In the book "The Gods Themselves", Isaac Asimov imagines humans living on the Moon would enjoy much greater freedom of movement on the Moon.

Each pair went up in unison; each pair rose and fell in a more complicated pattern. One pair kicked off simultaneously to cross the tube in a low parabola, convex upward, each reaching the handhold the other had abandoned, and somehow skimming past each other in mid-air without touching. That evoked louder applause.

The Earthman said, "I suspect I lack the experience to appreciate the finer points of skill. Are these all native Lunarites?"

"They have to be," said Selene. "The gymnasium is open to all Lunar citizens and some immigrants are fairly good, considering. For this kind of virtuosity, however, you must depend on babies that are conceived and born here. They have the proper physical adaptation, at least more than native Earthmen have, and they get the proper childhood training. Most of these performers are under eighteen."

"I imagine it's dangerous, even at Moon-gravity levels."

"Broken bones aren't very uncommon. I don't think there's been an actual death, but there's been at least one case of broken spine and paralysis. That was a terrible accident; I was actually watching— Oh, wait now; we're going to have the ad libs now."

This raises the question: How far could one fall on the moon before sustaining injuries? In an interview with a biomedical engineer, it is said that 4,000 Newtons of force would break a femur. Could a 170 lb (77 kg) person jump high enough on the Moon to sustain that much force on landing? How high would that person have to fall to experience that much force?

Terminal velocity during free fall on Earth is reported to be about 110 mph (49 m/s). That would most probably result in a fatal fall. How high would a person on the Moon have to fall to reach that on the Moon?

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    $\begingroup$ Even on earth, there is no simple answer to how far you can fall without injury. People have died in falls from a couple of meters up. People have survived falls from much higher. Your chances of injury and death depend very sensitively on how you hit. $\endgroup$
    – user687
    Commented Jul 16, 2017 at 16:46
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    $\begingroup$ Looking at this from a conservation of energy standpoint, assuming I'm jumping on flat terrain, all the energy of the fall has to be supplied by me jumping in the first place. If I'm landing with X newtons of force, I had to push off the ground with that same force. So, if I've somehow jumped high enough to hit the ground with 4000 newtons of force, I've already put that femur-breaking 4000 newtons through my legs by jumping. That's just in response to the quoted passage, though. Not the question. $\endgroup$
    – UIDAlexD
    Commented Feb 14, 2018 at 20:01
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    $\begingroup$ If a person jumps so high that its femur would break on landing, it would break during the first phase of jump anyway. The force necessary to accelerate before the jump is equal to the force necessary to decelerate if we assume no losses. If we accept losses of energy, the force for acceleration is even higher. $\endgroup$
    – Uwe
    Commented Feb 14, 2018 at 22:44
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    $\begingroup$ @Uwe only if we have the femur taking the force in the same direction as when jumping. It's quite conceivable that someone could jump - putting force into their legs in a manner that the bone can take (compression across the length) but then fall in a different manner (e.g. coming down on top of their leg but after their foot slipped), so that the majority of their weight is applied to the side of the bone where it is less able to withstand the force. $\endgroup$
    – Baldrickk
    Commented Feb 15, 2018 at 15:12
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    $\begingroup$ Amazing. I am here because I just read that exact passage in The Gods Themselves and that prompted me to search how far I could safely fall on The Moon. I love Stack Exchanges. $\endgroup$ Commented Jul 11, 2020 at 8:47

10 Answers 10


The simplest would be defining some arbitrary impact velocity that is at the limit of being fatal, and we then consider everything else (surface properties, subject's physique,...) except gravitational acceleration constant. We can also neglect air resistance to make it simpler, since we're more interested in a safe height to jump off on the Moon, than that on the Earth. Plus air resistance at such small heights wouldn't change our results much anyway.

