Mars' atmosphere is only .088 psi. Is that enough to cause heat at high altitude? Shouldn't you be able to fly close to the surface at orbital speeds like on the moon? How does a parachute work with such little air?

Mars Exploration Rovers: Entering the Mars atmosphere:

Mars Exploration Rovers: Entering the Mars atmosphere October 01, 2002

The aeroshell protects the rover from fiery temperatures as it enters the Martian atmosphere in January, 2004.

Credit NASA

Wikipedia Atmosphere of Mars:

The atmosphere of Mars is the layer of gases surrounding Mars. It is primarily composed of carbon dioxide (95.32%), molecular nitrogen (2.6%) and argon (1.9%). It also contains trace levels of water vapor, oxygen, carbon monoxide, hydrogen and other noble gases. The atmosphere of Mars is much thinner than Earth's.

Average surface pressure 610 Pa (0.088 psi)

Carbon dioxide 95.32%

Nitrogen 2.6%

Argon 1.9%

Oxygen 0.174%

Carbon monoxide 0.0747%

Water vapor 0.03% (variable)

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    $\begingroup$ Edited your question, dumbass :) It's a bit more suited to our site now. I think it was a great question and belongs on this SE, but the way you asked it was a bit off-putting. Should be better now. $\endgroup$ Commented Jul 10, 2020 at 1:24
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    $\begingroup$ @AntonHengst I was about to say something about name-calling before I saw OP's username haha $\endgroup$
    – Josh Eller
    Commented Jul 10, 2020 at 14:06
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    $\begingroup$ Should be noted that most meteors are seen at 76-100 km, where the atmospheric pressure is under 1 Pa. $\endgroup$
    – jamesqf
    Commented Jul 10, 2020 at 16:27
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    $\begingroup$ If your only goal is to reach the surface, from orbital speed, like a ballistic missile warhead, then yes you need a less heat shielding on Mars than on Earth (but not none). But if your goal is to reach the surface and slow down for landing, then you need to dissipate all that energy. On Mars the atmosphere is thinner so the braking will be a bit different, but ultimately just as "hot". $\endgroup$
    – tomnexus
    Commented Jul 10, 2020 at 17:02
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    $\begingroup$ Just to note, density is the relevant atmospheric metric here, as opposed to pressure. Average atmospheric temperature won't matter when you're hitting it at 1 km/s -- the air you're displacing is going to feel much hotter. $\endgroup$
    – Sam
    Commented Jul 10, 2020 at 19:42

5 Answers 5


There is an atmosphere on Mars.

An atmosphere ~1% as thick as Earth's sounds like it ought to 'basically not count as an atmosphere' but 1% earth's atmosphere is still quite a lot of gas! The pressure of Mars's atmosphere is about ten million times larger than outer space.

I can't find it right now, but there's a quote about the Martian atmosphere, that says something like "Mars has just enough atmosphere to be a problem. It slows you down, but not enough to make landing safe".

Atmospheric reentry on Earth is considered to start taking place at the Karman line, which is an altitude of 100 km. The air pressure at that height is about a millionth that of sea level, and that's when you start seeing the effects of re-entry. Since Mars's ground level atmosphere is 10000 times thicker than that, there's no way of avoiding going through a significant amount of atmosphere on your way to land.

The craft carrying the Curiosity rover entered Mars atmosphere at 5.8 km/s. That's about Mach 17. Even a rarefied stream of gas will impart a huge amount of kinetic and thermal energy at that speed.

