# 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: where is the new velocity at perigee, is the same radius as before the manoeuver, and is the new semimajor axis of the orbit, which means that 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 is higher;
• consequently, is smaller;
• and is bigger;
• so the new radius is bigger than the previous one (and in particular remains the apogee);
• if the new speed is lower than the previous one, is going to be smaller than , and possibly also smaller than , thus replacing it as perigee.

It could also be that the new apogee is , 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 and respectively, the two-impulse manoeuvre from the first orbit to the second one that minimises the total velocity increment is the transfer along the elliptical orbit of periapsis and apoapsis , 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 , with , and similarly the velocity of the external circular orbit is , with . We also have some other quantities of interest: , the velocity vector immediately after the first impulse, and , the velocity vector immediately before the second (and final) impulse, as well as , the angle between and and , the one between and .

We want to minimise the total delta-v, that is We can simplify the minimisation problem by using the conserved quantities that we already know: and also, because of the conservation of angular momentum, With them, we can write both and in terms of and : We can now rescale everything by the speed in the internal circular orbit , by setting , so that: where we have put in evidence in the new quantity that we want to minimise, that the two parameters we can play around with are and . Let us start by checking how varies in terms of : which attains minimum for , independently of the speed and consequently of . Now we can differentiate : This quantity is always positive since (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 such that we actually reach the required new radius . This happens when the elliptic orbit has apoapsis at height , that is when is orthogonal to and 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 , with denoting the semimajor axis.

Finally, we could also write the total delta-v required in terms of the ratio of the radii: ## 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 and subject to some external forces of sum (e.g. gravitational, atmospheric friction…). Let be the ejection velocity of the propellant (with respect to the rocket) in a (infinitesimal) time . We want to get the modulus 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 . In a time the rocket is subject to the following variation of momentum: where, in the first term, we have the difference between the momentum at time and the momentum at time . Since external forces are not impulsive, we have that . If we consider a first order approximation and forget about the second order term , we get , which projected along gives hence ,

with the mass of the rocket at time , and the mass at the end of the manoeuvre. This is called Tsiolkovskij equation or rocket equation and expresses the variation terms of the fuel required and the velocity of the exhaust.

It makes sense to introduce the following quantity, called specific impulse: , 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.

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 , with Newton as a unit, while thrust to weight ratio (TWR) is defined by , 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 m and apoapsis m and then perform a circularisation burn. The semimajor axis of the elliptical transfer orbit is 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 , 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.