Let's, for the sake of argument, assume we have a crash test dummy that won't break into pieces, if it impacts the ground no faster than 60 km/h. At Earth's mean sea-level surface gravitational acceleration (9.80665 m/s2) that comes out as a jump at a height of 14.16 m, and the free fall duration of 1.7 s before impact. I'm cheating by using an online free fall and constant acceleration calculator, but the maths for a free fall without air resistance go as follows:

$$v(t)=-gt+v_{0}$$ $$y(t)=-\frac{1}{2}gt^2+v_{0}t+y_0$$


  • $v_{0}$ is the initial velocity (m/s).
  • $v(t)$ is the vertical velocity with respect to time (m/s).
  • $y_0$ is the initial altitude (m).
  • $y(t)$ is the altitude with respect to time (m).
  • $t$ is time elapsed (s).
  • $g$ is the acceleration due to gravity (9.81 m/s2 near the surface of the earth).

We can derive everything else we need from these two equations, let me know if you require a more detailed answer in this part. Moving on, now let's see at which height of a jump we reach 60 km/h impact on the surface of the Moon. Moon's average surface gravity acceleration is 1.622 m/s2. Plugging that into our online constant acceleration calculator (inserting acceleration, start velocity of 0 km/h and end velocity of 60 km/h), we get:

  • Jump from a height of 85.63 m
  • Free fall time of 10.28 s

Recalculation for 110 mph (177 km/h) impact velocity post edit of the question:

  • Jump from a height of 745.41 m (2,445.58 ft) on the Moon, 123.29 m (404.49 ft) on Earth
  • Free fall time of 30.32 seconds on the Moon, 5.01 seconds on Earth

And for 116 mph (187 km/h), equaling 4,000 N force on impact of a 170 lbs (77.11 kg) object:

  • Jump from a height of 829.49 m (2,721.42 ft) on the Moon, 137.2 m (450.12 ft) on Earth
  • Free fall time of 31.98 seconds on the Moon, 5.29 seconds on Earth

Edit to add: Both these last impact velocities (110 and 116 mph) are over 80.85 mph (130.12 km/h) that I calculate (see comments for details, thanks @LorenPechtel!) is the terminal velocity on the Moon for a freeflying skydiver, if all other conditions remain constant (mass of the skydiver, drag coefficient, fluid density, projected area of the falling object) both on the Earth and on the Moon. I.e. this would mean equal atmospheric conditions, no spacesuit, and for a skydiver whose terminal velocity on Earth is 320 km/h (200 mph or 90 m/s), as given in Wikipedia examples of terminal velocity.

So if we accept that a human can survive an impact at given velocities (110 and 116 mph), Moon's gravity isn't sufficient to counteract atmospheric drag of 1 atmospheric pressure to such speeds to kill you on impact. I.e. the maximum survivable altitude to jump off in such conditions would be, theoretically, infinite. In reality, you'd then have to deal with the heat released during atmospheric reentry beyond terminal velocity, which becomes a different question altogether. And it would of course be unfeasible, the Moon doesn't have sufficient gravity and magnetic field to sustain Earth-like atmospheric pressure.

Solution to calculate lunar terminal velocity, with a known terminal velocity of the same object and same atmospheric density on Earth:

$$v_{(M)} = 0.4067 \cdot v_{(E)}$$

Where $v_{(E)}$ is terminal velocity for Earth's gravity. Derived from:

$$v= \sqrt{\frac{2mg}{\rho A C_d }}, \ x = \frac{2m}{\rho A C_d }$$

Where $x$ denotes our constant mass of the object, drag coefficient, fluid density, and projected area of the falling object. Thus:

$${v_{(M)}}^2 = \frac{g_{(M)}{v_{(E)}}^2}{g_{(E)}} \Rightarrow {v_{(M)}}^2 = \frac{1.622}{9.80665} \cdot {v_{(E)}}^2 \Rightarrow v_{(M)} = \sqrt{0.1654 \cdot {v_{(E)}}^2}$$

For a belly-to-earth (i.e., face down) free-fall position, again using Wikipedia quoted values, in Earth's gravity achievable terminal velocity is only 195 km/h (122 mph or 54 m/s). On the Moon, using our conversion value of 0.4067, this amounts to 79.3 km/h (49.6 mph or 22 m/s). I would still argue that this is not survivable due to lack of vegetation and unweathered (sharp) terrain on the Moon, but it is a food for thought. For a 170 lbs, this "only" amounts to an impact force of 1,710 Newtons. With some luck and landing into a deep pocket of lunar dust and no larger boulders in the way, it might be survivable. Still, that's in a face down position, so most likely not.

