Orbital Mechanics and The Martian (with math)
I'm not a rocket scientist, but I do occasionally watch them on TV. Today, I'm going to really take advantage of the First Rule of LobbySeven Commentary: I can write about what I want and you have no real way to stop me.
I read "The Martian" a good while ago (in one sitting, stranded at Toronto Pearson, if you want to know) and recently watched the film version. I liked them both. Now, the story was supposed to have a huge amount of attention to detail, and it did, without question. But, there was one part I have to take some question with. I have no problem with the orbital mechanics, but I when they had one of the spacecraft significantly change it's mission it got me thinking.
The original plan for the Mars Ascent Vehicle ("MAV") was to meet up with the larger "Hermes" in Low Mars Orbit (which I'll call "LMO"). This is a similar to the mission plan from Apollo, when the Lunar Module met the Command Module in Moon Orbit ("Moon Orbit Rendezvous"). Apollo missions would have been far more difficult this plan, because then "everything" would need to be brought down to the Moon and back up. Given the mission in The Martian had even more "stuff" it makes sense they'd have a similar plan.
However, when things go awry, the MAV is now supposed to meet up with the Hermes when it is conducting a fly-by rather than orbiting Mars. During a Mars fly-by, you are going a lot faster than when you are in LMO. Spaceships are build for specific purposes. Yes, there is always some margin of error in the design. But could a ship built to rendezvous in LMO reach the speed necessary to escape Mars orbit? Let's find out!
I like to think of "delta-v" as the currency of space travel. When a spacecraft takes off, it has a certain amount of delta-v, depending on factors like the weight of the spacecraft, amount of fuel carried and the efficiency of its engines. When a spacecraft performs any maneuver - takes off to orbit, transfers to another planet's orbit, descends to the surface - it "spends" some of this delta-v. In space, there is no way to get any back. The only thing you can do without using your delta-v is land on a planet/moon that has an atmosphere. When you slam into the air, it slows you down without using any fuel. Come to think of it, you could also "land" on something that doesn't have an atmosphere, but because you won't slow down first, I don't recommend it.
Because moving from one position to another requires a fixed amount of delta-v (more or less), we can draw a "map." Think of it as a road map, but distances are expressed as the amount of gasoline required to get from place to place. Unlike driving on I-94, in orbit you have nowhere to stop for more fuel.
The relevant points for our discussion are:
Mars Surface to Low Orbit, 3,800 delta-v
Low Mars Orbit to Intercept, 1,440 delta-v
These numbers are in "meters per second," which is the usual unit for delta-v. This means that the minimum speed at which the Hermes could have been traveling when meeting the MAV was 5,240 m/s. The "Trajectories of the Martian" picture at the top says 5,360 m/s, a bit higher, because the alignment was probably not perfect. So our question is: could a spaceship designed for 3,800 delta-v get to 5,240?
This brings us to my third-favorite equation of all time (Bayes is #1, we'll get to #2 later), the Tsiolkovsky rocket equation:
What does this tell us? It says that for a spaceship to go far, you want to shoot out as much its mass as possible, as fast as possible. Recall the situation faced in The Martian: we take a spaceship with a given delta-v, reduce the weight, and want it to go further. This gives us three equations and three unknowns:
I'll spare you the details, but trust me that you can reduce these equations to find the maximum final mass of the original Mars Ascent Vehicle that would would permit the change to work:
Don't worry too much about what it all means: I've incorporated it into a handy Google doc that you can find here.
Let's say that they included a buffer of 100 on the MAV as designed, bringing it to 3900. We know it fell just short of the required 5240 for rendezvous, so let's say it ended up with 5200 m/s. They said that the mass was reduced by 5000 kg, and there were also six fewer passengers weighing perhaps 420 kg between them. Unfortunately, I'm not able to find anybody crazy enough to publish fan fiction Mars Ascent Vehicle specs, so we need to guess what its exhaust velocity was.
The Apollo Lunar Module ("LM"), the closest vehicle we've actually built to the MAV, had exhaust velocity of 3,050 m/s; the Saturn V around 4,130. This would imply a MAV maximum empty mass of 12,800 - 14,400 kgs. Even assuming that in the future we built a rocket engine twice as efficient as the LM, it gives only 16,300 kgs. Given that the MAV was able to have 5,000 kgs of mass removed and still fly, these numbers seem unlikely. However, it bears noting that the LM had a "dry mass" of only 2,150 kgs.
And there is another issue: if the final mass were 16,300 kgs, using the "super efficient future engine", that means the MAV mass with fuel was around 31,000 kgs. Using Mars' gravitational constant of 3.71 gives a weight of 115kN. I would assume that taking off from Mars would require a thrust-to-weight ratio of 1.75 at a minimum (the LM was closer to 2.1). This means you would need 200kN of thrust, in your hypothetical super-efficient engine. The LM ascent engine had 16kN of thrust. So, your engine needs to be 12x as powerful, 2x as efficient and attached to a spaceship that can have 1/3 of its mass removed and still fly.
We haven't proven the thing impossible, but I was right to be skeptical. Any good MIT Course 16s want to weigh in? Either on my analysis or the overall feasibility?