questions about the ∆v diagram and aerobraking

I also had some excellent questions about this diagram:

Specifically, the places where aerobraking can occur to assist (decrease the ∆v cost). The first thing we need to understand is what these places even are.

Well, they aren’t all places, per se, and that’s where it gets confusing. They are orbits. GEO, for example, is a “geostationary orbit” meaning it’s an orbit at an altitude such that you are moving around the Earth at the same rate the Earth spins, meaning you hold your position apparently stationary over one place on the planet. Sri Lanka is fine; that’s where Arthur C. Clarke put his space elevator in Fountains of Paradise. Since there’s no air there and there’s no air on any direct path to its links (GTO, L4/5, and LEO) there’s no opportunity for aerobraking.

You would guess from this diagram that GTO is further away than GEO. But that’s an artifact of the diagram which is not showing distances at all. GTO is a “geostationary transfer orbit” which means it’s an elliptical orbit with one end in a potentially geostationary position. The other end will be much closer to the Earth:

Approaching the GTO orbit from C3=0 and using aerobraking to reduce the cost of that burn.

So we can see that if you were entering GTO from somewhere else (say, C3=0 which we’ll explain later), you could enter at the shallow end of the orbit and use the atmosphere to slow down enough to complete the burn for the orbit and wind up in GTO.

C3=0 is the escape orbit of the planet (Mars or Earth in this diagram). This is the velocity at which you are no longer orbiting the planet but rather are now orbiting the sun while in the rough vicinity of the planet (or not, but that’s where the transfers take place).

The blue dot is Earth. The little spaceship is your little space ship. It is no longer orbiting the Earth–its orbital velocity is past Earth’s escape velocity but less than the Sun’s. This is C3=0.

So if you wanted to burn into a GTO around Earth you need to slow down. Nothing’s stopping you from doing that so your vector passes through the atmosphere, using that friction to make the burn cheaper. However, going from GTO to C3=0 needs more velocity not less, so aerobraking is no help and that’s why the arrow only goes in one direction on the chart. It’s only useful to go through atmosphere when you’re slowing down.

Similarly, in “deep space” at your Mars transfer orbit you want to enter the C3=0 of Mars. While there’s no air there, you can plot your path such that you pass through some air on the way. Why would you want to though? If you have an elliptical transfer orbit you probably want to speed up to reach C3=0. I am guessing that they mean that you could be on a longer elliptical orbit for transfer such as:

Earth is blue, Mars is red, and you are clearly in way too much of a hurry to match up with Mars’ orbit so you are going to plan a path way past it and brake in the atmosphere as you pass.

 

In this case you want to slow down or possibly change direction and it seems like there’s an opportunity to use Mars to do it. I’m not sure I see exactly how that would help, but you certainly can pass through Mars atmosphere from a transfer orbit. I’m just not sure why.

Although the atmosphere at Mars is much much thinner, at high speeds and with a high cross section, it will still slow you down a bunch. The Odyssey mission, for example, used aerobraking to slow down. They burned from capture orbit (the orbit at which they just start to move slower than the escape velocity of Mars and so are now orbiting Mars instead of just the sun) to what they called an “aerobraking orbit” which was designed so that every time the craft passed through the short side of the orbit the vessel would slow down, slowly circularizing the orbit until it was suitable for the mapping work.

Over at the moon there’s no atmosphere at any of the adjacent nodes to lunar orbit so there’s no aerobraking. At the moon you have to do all the work. It’s also pretty cheap to land and take off from though!

Thanks to Pierre Savoie for the questions that spawned this

questions about the ∆v science

I got some excellent questions on the last few articles and the answers deserve some space, so here’s that space.

I’m about to reveal how clueless I am about these topics, but these are making me reflect on recent sifi fiction. Could weapon recoil provide ∆v significant enough to be strategic (assuming weapons use power that doesn’t steal from thrust capacity)?

This will become clearer in a later answer but the short version is: probably not. Ships are going to be very hard to move due to their mass. Additionally, you probably don’t want a weapon that costs you ∆v unless you point it in exactly the right direction: the odds of that being both the direction of your target and opposite your desired vector change is mighty small.

Could you use weapons as thrust in an emergency? You could probably use some weapons to rotate the vessel rather than apply ∆v. Rotating your ship is comparatively cheap! In fact any weapon with recoil probably has a compensating jet to avoid this. I believe this is the case with the Rocinante in The Expanse! From the entry on PDCs in the fandom wiki:

They also utilize thrusters on their rear to counteract the recoil of the firing cannon, that would otherwise knock the ship off course.

To be clear, though, it probably wouldn’t affect the ship’s course, but it would rotate the ship. And I suppose if you’re burning the drive while firing that would indeed knock you off course.

Or nearby detonations (does space conduct shockwaves)? It’s intriguing to weigh the ∆v cost of any sort of space confrontation or skirmish.

Since there’s no atmosphere in space (by definition) there’s nothing intrinsic to transmit a shock wave. There will be some shock from the expanding plasma of the explosion of course, but we normally call that “damage”! And maybe a little photon push as well. Nothing that you would want to use as thrust.

However! If the explosion is energetic and close enough (and ideally shaped for the task), you can indeed propel your space ship with nuclear bombs. This would be a very poor ad hoc solution to a problem, but not an infeasible design. Obviously there are significant drawbacks to the design.