  • $\begingroup$ Do you mean this answer? space.stackexchange.com/a/12328/2981 . $\endgroup$
    – GdD
    Commented Jul 10, 2020 at 7:57
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    $\begingroup$ note: Atmospheric reentry on Earth is considered to start at the Karman line $\endgroup$ Commented Jul 10, 2020 at 11:04
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    $\begingroup$ @user253751 Though Mars has an equivalent line, at around 80km above the surface, compared to the 100km for Earth. Not that different actually. For comparison, Venus is around 250km. See this article. $\endgroup$ Commented Jul 10, 2020 at 14:00
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    $\begingroup$ @user253751 by who? Not by the Apollo or Space Shuttle programs. Entry interface was defined to be 400k feet. $\endgroup$ Commented Jul 10, 2020 at 14:41
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    $\begingroup$ @user253751, since it's possible to fly on Mars, you can compute a Karman line for it: it's the altitude at which the stall speed for a reasonable aircraft equals orbital velocity. Turns out it's about 80 km. $\endgroup$
    – Mark
    Commented Jul 11, 2020 at 1:51

An atmosphere does not need to be very thick to heat up a reentry vehicle significantly. And it should be noted that reentry vehicles entering the Earth's atmosphere get heated up at a pretty high altitude with thin air as well. The point of maximum heating is also not necessarily the same as the point of maximum acceleration.

At near-orbital speeds, two important things happen:

  1. Each air molecule hits at, well, near-orbital speed, and that kinetic energy gets turned into heat as the gas is compressed.

  2. Since it's moving at near-orbital speed, it sweeps through a lot of air in a short time. At 5.8 km/s, that's 5,800 cubic meters of air every second for a 1 square meter cross-section reentry vehicle.

Based on this graph on Wikipedia, it looks like most probes entering the atmosphere of Mars get down to around 50 km before slowing down much. At that altitude, the air has a density of just about $10^{-4}\space\mathrm{kg/m^3}$ (compare $1.2\space\mathrm{kg/m^3}$ for sea level at Earth). For a spacecraft that gets to that altitude at 5.8 km/s (which Curiosity approximately did, and Pathfinder was much faster), you are sweeping through air carrying a kinetic energy load of OVER NINE THOUSAND kilowatts per square meter. (A well-designed aeroshell will only allow a fraction of this energy to heat up the surface of the heat shield.) This one second of air masses more than half a kilogram.

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    $\begingroup$ "No way nine thousand!" $\endgroup$
    – Milind R
    Commented Jul 10, 2020 at 19:45
  • $\begingroup$ 9 Megawatt per sqm seems somewhat off, otherwise solid explanation. $\endgroup$ Commented Jul 11, 2020 at 11:42
  • $\begingroup$ @ZsoltSzilagyi Looks about right, actually. Kinetic energy of 0.58kg object at 5.8km/s is 9.76 MJ. $\endgroup$
    – March Ho
    Commented Jul 11, 2020 at 18:11
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    $\begingroup$ It's the kinetic energy of air swept by the reentry vehicle. Not the actual heat flux incurred. $\endgroup$
    – ikrase
    Commented Jul 11, 2020 at 19:25
  • $\begingroup$ Yepp, I stand corrected. Intuition just fails at these magnitudes. $\endgroup$ Commented Jul 12, 2020 at 23:49

Already a lot of good answers, so just one additional aspect to the explanation:

When orbiting Mars, the vehicle has a lot of kinetic energy that has to be reduced to zero before landing. Coming from a low orbit, you have to somehow slow down from 3.5 km/s (8000 mph) to zero.

Over non-atmosphere bodies like the moon, the only chance to slow down is firing the engines in opposite direction. Of course, you can do so over Mars as well. Using your engines, slow down "outside" of the atmosphere, and the begin your thrust-controlled descent into the atmosphere and onto the surface. But that costs a lot of precious fuel.

A more fuel-economic way is to let the atmosphere do the braking. The downside is that your kinetic energy gets converted to heat in that process. You can't avoid that, there's no atmospheric braking without heat production.

And there isn't much freedom to choose different descent paths. An idea that comes to one's mind might be to brake so slowly that the resulting heat gets spread over a long period of time, thus not producing so high a temperature that a heat shield is needed. But that would mean to slow down more and more while staying in higher parts of the atmosphere. Good idea, but it won't work, as with less than the orbital speed you'll inevitably start falling down from the sky - you can't stay up there as long as you like.