Another thought though, that our subjects are acrobats that would probably achieve jumps of such heights with their own power. So they would first have to work against the same gravitational acceleration that will later try to kill them on impact. Point being, that if you can't jump on Earth to heights that would be fatal to land from, you won't be able to do that on the Moon either. So the frequency of injuries and fatalities should be much the same, assuming all other conditions are equal, and no severe bone density loss due to staying in roughly 1/6-th the Earth's gravity. Of course, bone density loss usually comes with muscle atrophy, so the ability to jump as high might decrease, too.

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    $\begingroup$ Great last point. That should be included in basic training for lunar colonists. It seems likely that the real danger would be damaging a space suit long before sustaining physical injury from a blunt force landing. $\endgroup$
    – TK-421
    Commented Jun 3, 2014 at 3:40
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    $\begingroup$ The last point is the most important one, the amount of upward force you exert when jumping, is the amount of of force you will feel when you hit the ground again, and it is the same regardless of the gravitational acceleration. Gravity just determines how long you will be up in the air, not how hard you will hit (in this case). $\endgroup$
    – BenVlodgi
    Commented Jun 3, 2014 at 12:15
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    $\begingroup$ Update: If I'm reading this: en.wikipedia.org/wiki/Stokes%27_law right terminal velocity should scale linearly with gravity and this: physics.stackexchange.com/questions/30157/… seems to say with the square root. Anyone with more math knowledge able to help? $\endgroup$ Commented Jun 5, 2014 at 3:52
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    $\begingroup$ Terminal velocity scales with square root of g. See en.wikipedia.org/wiki/Terminal_velocity#Physics Also with inverse square of density. $\endgroup$
    – HopDavid
    Commented Jun 6, 2014 at 15:39
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    $\begingroup$ @TildalWave For long falls I was picturing something like a cover over a crater to make a shirtsleeve "outdoor" environment for a city. Place a clear cover over the city, put 33 feet of water on top of that and then another cover to keep it from evaporating. You don't need super-strength materials, it's supported by the air pressure. A UV shield and radiation shield in one. An inspector falling from the scaffolding would have a considerable fall in atmosphere. $\endgroup$ Commented Jun 6, 2014 at 23:26

Going away from the scientific answers, and more towards the psychological answers, I'd like to approach how one makes a mistake bad enough to break a bone. The first step involves getting enough energy to break the bone with a fall in the first place. The body should naturally do a reasonable job of balancing the muscle strength needed to do tasks against the bone strength needed to oppose the muscles. This is why you see little fear of broken bones on the ISS -- their muscles and bones atrophy in harmony. However, when they come back to earth, their muscles are forced to push themselves to the limit to fight gravity, and the bones haven't been given an opportunity to catch up. Think similar to forgetting to stretch before physical activity.

So to get the energy to break your bones, you'd probably have to rely on something other than jumping. A tall fall could do it, just like on earth (only taller). This is where the psychology comes in.

If you have not had enough time to get used to low gravity rules, your brain may not realize it is in trouble until too late. You may get cocky, and think you can take the 1km fall because the 900m one wasn't so bad. The brain has a lot of safeguards to prevent us from hurting ourselves.

The most likely culprit would be mass vs weight. If someone had not gotten used to low gravity (such as an Earthlander visiting the Moon), they may drastically underestimate the inertia of an object because it "doesn't weigh much." This may cause them to carelessly put themselves in a position where they must stop a high-mass object with their body, and fail to realize how much trouble they are in until it is too late.

This has happened at least once in our spacewalks. An engineer designed a procedure that involved powering down a spinning dish and then stopping it to do maintenance. The earthbound engineer accidentally assumed that the dish would have little inertia, because they were weightless. The spacewalker found it was virtually impossible to stop the disk. They ended up finding a smooth ring on the dish, and appyling friction for several minutes to get the rotational rate down to where the spacewalker had the muscle (and bone) to actually stop the rotational inertia of the disk.


