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I was told that if you theoretically jump down from a three-floor building (assume 10 m) on the Moon surface, you will equally break your bones as in if you were doing that on Earth. I do not know why would this be true: the acceleration due to free fall on Moon is significantly smaller, thus the final v shall be smaller than v on Earth. Would you be perfectly fine if you jumped down a 10 meter building on Moon?

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    $\begingroup$ Duplicate of space.stackexchange.com/q/4678/6946 $\endgroup$ Commented Mar 16, 2015 at 12:33
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    $\begingroup$ Not sure this is a duplicate... the answers on the other question are so long and detailed that I didn't see if the answer here should be yes or no. I vote to close this one open, in the hope that someone will provide a simple table with height equivalences for Earth and Moon. $\endgroup$
    – mins
    Commented Mar 16, 2015 at 18:13
  • $\begingroup$ @SteveJessop that answer assumes that a crash test dummy won't break apart at 60 km/h, and goes from there. Obviously not applicable and not the same answer. $\endgroup$ Commented Mar 16, 2015 at 21:00

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It is certainly not the same.

Gravity on the Earth's surface is 9.81 m/s2. Assuming no upward or downward initial velocity, this means your velocity at ground level falling from 10 m is about 14 m/s (sqrt(2 * 9.81 * 10)).

Gravity on the Moon's surface is 1.625 m/s2. The same fall will result in final velocity of about 5.7 m/s (sqrt(2 * 1.625 * 10). This is equivalent to a fall on Earth from about 1.6 m (5.7 ^ 2 / 2 / 9.81).

Assuming you don't land on your head, you should be okay.

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    $\begingroup$ Another way to think about this is that since the moon's gravity is 1/6 of Earth's you fall 6 times the distance to achieve the same velocity. Most could jump 2 meters on Earth without injuring themselves, so they could jump 12 meters on the moon with the same result. $\endgroup$
    – GdD
    Commented Mar 16, 2015 at 10:58
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    $\begingroup$ Interestingly, if 'you' are (say) a spider (in a miniature space suit of course) the landing on the moon might be harder. On earth, a spider would reach terminal velocity in well under one meter. On the moon, with no atmosphere, it would carry on accelerating. $\endgroup$
    – abligh
    Commented Mar 16, 2015 at 14:32
  • $\begingroup$ Unless the suit had a port for the spinnerets, and the spider had somewhere to attach her silk... $\endgroup$ Commented Mar 16, 2015 at 17:26
  • $\begingroup$ At further distances, the terminal velocity issue is also true for humans. The moon does actually have an atmosphere, but it is negligible for any practical purpose. Terminal velocity on Earth is about 54 m/s (Reached from a height of about 150 m). There have been cases where people actually fell out of a plane without a parachute and survived due to this. With a parachute, this is more fun than scary. On the moon, you can go much faster and a parachute won't help. $\endgroup$
    – neelsg
    Commented Mar 17, 2015 at 7:26
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If you come from Earth and make the jump, you'll be okay as the other answer says. But if you were to live full time on the moon or in space, your body would adapt to the weaker gravitational field, resulting in muscle mass loss and other effects - if these included changes to the bone structure, such a jump could be dangerous after six months or so.

It would be worth reading about the physiological changes to the MIR cosmonauts and ISS inhabitants to understand better what happens after a few months to a year - and how exercise programs mitigate the effects - but extending the lessons to multiple years or a lifetime in reduced gravity would probably be speculation.

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  • $\begingroup$ Any source for these “six months or so”? I know plenty of zero-gravity data has been collected, but how do we extrapolate to Moon gravity? $\endgroup$ Commented Mar 16, 2015 at 14:35
  • $\begingroup$ None with respect to lunar gravity. What I've read comes from "The Story of Space Station Mir" by David Harland. Sorry for not digging out page references but someone has borrowed it atm. amazon.co.uk/The-Story-Space-Station-Mir/dp/0387230114 $\endgroup$ Commented Mar 16, 2015 at 17:24
  • $\begingroup$ This is a good point $\endgroup$
    – neelsg
    Commented Mar 17, 2015 at 7:31

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