So, the NASA engineers decided to make use of the atmospheric braking instead of doing a completely thrust-controlled descent, and to incorporate a protection against the inevitable heat.

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    $\begingroup$ Another way of looking at things would be to say that one could design a streamlined spacecraft that would minimize the amount by which it heats up before impacting the planet surface, but it would then have to be insanely strong to survive the impact. $\endgroup$
    – supercat
    Commented Jul 10, 2020 at 19:29
  • $\begingroup$ The only way to slow down presently is rockets. With sufficient development you could land on a maglev track on the moon, or by capture by a vehicle riding the maglev track on the moon (which permits your periapsis to be above the terrain in case you make a mistake.) $\endgroup$ Commented Jul 11, 2020 at 2:34

The 'plasma blackout', where the spacecraft is covered in a blazing ball of fire and the heatshield plays the most important role, is between 90 and 55km altitude (300k-180k ft):

enter image description here

Pressure corresponding to these altitudes on Earth is between 0.1pa (0.001 millibar) and 50pa (0.5 millibar):

enter image description here

On Mars, these pressures occur respectively on 80 and 23km altitude - carbon dioxide atmosphere playing a role in different layout of pressure layers, but that still means the pressures where the heatshield is important lie well above Mars surface and a heatshield works just as well as on Earth, braking the spacecraft from space travel speeds to subsonic.

enter image description here

It's the parachutes that do really lousy job , requiring rocket skycranes or at least inflatable airbags on the payload for landing.

Graph sources: plasma Earth Mars


There is an atmosphere, as your screenshots point out. In fact, NASA is currently operating the MAVEN spacecraft (Mars Atmosphere and Volatile Evolution Mission) and ESA the Trace Gas Orbiter (TGO), both of which study Mars' atmosphere.

It's often said of Mars' atmosphere that "it's just thick enough to burn your spacecraft, but just thin enough to barely slow you down." Therefore, probes must use active systems to slow down the spacecraft before it reaches the surface. These range from retrorockets (like on Curiosity and Perseverance) to airbags (like on Spirit and Opportunity).

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    $\begingroup$ In actuality, it does a lot more than "barely slow you down". The problem is that while they can easily shed the great majority of their speed by using the atmosphere, craft above a certain size still have a very hard time decelerating to less than the speed of sound, which was considered necessary for starting up landing rockets. That's why SpaceX's development of supersonic retropropulsion with their Falcon 9 boosters is such a big deal. $\endgroup$ Commented Jul 10, 2020 at 1:30
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    $\begingroup$ @ChristopherJamesHuff Why is supersonic retropropulsion difficult? $\endgroup$
    – ikrase
    Commented Jul 10, 2020 at 6:19
  • $\begingroup$ @ikrase that's really another question, but I think it has to do with igniting the rockets while effectively in a wind blowing faster than mach 1 into the nozzles. Makes lighting a match in a hurricane seem routine. $\endgroup$ Commented Jul 10, 2020 at 7:29
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    $\begingroup$ @ikrase well...judging from SpaceX's experience with it (the reentry burn being a part of their recovery process that's worked perfectly from the first attempt), the biggest obstacle was people saying "we've never done that before". It's hard to model or ground test, but it's honestly an indication of the stagnation in aerospace that nobody bothered to attempt it before SpaceX. Look at the ExoMars parachute issues for an example of why someone should have been doing this research... $\endgroup$ Commented Jul 10, 2020 at 13:22
  • $\begingroup$ @ChristopherJamesHuff I agree, I don't see why subersonic retropropulsion should be difficult. Starting an engine at right angles to the wind could be difficult. But once you've orientated the engine facing into the wind, there won't be a hurricane blowing across the nozzle. There will be a backpressure from the wind but flow at the combustion chamber throat will be stagnant. $\endgroup$ Commented Jul 12, 2020 at 1:56

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