# Would it be feasible to decelerate a crewed vehicle from ~25 km/s only using the atmosphere of Mars (in the context of an "express transit")?

I boldly assumed it would some day be possible to accelerate a crewed vehicle from a refueling station (LEO, GEO, Moon orbit?) to ~35-40 km/s (with several stages) relative to Mars, so it would make the 70-100(?) million km journey at an average velocity of ~ 30 km/s and encounter Mars with ~25 km/s(?). That would make the transit sort of acceptably short, roughly 1 month long.

Since the escape velocity from Mars is ~5 km/s, the craft would need to be decelerated by at least 20 km/s and since Mars conveniently has an atmosphere I wanted to ask if it would be possible to get the job done purely by atmospheric braking. I realize that this $$\Delta V$$ of 20 km/s would need to occur in one pass (the remaining 5 km/s would be reduced in a second atmospheric pass to either enter an orbit or land on Mars).

This is about twice the $$\Delta V$$ of the returns from the Moon, so it would be quite brutal, I assume. If this is definitely impossible the classical way, would it be helpful to use a sacrificial parachute or other device that gets deployed initially, in extremely thin atmosphere, before the main atmospheric braking occurs?

• Thanks @BrendanLuke15 for editing including tags - I seem to have trouble finding appropriate tags: I was looking for "crewed" but didn't find anything. Dec 30, 2021 at 13:01

Nope, sorry.

Completely disregarding the (enormous) heat factor of the aerobraking.

The requirement specifies manned.
The path through Mars' atmosphere will be very nearly a straight line.
A bit of trig tell us that the path through the atmosphere (up to 5% of surface pressure) is only some 570km long.

Decelerating from 25km/s to 5km/s in 570km, requires an average deceleration of -526.32m/s2
That's an average of 53.65g (for 38 seconds only)

Your astronauts will be strawberry jam.

P.s. If you can drop that initial speed to 15km/s, the deceleration drops to 17.9g (58 seconds)
Still very painful, but potentially survivable with some elaborate support structures (we're talking liquid immersion g-bath, and the like)

• @BrendanLuke15 took the simple assumption of atmosphere must be 5% or greater of surface. Threw that through a chord vs sagitta of a circle segment calculator. I also tried calculating the drag coefficient needed to achieve such decelerations in Mars atmosphere, and the numbers went nuts. craft with ballute 400x its surface area! Plasma temperatures of 60000k! nutz figures, i tell you! Dec 29, 2021 at 17:04
• Apollo was half as fast, and had a much longer way to break - being only a bit above escape velocity gives a nice curve to the flight path - but that won't happen at 5x escape velocity. Dec 29, 2021 at 17:08
• @asdfex Apollo has 1/5th the energy to scrub, and something like 6x the time to do it in. So, at a first ballpark estimate, 30x as "hard" as an Apollo reentry. Dec 29, 2021 at 17:19
• Note that it'd "only" take ~18 gravities to follow the curve of Mars at 25 km/s. You wouldn't be able to brake purely with the atmosphere, but a combination of atmospheric braking, negative aerodynamic lift, propulsive augmentation of said lift, and propulsive braking could result in a rough but survivable braking maneuver that's far more efficient than scrubbing 25 km/s propulsively. Certainly far more difficult and riskier to your health than just packing an additional month or two of consumables, though... Dec 29, 2021 at 18:55
• @FelixTritschler It was meant as an observation of how superficially complex questions can actually be quite simple with a bit of reduction. It wasn't a criticism of the question or the answer.
– J...
Dec 31, 2021 at 15:33