I know that this is an old topic, but I just thought I would throw my two cents in. The notion that someone can jump 21 times higher on the moon than they could on Earth is not wrong in theory. It is reasonable - assuming ALL OTHER variables are exactly the same (they aren't in reality, but for the sake of argument, let's assume they are). I can demonstrate this, and I apologize for how convoluted this will be, but those saying that the physics doesn't add up are not really correct. Allow me to show why.

Consider an astronaut with a mass of 80 kg who can jump 18 inches on Earth (0.457 meters) with his full range of motion. Some force is exerted (on the ground really) for some amount of time. In other words, imagine the astronaut bends over - ready to take his jump, and as soon as he starts his jump, start this clock. Stop the clock when his feet leave the ground. This is the amount of time his force is applied (this will be important later). Once his feet leave the ground, he has some velocity upward. High enough that he reaches 0.457 meters. I've actually timed myself jumping and consistently get between 0.25 and 0.35 seconds for the time here, so I'll just use 0.3 seconds. First, calculate his "initial" velocity - that is, his velocity the moment his feet leave the ground and a force is no longer being applied. This can be calculated as follows :

where g = 9.8 m/s² (since we are talking about a jump on Earth here - in Earth's gravity)

h = height

v = √(2gh) = 2.99 m/s

I've estimated a reasonable time for the amount of time a force was applied - this was about 0.3 seconds. We can then calculate the acceleration :

a = dv / dt = 2.99 / 0.3 = 9.97 m/s²

From this, we can now calculate the force :

F = ma = 80kg * 9.97 m/s² = 797.6 Newtons.

This is actually the NET force (which is probably where a lot of people go wrong when they are trying to calculate jump heights between the Earth and moon). Obviously, we know that there is a gravitational force on the astronaut as well, so any acceleration is the result of a NET force - that is, this 797.6 Newtons we just calculated is that force which is opposite in direction to gravity, but IN EXCESS of it, so we need to know the gravitational force :

W = mg = 80 * 9.8 = 784 Newtons.

This means the actual force the astronaut exerts on his jump is 1,581.6 Newtons - 784 N of which overcomes gravity, the remaining 797.6 N is the net force upward.

This force will be the same on the moon. Astronauts do not get stronger or weaker. Assuming the astronaut jumped as hard as he could on Earth, his hardest is the same on the moon, so we need to know how this force propels him on the moon. Again, we need to know the NET force on the moon, so how much of this 1,581.6 Newtons is in excess of lunar gravity?

W = mg = 80 * 1.6 = 128 Newtons

So this means that the net force upward on the moon for our astronaut is 1,453.6 N.

How does this net force accelerate our astronaut (I'm not considering other factors - like the space suit - just yet).

a = F / m = 1,453.6 / 80 = 18.17 m/s²

With full range of motion, our astronaut is able to apply his force for 0.3 seconds, so the initial velocity on the moon (again, his velocity the moment the force is removed and his feet leave the ground) is :

v = at = 18.17 * 0.3 = 5.5 m/s

How high will something go in lunar gravity with an initial velocity of 5.5 m/s?

h = v² / 2g = 30.25 / 3.2 = 9.45 meters.

So, this is actually pretty simple physics - just considering all the factors a lot of people leave out, but if you assume all other factors are the same, an astronaut who can jump 0.457 meters (18 inches) on Earth can potentially jump 9.45 meters (31 feet) on the moon. This is close to 21 times higher as the original comment references. It's sound physics. It's just not very good at considering all factors.

One MAJOR factor not considered here is the extra 90 kg of mass the astronaut has on his back when he is on the moon! I'm not sure how you can forget about this, but if someone has done the math right and concluded they should be jumping 21 times higher, they forgot to add this mass into the equation. Let's do it (starting from the acceleration on the moon using 170 kg instead of 80 kg :

a = F / m = 1,453.6 / 170 = 8.55 m/s²

And the initial velocity now :

v = at = 8.55 * 0.3 = 2.6 m/s :

Now, the height :

h = v² / 2g = 2.11 meters.

Already, we are getting closer to observed reality, but there's more factors to consider (more than I will mention - these are just the big ones). We are assuming that the astronaut has the same range of motion while wearing that cumbersome suit as he did on Earth. This is most certainly not the case. So what does this change? Well, he can't bend over as far while wearing a suit. If he can't bend over as far, then clearly the distance he travels between his fully crouched position and fully extended position is shorter. Since the force is only applied during the time it takes to go from crouched to extended, then clearly, the time his force is being applied is shorter. How much shorter is certainly up for debate, but there's no doubt that it would be significantly shorter. Let's suppose it is 2/3rds of what it would be on Earth. I think this is actually quite generous. I think it would be at least half the original time, but let's use 0.2 seconds - adjusting from the point where we calculate the initial velocity above (through our adjusted mass calculations as well) :

v = at = 8.55 * 0.2 = 1.71 m/s

Now, the height :

h = v² / 2g = 0.91 meters.

Now, we are getting into something which is more consistent with what we see. There are more factors, but the point of this exercise was just to point out that any claims that the way the astronauts are jumping on the moon is in any way inconsistent with physics is nonsense. Even when those making the claim understand the physics concepts enough to calculate something accurate, it doesn't mean they have considered all the factors - and the person who says they should be jumping 21 times higher certainly wasn't considering all the factors.

I think it's been mentioned, but all of this assumes that the astronauts are going around jumping as high as they can with every jump, and that is not the case at all, so that right there might be the biggest factor into why they do not jump as high as some expect. One last thing to consider - which I believe Jay brought up (that I was unaware of - thanks, Jay!) is that the joints in the suits themselves would provide resistance, so this would further reduce the jump height as not only would the astronauts' jump force need to overcome lunar gravity, but whatever resistance the joints in the suits themselves provided as well - reducing our upward net force that much more.

EDIT (because I wanted to add this for fun!) :

One other thing I didn't consider in my original comparison was that our original astronaut who accelerated upward at 9.97 m/s² on Earth - who would in turn accelerate upward at 18.17 m/s² on the moon (without a suit and with the same range of motion), would not actually jump as high as 21 times higher on the moon simply because the higher acceleration would make him leave the ground sooner, so the force would be applied for less than 0.3 seconds. Assuming that the distance traveled during the force is about 0.5 meters (I measured on myself), at 18.17 m/s², this distance is traveled in just 0.23 seconds (not 0.3), so to adjust that -

v = at = 18.17 * 0.23 = 4.2 m/s

h = v² / 2g = 5.5 meters.

So, in reality, an 80kg person who has a 0.457 meter vertical on Earth has the potential to jump 5.5 meters, so just 12 times higher. I hope someone enjoys this mess!