How does starship mass impact ∆v strategy? I’m probably wrong but I’m assuming greater mass requires greater thrust to achieve the same vector, and more thrust requires more fuel or efficiency, which all rolls into a single measure of total ∆v capacity.

I’ve only put the thumbnail here because it’s massively detailed and you should go to Winchell’s site to read about it and download the whole thing.

You are absolutely correct! For any given drive capability you can calculate the ∆v of the whole system by estimating the proportion of reaction mass (the mass you store only so you can shoot it energetically out the back) to the payload mass (the mass you have to keep). This is because of the “rocket equation” which I’m not going to go into, but Winchell Chung has an awesome chart showing ∆v for any given hypothetical drive type and any given mass ratio! It’s the basis of the game design that’s emerging here.

So yes, one of the joys of using ∆v as the core resource is that it encapsulates all kinds of information about the ship.

But would reducing your mass increase ∆v capacity through fuel efficiency?

The Marie Therese before ejecting the spin-grav luxury cabins and swimming pool.

Essentially yes! Let’s say you were the players in my last Diaspora campaign and were escaping in a luxury liner with a huge rotating spin-gravity living space. And you’re being chased. Ejecting that useless mass (which was huge) would change your r-mass:p-mass ration substantially, and give you a ton of spare ∆v.

Mass also impacts gravitational vectors, right?

Nope. The force you experience is dependent on your mass, but the acceleration you experience is not — it’s pretty much 9.8 m/s² for everything on Earth (but an elephant gets a lot more harm falling from a height than a mouse does — that’s the force). But not everywhere, and certainly not at different altitudes. But that’s way more detail than we need.

Does anything in space provide opportunity for aerobraking other than atmospheres?

Atmosphere is all I can thing of. Most interstellar gas clouds are way too tenuous to be interesting at this scale. Doesn’t mean you can’t invent one though!

Renewable sources of ∆v reserves look increasingly important (e.g. rechargeable solar vs consumable fuel?).

Well, basically your ∆v is going to be based on how much and how fast you can throw something out the back. So renewable is pretty tough unless you have a way to convert energy into mass. At least for rockets anyway. Remember for a rocket we’re not talking about power (although you might need some power to run the rocket, and that power supply will have its own energy needs which might include solar) but rather reaction mass.

Many thanks to Adam Minnie for the questions!

aerobraking

Okay let’s look at aerobraking. This is a way to get free ∆v for deceleration We’ll look at the principle and then talk about its utility. First recall our gravity example but I’ll add an atmosphere to the planet.

Now the atmosphere is a source of friction and friction reduces your velocity. As you bang into gas molecules, you slow down. We can treat this as a third vector in the system, opposing our direction of travel (our initial velocity vector).

Back when we were first talking about vectors, we said that you add vectors by arranging them head to tail and finding the hypotenuse. This was a lie, of course, as all science education advances by correcting the lies told in earlier lectures. In fact you just connect the last head with the first tail, like so:

Oops. I crafted this example without figuring out ahead of time what it would do and it looks like this atmosphere is too dense for our maneuver! When science fiction stories talk about a “degrading orbit” this is what they mean if we’re being charitable: eventually atmosphere will drag you in if you’re too low.

However, that doesn’t have to be a crash! That could be a landing! In which case we saved the atmospheric vector from our ∆v resources. We totally meant to do that! Of course that’s not what that diagram represents since that enormous vector is our velocity and it does not look like a safe landing speed. Or angle. But that’s the principle and however much additional ∆v we have to spend to land safely, it’s reduced by the amount of drag provided by the atmosphere.

So let’s consider a hypothetical interception scenario — you are fleeing through space and the cops are after you. You have, let’s say, 7 units of ∆v and spend 2 planning a course to a safe planet. It’s months away and you’re committed and you now have only 5 units of ∆v. You’re saving some for slowing down at the far end — you need 3∆v to make orbit at your destination.

You: 5∆v

The cops spot you and match your course with 2 ∆v of their own. They have fast interceptors (high thrust for tactical corrections) but less reserves so let’s say they start with 5∆v. Now they have 3.

You: 5∆v

Cops: 3∆v

You spot the cops on your infrared telescope and track them for a couple of days, identifying their planned intercept point. You have a bunch of options. You have more total ∆v and if you know this you could burn more than they can afford to to change your course and correct back. If they spend any more than they have they won’t have the ∆v to slow down and stop somewhere to refuel. But if they aren’t tricked and don’t burn, then when you correct back they will still be on target.

You could also fake a course to a totally different location, spending maybe 3 more ∆v leaving you with 2. If the police correct for that they will be totally committed (they won’t be able to slow down so they are clearly going to try to kill you in a flyby and then count on other cops to save them from leaving the solar system) but you have some spare.

You: 2∆v

Cops: 0∆v

Maybe it’s not enough to make orbit around your original destination — you’ll be going too fast. But if you’re equipped with heat shields for aerobraking, you can steal 1∆v from the atmosphere and get the 3 you need to make orbit despite going way too fast!

Or maybe you won’t slow down! Maybe you’ll steal 1 ∆v from slingshotting your destination on order to head somewhere completely else!

You: 3∆v

Cops: 0∆v

Finding spare ∆v in the system geography is how you exceed your space craft’s specifications, and it’s a skill your character might have. Yes, we’re almost talking games now.

Since this is science fiction, what other sources of ∆v might be lying around a high tech industrialized star system?