Without going into detailed calculations, I recall that the gravitational acceleration on the Moon's surface is about $\frac16$ of what it is on the Earth's surface. Now by the simple formula that work is force times distance, for an object to acquire the same kinetic energy (and therefore the same velocity) as falling from a height $h$ on Earth, it should fall from a height about $6h$ on the Moon. (This is ignoring friction, which is certainly valid on the Moon, and valid for moderate (yet fatal) heights on Earth.)

The damage done on hitting the ground is mostly a function of the velocity of impact, so whatever height makes you uncomfortable on the Earth, six times that heigh should make you uncomfortable on the Moon. You will have longer to prepare for impact on the Moon; in fact six times as long, since with the final velocities being equal, so are the average velocities (half the final velocity when starting stationary), and falling from height $6h$ on the Moon will take $6$ times as long as falling from height $h$ on Earth.

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    $\begingroup$ This is the best answer. The accepted answer unnecessarily puffs up this trivial calculation into many lines of math. $\endgroup$
    – user687
    Commented Jul 16, 2017 at 16:43

I'm starting to sound like a broken record, but I agree with Tildal's post (yet again). There is one more thing I would add that concerns an Earth native who recently travels to the Moon.

Jumping ability comes from strength in muscles and tendons, while bones support weight. It is possible to jump on Earth and sprain an ankle and such if one jumps too high and/or lands incorrectly.

Good landings come from apropriate training, coordination, etc. As a counter point, one who just came from Earth's higher grav might be able to jump very high on natural strength alone and not be able to land as easily as they jumped. The more dense bones might be able to take the force, but I am concerned with the ability of the tissues to absorb the shock and take the damage.

When you combine the speeds that Tildal writes about with a solid ground and number of factors could contribute to injury. It could be possible that an untrained individual may not be able to react quickly enough to the increase in speed or not be able to accurately estimate the distance to the ground (much like when jumping out of a plane on the Earth, the trees come up rather quickly). Any number of issues could arrise that could cause serious damage. I don't see people dying per se, but with speed and a solid ground, one could hurt themselves much more easily than they would on Earth with a similar jump.

In relation to the book (great book btw), these acrobats are trained and Lunar natives or at least they have been on the Moon long enough to become skilled with movement with the lower gravity. It is possible (with the aid of some suspension of disbelief) that they are able to react more quickly, and perform movements that help mitigate or disperse any force applied, much like a parachute landing fall (plf) can save someone from falling a great height by rolling the force in a different direction.

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    $\begingroup$ I like assertion about estimating distance. With nearly no atmosphere on the Moon (what's usually referred to as an exosphere since the few particles per cubic centimeter don't behave like normal gas does), there isn't any atmospheric scattering of incident light and even on the Apollo photographs we can see the distant ridges as sharply as features closer to the camera. And because of the same reason and lack of shadow gradient, it's sometimes difficult to tell, if what you're landing on is a dent or a bump. $\endgroup$
    – TildalWave
    Commented Jun 4, 2014 at 9:14
  • $\begingroup$ A jumper will return to a level surface with the same velocity he departed. A jumper's muscles still exert the same newtons over the same period of time, so there's no increased speed. $\endgroup$
    – HopDavid
    Commented Jun 6, 2014 at 15:42
  • $\begingroup$ The force exerted by muscles designed for full gravity will generate a higher acceleration when performed at the Moon's 1/6th gravity. This may be incorrect, but the speed at which you jump will be near 6x faster. If someone had a vertical jump acceleration of 4m/s on Earth than their equivalent measure on the Moon would be 24 m/s. As you said the landing velocity will be the same as the take off. As I've used a parachute with around 7 m/s and knowing that the ground comes up rather fast, I'd assume that 24 m/s might not be easy to react to. At least that is my interpretation. $\endgroup$
    – Alexinawe
    Commented Jun 6, 2014 at 16:06
  • $\begingroup$ @Alexinawe If a jumper lies in a wagon with his feet against a vertical, gravitational acceleration towards the wall is zero. Using your model that jumping speed is inversely to gravitational acceleration, the jumper should be able spring away from the wall at infinite speed. But the speed is limited by how fast the legs can extend themselves. It takes time for muscles to contract. $\endgroup$
    – HopDavid
    Commented Feb 24, 2015 at 0:26
  • $\begingroup$ @HopDavid: make it an ice rink, since friction against the floor of the wagon might seem "like" an opposing gravitational force to someone whose intuition is already leading them astray as regards Earth vs. Moon. Or the wall of the ISS. $\endgroup$ Commented Feb 24, 2015 at 0:53

