# Does going to Mars take the same energy as making a rendezvous with something going to Mars?

I think I found a mathematical error in the 2015 movie The Martian1. Warning, there will be spoilers.

The final plan they settle on to rescue Mark Watney is to have the Tiayang Shen rendezvous with the Hermes, which will perform a gravity assist off of Earth, sending it back to Mars, where Watney will meet it in orbit to return to planet Earth.

But wait, wouldn't it take the same amount of delta-v for the Tiayang Shen to meet the Hermes as it would for the Tiayang Shen to just go all the way to Mars on the same trajectory? If the rendezvous occurs with the Hermes on a Hohmann transfer trajectory between Earth and Mars, and physics 101 says you need to match velocity and position to rendezvous, then the Tiayang Shen must propel itself to an Earth-Mars Hohmann transfer trajectory as well. Perhaps it joins at the cheapest anomaly along the orbit, but it still needs to enter an orbit which would bring it to Mars, Hermes or not. The only difference is slowing down upon reaching Mars, which Hermes may be capable of with its ion engines, but the Tiayang Shen may not be, given limited delta-v. Not to mention that they were originally planning to aerobrake to the surface of mars anyway, so additional maneuvering fuel used after entering orbit could be as low as zero.

Why would having the Hermes speed up the journey? Why did they need to send it back at all given the Tiayang Shen?

• I think Taiyang-Shen is the rocket, not a ship, so it's not capable of flying to Mars in a controlled manner (I don't think it even has a restartable upper stage). Plus, had there been a purpose built cargo ship to the Mars, Taiyang-Shen probably isn't powerful enough for it (Taiyang-Shen isn't designed for Mars cargo mission in the first place). In the end the only choice is to let Taiyang-Shen only carry the cargo container and let Hermes grab it. Aug 4, 2021 at 11:28
• You are correct. If you rendezvous with a spacecraft going anywhere, then you will be going there together until the next time one or the other of you fires thrusters. I would not say that it proves a "mathematical error" in the movie though: Hermes was capable of bringing Watney back home again. Tiayang Shen, maybe not so much. Aug 4, 2021 at 16:09
• Hermes wasn't on a Hohmann transfer trajectory. It had an ion drive. Aug 5, 2021 at 12:48

Yes, it takes roughly the same amount of delta-v, your analysis is sound and good. But no, there is no error in the book because that's the premise of that part of the story: we can either reach Mars with food and supplies to last Watney until Ares 4 comes around and picks him up OR we resupply Hermes and they pick him up now(ish).

The difference is not in the time it takes to get there, the difference is that one of those mission profiles can bring him back and the other can't.

It was very unfortunate that, at the time, there was one and only one booster in all of earth capable of the feat, so they had to choose.

• Additionally, the resupply of Hermes doesn't require an entry-descent-landing module or long-duration spacecraft. Aug 5, 2021 at 4:17

You are correct. There are, however, a few differences you missed.

1. Hermes actually was going to accelerate using an ion drive type system to get to Mars, thus it wasn't actually on a direct path to Mars.
2. Landing something on Mars is hard. At a minimum a heat shield and parachute would be required, even to have a less than perfect landing. Those can be removed when one isn't actually landing on Mars itself.
3. The mission profile for the return mission was different Mark would have to survive on Mars for a few years until the right orientation happened, not to mention the Hermes being there to take him home again, all of which would take years for that profile, requiring a fair bit of food. The other mission could be accomplished much quicker.

All in all, the explanation given was good. Either mission was possible, but they had to commit to one or the other.