# At what g is terminal velocity not terminal?

How weak would gravity need to be in order for a human to reliably survive the terminal velocity of falling through air?

(Context: watching scifi on a space station with a variety of artificial gravities, it occurred to me that medium-strength gravity would have some advantages; also I note that insects already seem to have this luxury on earth, with their small masses and high air resistances...)

• I imagine the answer is low enough to make walking impossible, but I don't know for sure. – Nathaniel Jun 15 '12 at 17:59
• Related question that I thought .. what is the biggest size animal that can survive a terminal velocity drops. Ants can do it, humans can't. Where is the transition? – John Alexiou Jun 16 '12 at 1:58

References for the terminal falling velocity of a skydiver give numbers in the range of $54 m/s$ to $76 m/s$. I expect that the real range is even larger, since it's strongly affected by the orientation and body position of the skydiver.

For normal atmospheric drag on this scale, we have the fairly good approximation that force is proportional to velocity squared, $F_{air} \propto v^2$. For normal terminal falling conditions, the upward air force is exactly equal to gravity, and the air force is a function of velocity exclusively because we're assuming the same air composition for now.

You ask about a fall that is "reliably survivable", I should note that this would still NOT be a comfortable fall. In low gravity environments, you could expect some gnarly bounces to top it all off. I don't have a good reference for this, but I would anecdotally put it in the neighborhood of a $40 ft$ fall, which would equate to about $15 m/s$.

I will refer to the air force during a normal terminal velocity fall on Earth as $F_{air}$ and introduce $F_{air}'$ for the force on the new planet. I'll do some simple equivalencies to get an answer for the new gravity needed.

$$m g = F_{air} = (\text{const}) v^2$$

$$\frac{F_{air}}{F_{air}'} = \frac{v^2}{v'^2} = \frac{g}{g'}$$

$$\frac{v}{v'} = \frac{60 m/s}{15 m/s}$$

$$\frac{g}{g'} = \left(\frac{60}{15}\right)^2 = 16$$

So my answer is quite simply that gravity would have to be 1/16th as strong as it is on Earth, or $0.6 m/s$. Are there any bodies in the solar system quite like this? Wikipedia is helpful here. Several bodies come close, like Pluto, Eris, or Triton, but none of them have much of an atmosphere. It is fun to think about, but I doubt that an atmosphere of such a high density with such a low gravity will be found in our local celestial neighborhood.

Walking would be difficult if such a planet existed, but not impossible. The moon is 1/3rd the gravity of Earth, so this hypothetical planet would be roughly 5 times less gravity than the moon. It would be very bouncy, but still very different from zero-gravity.

• Your figure of $0.6m/s^2$ doesn't match the moons you've selected; off by a factor of 10. Pluto would be a better example. – Mark Beadles Jun 15 '12 at 20:13
• @MarkBeadles This was apparently a careless mistake where I was reading from the column with the wrong units. Edited to hopefully be correct now. – Alan Rominger Jun 15 '12 at 20:35
• Walking on the moon is already quite difficult. You have to break into a run if you want to get anywhere - see clavius.org/gravleap.html . You could probably get around in 1/16th g, but not by walking in any normal gait. (I only mention it because the OP mentioned choosing the strength of gravity on a space station - I would have thought ease of walking would be a major consideration there.) – Nathaniel Jun 15 '12 at 21:28
• @Nathaniel You're entirely right. I tried to avoid saying it would be easy. For low gravity environments less than 1/10th Earth gravity, it might feel more like zero gravity than a gravity environment, except for the fact that objects still collect on the surface. You might not be able to do much resembling normal walking – Alan Rominger Jun 17 '12 at 15:24
• Also, note that the air pressure is going to be dependent on the size of $g$ (pressure is determined by the weight of overhead air), so a smaller $g$ will imply less pressure, thus, less air resistance, so this is probably an overestimate. – Jerry Schirmer Sep 25 '12 at 17:45

The smallest body that we know has an atmosphere is Titan, which has about 1/7 of Earth's surface gravity ($1.4\ m/s^2$) but an atmospheric pressure of 1.45 Earth's ($146.7\ kPa$). Using $pV=nRT$ where $T=95K$ and a mean molar mass of Titan's atmosphere being $28.6\ g/mol$ allows us to calculate an atmospheric density of $5.87 kg/m^3$, greater than Earth's.

So since there is less gravity but more atmosphere, I decided to look at the specific question of whether a terminal velocity fall on Titan is survivable. The answer was enlightening.

Using the formula for terminal velocity $$V_t = \sqrt{\frac{2mg}{\rho AC_d}}$$

where (using reasonable estimates for the human coefficients): $$m_{human} = 75\ kg\\ g_{Titan}=1.4\ m/s^2\\ C_{d\ human}= 1.0\\ \rho_{Titan}= 5.87\ kg/m^3 \\ A_{human}= 0.75\ m^2$$

we get the interestingly low figure of

$$6.9\ m/s$$

Such a fall would be survivable.

Of course, the extreme cold and unbreathable atmosphere would not be, but at least the fall wouldn't kill you.

• According to this formula, Venus has even lower terminal velocity of 5.5 m/s. – MadBender Aug 9 '13 at 18:15
• Of course, the air pressure on Venus would crush you on the way down. But at least you wouldn't splat too much after hitting the ground. – NeutronStar Feb 7 '14 at 1:47