First of all, let's ask how far does a fall from Earth have to be to have injury? For simplicity, I'm going to assume a health middle aged adult, the typical range for astronauts. Looking around, the height of serious injury on the Earth appears to be 7 m, or something relatively close to that. In fact, the height might be even lower.

I'm also going to assume that the person was well conditioned. We know that bone mass is lost in zero gravity, and I'm going to assume that the person falling still have an Earth bone mass. As bone mass is lost, the chance of injury increases dramatically.

So, a 7 m fall from Earth will end up with a velocity of about 11.7 m/s. What distance is required to have that speed on the moon? The answer is about 43 meters. Of course, that's the serious injury level, and in fact the fall distance might well be less than that, as you lose bone mass from being in low gravity for so long. But this should be a good starting point at least.


How far could one fall on the moon before sustaining injuries?

As on Earth, it really depends how you land. High-jumpers land on crash mats, therefore it seems reasonable to suppose that on Earth it's possible to jump high enough to sustain injuries on landing, anyway if you choose to land on the back of your shoulders and head. I certainly would not choose to leap head-first at the ground, I reckon that would mess me up ;-)

As others have pointed out, the strength of gravity doesn't make a difference here, you land with whatever speed you can jump at, so the same would apply on the Moon. You can't jump so high that you'll injure yourself if you land well, because humans are well-equipped to jump and land. Our legs can handle take-off speed in reverse. Any of our potential ancestors whose legs couldn't do that, probably broke their legs early and never reproduced...

It might be harder to jump into the air and land on your feet on the Moon than it is on Earth. Aside from being unfamiliar with low gravity, the jump takes longer (6 times longer), so you have to jump more accurately to avoid rotating significantly in the air. So take it a bit easy before you go bounding around the place, your risk of landing on your head is increased compared with Earth.

What about heights we can't jump to? On earth, a 4m fall onto something hard is pretty nasty if you don't land properly, and it's difficult-ish to land properly. This is why you don't jump out of an upstairs window onto a concrete drive unless there's something even worse in the house, like a fire. Since the Moon has 1/6 the gravity, and energy is equal to force times distance, you'd hit the ground after a 24m drop with the same kinetic energy (and hence the same velocity). This should therefore be considered a dangerous height, just like upstairs windows are dangerous.

The breaking force for the femur, though, is liable to be achieved through unfortunate landings rather than sheer speed. People break legs falling off swings or ladders, never mind upstairs windows. The issue is what's the maximum force put through the bone at any instant during the landing (and the angle the force is applied: bones are stronger in compression than they are in torsion). A landing will bring you to a halt, and a good landing does so over as long a period as possible, with as close as possible to constant force, so as to get the max force as low as possible.

The actual process of landing with a given velocity is roughly the same on the Moon as on the Earth. The hardness of the surface, the angle and posture you hit it, and your ability to bend yourself in order to prolong the landing, all contribute. A bulky vacuum suit might make it harder to land "well" on the Moon than it is on Earth, especially if it restricts your flexibility at the knees and hips.

49 m/s

That's easy in a vacuum - velocity of 49m/s requires kinetic energy ($1/2 m v^2$) of 1200 Joules per kilogram, which requires a height of 122m on Earth (where $g$ is about 9.8, although there's nowhere on earth you can fall that far in a vacuum), and 6 times that on the Moon, 730m or so.

Note that on Earth you can increase terminal velocity by falling head-first or feet-first, the figure you give is for a skydiver pose. In a vacuum there is no terminal velocity and your posture is irrelevant.


