# What is our estimated running speed on Moon's surface?

I was wondering if we have the chance to run on the Moon's surface, how would you expect it look like? I expect our velocity will increase for the same work we do on Earth, but not sure if this will be multiples in term of gravity variations.

How do you think our maximum speed would reach?

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I don't know about the physics, but I am fairly sure that you won't be able to run faster than you can run downhill on earth, without stumbling. –  Glen The Udderboat May 14 '13 at 13:42
Well nonagon. I think if we analyse our motion we will realize it is seriously affected by gravity (pulses of jumps). I understand that you have already excluded jumps but how can we run by then? –  Tariq May 14 '13 at 13:45
There are a few videos of astronauts running on the moon, e.g. youtube.com/watch?v=wo3-fuYKWB4 and youtube.com/watch?v=HKdwcLytloU . Note that they have to adopt quite different gaits from on Earth. However, they were encumbered by space suits that were heavy and stiff, and they were probably moving fairly carefully in order to avoid the danger of a puncture. It's difficult to say how much faster a human could run in a more optimal suit, but it's almost certainly slower than Earth. –  Nathaniel May 14 '13 at 13:46
I think the fastest you could go is the escape velocity on the moon--after that you would fly off :) –  JoshRagem May 14 '13 at 13:49
lol @ JoshRagem. Indeed I was waiting for an encouraging speed before I submit your analysis. –  Tariq May 14 '13 at 13:51

Maximum speed can be ~ (because friction etc. are slowing you down and the moon isn't a perfect sphere)

$\frac{v^2}{r} = \frac{GM}{r^2}$ $-$ $(1)$ where $r$ is radius of moon & $M$ is mass of the moon.

EDIT : Since then in this case your normal reaction is $0$ , means you are almost flying , speed up a little bit more and you will fly off the moon along a tangent line onto a bigger radius and now you are flying actually along the moon like a satellite , and not running .

if you are running on the surface of moon provided you have that kind of energy in you . Gravity won't effect directly as gravity is already perpendicular to your direction of motion always ,but friction will be less as friction is $\mu_k N$ and N will be less since gravity force is less , otherwise no effect , I am assuming you aren't trying to jump , if you try to jump then your projectile range will be more than on earth , as gravity is less , you'll be having more time in air and as there's no atmosphere , no viscous force by air .

So in net you'll reach a destination in fewer steps and quickly because dissipative forces are less and will require less energy consumption to reach a particular destination .

But a slight decrement will be caused in your projectile range since Buoyant forces will also not be present because of almost $0$ atmosphere .

However , beware of all the craters :)

EDIT : the formula $(1)$ comes from the fact that if you are running on the surface means your distance from centre isn't varying that much . And the moon is pulling you towards the centre providing you the necessary centripetal acceleration to allow you to turn along the moon surface .

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Can you please elaborate on the maximum speed formula? Where did that come from? You go on to mention friction but I don't think friction is a significant factor for running in an atmospherless environment. What is a "dissipative force"? Inelastic collision perhaps? –  Brandon Enright May 15 '13 at 6:48
@BrandonEnright I have edited the answer . –  nonagon May 15 '13 at 6:51
That's somewhat like saying "you can't run faster than escape velocity". An upper bound that isn't even close to reachable isn't very useful. –  Brandon Enright May 15 '13 at 6:54
No this isn't escape velocity . It is completely different . It is the velocity that suppose you are tied to a string then in order to move along a complete circle , you have to maintain a certain max speed , otherwise string will break . And also in this the normal reaction is $0$ , means you are almost flying . If you go any faster , you will fly along tangent to the moon. –  nonagon May 15 '13 at 6:56
"escape velocity" was a poor choice of words. I meant orbital velocity for an object near the surface of the moon. Running any faster puts you in an orbit. That's an absurdly unrealistic upper bound for a human running. –  Brandon Enright May 15 '13 at 6:59

There are a few videos of astronauts running on the moon, e.g. here and here. Note that they have to adopt quite different gaits from on Earth. However, they were encumbered by space suits that were heavy and stiff, and they were probably moving fairly carefully in order to avoid the danger of a puncture.

It's difficult to say how much faster a human could run in a more optimal suit, but it's almost certainly slower than on Earth. Prior to the moon landings, NASA did some research where they simulated Lunar gravity by suspending people from ropes. The guy in this video gets up quite a speed by leaning forward and flailing his arms around like crazy, but he still can't keep up with someone sprinting under Earth gravity.

(It might not be very obvious from the above video how the gravity simulation works. The trick is that he's actually lying on his side. You can see it when they set it up at the start of this longer video, which also demonstrates things like jumping high into the air and climbing a pole one-handed.)

It's just about conceivable that this running speed could be improved upon through the use of things like weights or even stilts. I don't know whether research into such "augmented" locomotion under Lunar gravity has been done. (And if it was, I'm not sure whether it would really count.)

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I think you have to lean forward and jump hop (like astronauts did). Now if the lean angle is too great then you would slip. The limit is $\tan \theta \le \mu$.

Now with a jump of max (leg extension) speed $v$ at an angle $\theta$ from vertical kinematics have that the time per hop is

$$t = \frac{2 v \cos \theta}{g} = \frac{2 v}{g \sqrt{1+\mu^2}}$$

Also, the range per hop is

$$s = \frac{v^2 \sin\theta \cos\theta}{g} = \frac{2 \mu v^2}{g (1+\mu^2)}$$

making the average horizontal speed

$$v_{ave} = \frac{s}{t} = \frac{\mu}{\sqrt{1+\mu^2}} v$$

Also the hop height is $h = \frac{1}{4\mu} s$.

Now the maximum leg extension speed depends on your mass (and payload) and the maximum power output of your leg muscles. Note that since $h \ll R$ the surface curvature effects and gravity non-linearity are negligible.

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In a "perfect" space suit probably only 4-6 mph!?! In a dome on grass or something mayB 10 even 11 but remember that U would not only B moving horizontally but vertically as well which would slow down even the fastest runners tho U should B able 2 sustain these speeds 4 much longer due 2 the minimal "weight" aspects..:)

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