How would I calculate the terminal velocity of Mars?

What is the terminal velocity of a balloon entering Mars' atmosphere?

Would a balloon pop if dropped from space?

  • 1
    $\begingroup$ The atmospheric pressure of Mars is about 0.636 kPa, that is 0.63 % of atmospheric pressure on Earth. At a height of about 35 km, the pressure is the same at Mars ground. We should be able to build a glider for 35 km height first. $\endgroup$
    – Uwe
    Sep 23, 2018 at 15:17
  • 2
    $\begingroup$ Felix Baumgartner exceeded sound speed in his jump from 39 km height. So the terminal velocity at Mars surface should be supersonic too. $\endgroup$
    – Heopps
    Sep 23, 2018 at 16:26
  • $\begingroup$ Terminal velocity applies to falling objects. Is this balloon falling? $\endgroup$ Sep 23, 2018 at 17:26
  • $\begingroup$ @OrganicMarble If just "dropped" it would have this nasa.gov/larc/expert-panel-assesses-inflatable-spacecraft-tech on the bottom of the bottom possibly a streamer to keep it down right once terminal velocity has be reached and cooled to temp expand the balloon to a buoyant size. The balloon would fit inside this vacuum. ichef.bbci.co.uk/wwfeatures/wm/live/1280_640/images/live/p0/1t/… $\endgroup$
    – Muze
    Sep 23, 2018 at 18:00
  • $\begingroup$ @uhoh updated question $\endgroup$
    – Muze
    Sep 25, 2018 at 19:32

1 Answer 1


How to calculate terminal velocity in general:

$$ V_t = \sqrt\frac{2W}{\rho C_d A} $$


$V_t$ = terminal velocity
$W$ = weight (mass times local gravity)
$C_d$ = the coefficient of drag of the object
$\rho$ = atmospheric density
$A$ = frontal area of the object

Comparing Mars to Earth, weight is $\approx 0.38$ and atmospheric density is $\approx 0.0167$ that of Earth, so terminal velocity is $\sqrt{23} = 4.8$ times faster on Mars. (Assuming $C_d$ doesn't change, which it would, but this is close.)

  • 3
    $\begingroup$ A reasonable choice for Cd assuming a spherical blimpish structure would be 0.5, otherwise there are various tables a search engine will find. Strange things happen to Cd above mach 1 but this will at least get an approximation. $\endgroup$ Sep 23, 2018 at 23:12
  • $\begingroup$ @GremlinWranger Subsonic Cd can be worked up to about flat-plate drag with Cd=1. As "Cd is free" (more or less) it seems a shame to not get about another 30% decrease in terminal velocity that Cd=1 offers compared to Cd=0.5. $\endgroup$ Apr 23, 2020 at 11:06
  • $\begingroup$ @Hobbes I was delighted to see someone produce the terminal velocity equation (aka drag equation rearranged). That has been great fun over the years in calculating a surprising variety of things. || THey do not make it intuitively obvious on that page, but you can derive the equation by calculating the energy conveyed to the mass of gas displaced per unit time and accelerated to Vfall. That provides the correct result for Cd = 1. $\endgroup$ Apr 23, 2020 at 11:10

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