Although this question has many answers already, I thought I'd add a more general answer

How far could a human fall in a pressurised environment on various solar system bodies?

I'm imagining that there are multi-story habitats on various bodies in the solar system, all pressurised to 1 atm. I'm also imagining that these habitats have a 'lift shaft' of sorts that people can use to get from higher to lower floors. So the assumptions include a deliberate fall and a hard surface for landing. What then, are the limits to such a fall shaft?

First, let's start with what falls you can survive with and without injury. As many others have pointed out in many places, people can stumble and fall a foot and die, and there are examples of skydivers with failed parachutes surviving a landing. Nevertheless we can make a qualitative scale of the effect on the human body of a typical fall.

Apocryphal and general information exists on which distances people can fall and survive, also people generally understand how far they can fall on earth and not get hurt, whereas the velocity necessary to hurt is less clear. By taking distance and running it through the formula $v = \sqrt{2dg}$, we can find the velocity necessary to inflict injury.

This reddit discussion suggests an upper fall limit of about 10-12 feet (3-3.6 metres) for a parkour practitioner landing without rolling, and 16-21 feet (4.8-6.3 metres) with rolling. I imagine a fit but not necessarily trained human could comfortably fall 1-2 metres.

This stackexchange thread suggests an almost certainly fatal range around 9-12 metres.

From this information, I'm going to settle on some (somewhat arbitrary) numbers.

  • Highest comfortable fall: 2 metres (6.2 m/s)
  • Highest possible fall without injury: 3.6 metres (8.4 m/s)
  • Reliably fatal fall: 10 metres (14 m/s)

falling velocity vs deathiness[3]

How far can you fall on other bodies? / Which bodies can you fall indefinitely on?

There's an easy maths substitution to make. The terminal velocity for a skydiver on earth (belly down) is $$v= \sqrt{\frac{2mg_e}{\rho A C_d }} \approx 55 ms^{-1}$$ and since it's only the surface gravity that's changing between different solar system bodies, the terminal velocity $v_t$ can be calculated by $$v_t = 55\sqrt{\frac{g_b}{g_e}}$$

Where $g_b$ is the surface gravity of the solar system body and $g_e$ is the surface gravity of earth.

Assuming velocity increases linearly with time until terminal velocity is reached, here is the info for various solar system bodies.

table of how far you can fall without dying on various solar system bodies[4]

Some takeaways: Falling long distances is surprisingly lethal on all planets and most of the larger moons. This is due to how slowly the $\sqrt{g}$ term decreases. For objects larger than Enceladus, my fantasies about an unlimited-height empty lift shaft for falling are shattered.

The comfortable fall distance on the moon (12 m) feels surprisingly low, although 12 metres tends to look like a lot more when you're standing at the top.

If you put padding on the landing zone, you could probably get away with arbitrarily high fall shafts on Ceres, Rhea and Vesta

  • $\begingroup$ Excellent addition! Wishing I could upvote more than once! $\endgroup$
    – TK-421
    Commented Nov 8, 2019 at 5:07
  • $\begingroup$ A thought has occurred to me. We would not be good at judging survivable falling distances on other planets. At the top of a ledge on earth, we are very good at distinguishing between a 2 metre drop and a 6 metre drop. Between an easy 12 metre drop on the moon and a potentially serious 20 metre fall? I doubt we could judge it as well. $\endgroup$
    – Ingolifs
    Commented Mar 15, 2020 at 6:23

I would like to add that there is no real definite "fatal" velocity at the point of impact.

It is much easier to suffer a fatal fall resulting from damaged life-support equipment in a spacesuit on the Moon than it would be to suffer a fall which was fatal due to injuries sustained from the fall itself.

