_f} /{r_i} \right)}} .  \end{eqnarray*}

How to get in orbit

The rocket equation

Why do we want to reduce delta-v anyway? Let’s consider a rocket flying with velocity \vec{v} and subject to some external forces of sum \vec{F} (e.g. gravitational, atmospheric friction…). Let \vec{v_e} be the ejection velocity of the propellant (with respect to the rocket) in a (infinitesimal) time dt. We want to get the modulus \Delta v of the corresponding velocity variation of the rocket. We set the mass of the rocket (including engine mass and fuel mass and payload mass) to m. In a time dt the rocket is subject to the following variation of momentum:

    \[[(m - \mathrm{d} m)  (\vec{v} + \mathrm{d} \vec{v}) +       \mathrm{d} m (\vec{v} - \vec{v}_e)] - m \vec{v} = \vec{F}       \mathrm{d} t,\]

where, in the first term, we have the difference between the momentum at time t_0+dt and the momentum at time t_0. Since external forces are not impulsive, we have that \vec{F}dt\to 0. If we consider a first order approximation and forget about the second order term dmd\vec{v}, we get m \mathrm{d} \vec{v} = \vec{v}_{e} \mathrm{d}       m, which projected along \vec{v} gives

    \[\mathrm{d} v = - v_e  \frac{\mathrm{d} m}{m},\]

hence \Delta v = v_e \log \left( {m_0}/{m_f} \right),

with m_0 the mass of the rocket at time t_0, and m_f the mass at the end of the manoeuvre. This is called Tsiolkovskij equation or rocket equation and expresses the \Delta v variation terms of the fuel required m_0−m_f and the velocity of the exhaust.

It makes sense to introduce the following quantity, called specific impulse: I_{\text{sp}}=v_e/g, with g the gravitational acceleration at sea level (9.81 on Earth). The specific impulse (which has seconds as its unit, at least in the most common formulation, given above), is a measure of the efficiency of the engine used (namely it depends on the engine and the fuel used). As it is clear from the previous equation for the delta-v, the higher the specific impulse is, the better.

EngineI_{\text{sp}} (vac)massthrustTWR
Rocketdyne F1263 s8400 kg6.77 MN82.27
R25452.3 s3526 kg2.28 MN65.91
Merlin 1D340 s490 kg0.8 MN150
RD 180338 s5480 kg4.15 MN78.44
Hercules SRMU286 sn/a7.5 MNn/a
PPS 1350 1.5 kW1650 s5.3 kg0.088 N0.0017
Some rocket engines

It might be worth noting that usually rocket engines have different values for the specific impulse depending on being or not in the atmosphere (compared to being in the vacuum). Also worth stressing is that this rocket equation holds for engines that (approximately) fire in a single impulse, and does not hold for engines such as electrical ones (see the PPS 1350 in the table), which have very low thrust, even if they have a huge Isp.

Moreover, specific impulse it is not necessarily a good measure of comparison for engines, not just for problems as the one above (thrust vs isp, instant vs long time), but also because an engine with a very high Isp might have a huge mass, thus using a lot of its power to move itself, so it is better to compare the work provided by each engine. Reminder: thrust is given by \vec{T}=\vec{v}{dm}/{dt}, with Newton as a unit, while thrust to weight ratio (TWR) is defined by TWR=T/(mg), and is a pure number.

Using delta-v to compute the cost of manoeuvers is particularly good, because it already includes the mass of the vehicle and the kind of motor in it, so it can be computed independently of the vehicle chosen (and can actually inform the choice of the rocket).

Launch to orbit

Suppose that we are on the surface of the Earth and we want to get into a circular orbit 250 km above the surface. This means that we have to launch on an elliptic orbit of periapsis 6.378\times 10^6 m and apoapsis 6.628\times 10^6 m and then perform a circularisation burn. The semimajor axis of the elliptical transfer orbit is 6.503\times 10^6 m, and we have to launch at a speed of 7995 m/s, arriving at the apoapsis with speed 7694 m/s, which we have to raise to 7769 m/s to circularise. So we need a delta-v of 8070m/s.

This assumes, however, that we are launching horizontally, and does not take into account atmospheric drag, gravity drag (the fact that we need to get off the ground, hence requiring a TWR greater than 1), but also, if we are launching from the equator eastwards, the 464 m/s kindly donated by Earth’s rotation. In reality, achieving such an orbit requires around 9400 m/s of delta-v. One could also argue that, in this case, we cannot really use the rocket equation, as the firing of the engine is not an impulse.

There is also another issue: if we were to use (ignoring thrusting issues and the lower performance at sea level) a Merlin engine to launch from Earth, from the rocket equation we have that m_i/m_f=\exp{9400/3335}\approx 16, which means that the propellant mass is 15 times the dry mass (that is the payload, the engine, the tanks, the instrumentation…). This is not feasible from an engineering point of view.

What happens in a real launch is staging: not all the dry mass is brought to orbit. We split the launcher into different parts that activate sequentially. As soon as one is burnt out, it is discarded. The delta-v contribution of each stage is summed up, to give the total delta-v balance of the whole stack.

As an example, let’s see what happened in the Apollo lunar missions. They launched with a three staged launcher, the Saturn V.

Stagem_i (stage)m_f (stage)m_i (vehicle)m_f (vehicle)\Delta v
12,290,000 kg130,000 kg2,945,200 kg785,200 kg3,409 m/s
2496,200 kg40,100 kg655,200 kg199,100 kg4,918 m/s
3123,000 kg15,200 kg159,000 kg51,200 kg4,678 m/s
payload36,000 kg
Delta-v balance for a lunar Saturn V-Apollo mission

It is interesting to notice that even with two stages, the Saturn V could not make it into orbit, and had to light up its third stage. At the same time, most of its delta-v was spent to get to orbit, and the remaining delta-v in the third stage was not sufficient for the rest of the mission: it only put the Apollo stack into lunar transit orbit. Then it was the turn of the engine on the Apollo Service Module (here packed inside the payload) to circularise the orbit around the Moon and take the crew back to Earth afterwards.

Beyond basics

There are many more interesting topics to cover in Orbital Mechanics: how to manoeuvre if we do not just want to get in a certain orbit, but also get there at a certain time, for example to rendezvous with another spacecraft (or a planet). This adds layers of complexity, together with issues coming from the fact that not necessarily we are in the luxury position of disregarding inclination changes.

Among the interesting things that could be discussed, there is the Oberth effect, stating that the same manoeuvre performed at apoapsis is less efficient than performed at periapsis, which is strictly connected to the (powered) gravity assists. Also, while Hohmann transfer has been proven above to be optimal, in terms of delta-v, this is only true if we restrict ourselves to two-impulse manoeuvers, whereas things might change if we allow for three impulses.

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