In an answer to a similar question, the Karman Line for Mars is stated as being at 125km compared to 122km for Earth. However, Mars has an atmospheric pressure about 1/10 that of Earth, its differences in mass, rotational period and gravitational acceleration not withstanding.

Why is it higher than for Earth?

EDIT: D'Ohhh!!! Just reread the original prior question again, as per @Nakedible. But I still can't see the justification for the 125km figure.

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    $\begingroup$ pretty sure it's gravity — Mars has ~0.4G, so it doesn't pack the atmosphere as close as Earth does. $\endgroup$
    – radex
    Commented Jan 24, 2016 at 7:38
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    $\begingroup$ Actually, the Kármán Line for Earth is about 100 km, not 122 km. $\endgroup$ Commented Sep 28, 2016 at 23:18

1 Answer 1


The Kármán line in Venus is around 250 km high, and in Mars about 80 km (reference). An answer on the question you have linked places it at 88 km on Mars. So I believe you have just misread the answers.

However, what is discussed is that entry interface altitude for Mars starts at 125 km, where as for Apollo program it was 122 km (400,000 ft). This is because even though the Kármán line is much lower on Mars, the lower gravity on Mars makes the scale height for Mars atmosphere much higher. This means that atmospheric drag starts to have an effect on spacecraft at around the same height as it does for earth.

I don't know if there is any specific rationale for those exact values (125 km vs 400,000 ft) of entry interface altitude though - I think they are just nice round numbers that are sufficiently above where the actual effects start.

A rough Mars atmosphere model can be seen here:

Mars Atmosphere Model

  • $\begingroup$ D'Ohhh!!! Just reread it again. But I still can't see the justification for the 125km figure. $\endgroup$ Commented Jan 24, 2016 at 7:53
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    $\begingroup$ 125 km was the entry interface used by Mars Science Laboratory for the landing of Curiosity. That is the reason it is quoted, but like said, I'm not sure if there is a specific reason for that exact value. $\endgroup$
    – Nakedible
    Commented Jan 24, 2016 at 7:56
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    $\begingroup$ Also note a significant diurnal effect on the Martian atmosphere. Its atmosphere (due to lower gravity and composition, being mostly carbon dioxide which has higher heat capacity than air in Earth's atmosphere) swells up a lot more on the days side compared to its night side, than it does on Earth. Then there's also typical entry geometry. Since Mars is in higher altitude heliocentric orbit than Earth, it will mostly be showing us its day side for direct line of sight communications during EDL. And entry angle will be steeper because of smaller radius (i.e. interface margin will be higher). $\endgroup$
    – TildalWave
    Commented Jan 24, 2016 at 8:13
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    $\begingroup$ Yes, 400,000 feet and 125 km were picked to be a) just high enough that the atmosphere is not a measurable factor in the flight path or dynamics, and b) to be nice round numbers. To be precise, the 125 km entry interface is not 125 km altitude over your landing site, but rather is always exactly 3522.2 km radius (distance from the center of Mars), which is 125 km above the Mars "reference" radius of 3397.2 km. This avoids having your entry interface depend on your landing site. $\endgroup$
    – Mark Adler
    Commented Jan 24, 2016 at 18:09
  • $\begingroup$ The above comment deserves a ton of upvotes as it is not just conjecture, but an answer to the rationale straight from the horse's mouth. No more guessing. $\endgroup$
    – Nakedible
    Commented Jan 24, 2016 at 21:29

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