In most cases of extreme impact like this, the mechanism of death is usually not the force of the impact itself, but the subsequent inflammation or hemorrhaging afterwards crushing vital organs such as the brain or, if the injury causes a laceration or avulsion, exsanguination (fatal blood loss) may be the mechanism of death instead. Since the mechanism of death is usually inflammation or hemorrhaging, it is difficult to determine exactly what impact velocities will yield fatal levels of the aforementioned simply because different individuals react to the same injuries differently. What would be a survivable injury for one person may not be survivable for someone else, and sometimes freak incidents occur which defy traditional explanation. An intoxicated individual is, according to an NCBI-sponsored study[1], more likely to die in most cases in an injury except in cases of brain damage.

For example, both skydiver Brad Guy and his instructor survived a 14,000-foot fall with a parachute that did not deploy properly, despite impacting a lake in a golf-course (water is highly incompressible and often acts as a solid when struck by an object traveling at high velocities.)

However, elderly people, children, or those with unfortunate conditions such as aneurisms or osteoporosis may not survive the impact associated with even a tumble from their height, much less a tumbling fall from 14,000 feet.

The damage from a fall also depends on how the body impacts the ground[2]: landing flat on one's back onto a flat surface is typically more survivable than landing at an angle simply because the force is evenly applied to a large surface on the body, thus there pressure applied to any one particular area of that body is less, and there are no compressional forces applied down the length of the spine which would result in complications such as a neck or lower-back compression fracture.

Landing on one's front is usually less preferable than landing on one's back simply because there are two prominent extrusions on the front of the body which there are not on the back of the body and which will cause the force to be applied unevenly: these are the facial structures and ribcage which, when damaged, have the risk of damaging the body's respiratory apparatuses such as the trachea, lungs (through puncturing from the ribcage) or jaw/nose damage leading to aspiration of blood and subsequent choking (there are martial artistry techniques which avoid landing on the chest by absorbing the force from moderate to large falls on the forearms and feet instead; I would not attempt a front-fall from 14,000 feet).

Landing on one's head is often injurious and has a more serious risk of injury from heights which most people consider insignificant, not only because of the threat of head injury or cranial fracture, but also because the pressure of the fall is being applied in a manner that compresses the spine, which can lead to spinal damage or even a broken neck. Side-impact falls are also dangerous because they cause the laceration and compression of internal organs, and since the force is distributed on a much-smaller cross-sectional area, resulting in a higher pressure along the side.

And none of these take terrain into account.

1 https://www.ncbi.nlm.nih.gov/pubmed/24351358
2 http://medind.nic.in/jal/t13/i1/jalt13i1p47.pdf

What would be far easier to know as a general threshold would be the height at which spacesuit equipment, such as respiration equipment, would suffer catastrophic damage. While I'm not an aerospace engineer and have no idea about equipment limitations of spacesuit respirators, I am confident that NASA has rated its equipment to determine how much force and pressure the equipment is able to withstand. From that, it would be simple to extrapolate a rough height from that piece of equipment would no longer be able to survive hitting the ground attached to an object equal to the weight of the spacesuit plus its human occupant on Earth, and then extrapolate that to that weight on the Moon.


Earths gravity is 6 times stronger than the moon. You can jump from a height 6 times higher on the moon. Most people would not want to jump from a height of more than 2 metres on earth. Therefore you could jump from a height of 12m on the moon.

Estimated injuries from various heights on the earth and the moon:

  • Broken limbs: 10m or 60m
  • Severe injuries: 20m or 120m
  • Possible fatality: 30m or 180m

Keep safe.

  • $\begingroup$ Wow! Splashdown velocity is $v=\sqrt{as}$, resulting that you are right. Sixth gravity means 6 times more possible height with the same consequences. $\endgroup$
    – peterh
    Commented Jan 4, 2021 at 16:18

I don't think any of the other answers have pointed out that you come down with the same speed that you went up with. On earth, I can't jump so high that I can't safely land, and so on the moon I wouldn't be able to either.

The difference with the moon is that I would be in the air a lot longer. If I overrotated and landed on my head I could die, just as I could die if I intentionally flung myself head down onto concrete on earth.

  • $\begingroup$ The top question (by tildalwave) addresses this at the end of his answer. I don't blame you for not seeing it though, there's a lot of content on this page you'd have to wade through. $\endgroup$
    – Ingolifs
    Commented Nov 9, 2019 at 5:26

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