Jump height on the moon

I’ve done a cursory search of the boards, and I cannot believe this hasn’t been examined yet.

Mrs. KCB is a third grade teacher, and is doing her annual “moon” module (making craters in the classroom and comparing impactor size vs crater size, multiple observations over a lunar cycle, neat stuff we never did…). One of her math tie-ins is calculating weight on the moon (earth weight / 6). She is looking to add another segment on calculating vertical jump height. One of the sources she found on the InterWebs said jump x 11. Her science text is saying jump x 6. Another web site gave jump x 22. She came to me with the question, and I thought there should be at least an exponent or a constant somewhere in that calculation - simply multiply by a small whole number can’t be correct. Furthermore, a child who can jump 2 feet to be able to catapult themselves 44 feet high doesn’t pass my sniff test. Then again, I’ve never been to the moon, so I can’t speak from personal experience.

Are we making too much of this? Is the conversion really that simple? What’s the actual formula?

The peak of the jump occurs when the kinetic energy of the jumper is zero and their potential energy equals the kinetic energy of the jumper just as they leave the surface. So, if we assume that the energy the jumper can put into the jump is the same, and they have the same mass on the Earth and the Moon (ie no spacesuit), we simply need to see what difference changing the force of gravity makes to the potential energy.

The potential energy is: U = mass * force due to gravity * height.
So, for a constant mass and energy: height = 1/gravity

So, since the Moon has one sixth the gravitational force of the Earth, the answer would be six times the height.

Yup. Things start getting weird for, say, medium-sized asteroids, where it starts to become possible to jump into orbit or close to it. But the Moon is nowhere near that point, so the simple linear increase is still valid.

The key here is that high jump records are not set by jumping in the vertical orientation, but by aligning the jumper’s body to minimize the change in center of gravity (aka the Fosbury flop). A good high jumper can clear a bar without the center of gravity ever being above the bar. A high jumper who can clear a seven foot bar is not going to jump 42 feet on the moon. He probably raised his center of gravity by 4 feet or so, so figure a 24 foot increase in that for a high jump around 27 feet.

I do have to sympathize with the OP. Much of this stuff is not so intuitive. I still scratch my head when I try to compare things whose strength weakens proportionally to the distance, vs. those which are proportional to the square of the distance, and others which are proportional to the cube of the distance. It dizzys me so much that I can’t think of any good examples right now.

OP did not ask about the high jump; he asked about vertical jump height.

I knew that, but the second post had the correct answer, so there was nothing more to say on that. I just stated that because I wanted to make sure someone writing a story didn’t decide a high jump over a 6 story building was possible, because I’ve seen that before.

I have no idea where the jump X 11 and jump X 22 numbers in the OP came from, maybe someone conflated long jump distances with vertical jump height.

Well, wait. As this guy points out, there’s an extra part to the original assumption. You jump by flexing your knees and pushing upwards, so the energy put into the jump goes in part to lifting your center of mass during the jump, and part to accelerating your center of mass. So if you assume that the energy the jumper can put into the jump is the same, then the jumper on the moon has a greater velocity when his feet leave the ground than the Earth jumper, and the actual answer would be something more than six times the height (eleven, according to the linked article). That also assumes thatthe “height” of a jump is measured from where your feet leave the floor.

If the same man is traveling at the same speed as he leaves the moon’s surface, he would indeed go 6 times as far. However, I believe his speed would be greater on the moon. To jump, his upward force must overcome inertia as well as the downward force of his weight. Hence, on the moon, more of the jump energy would be used to accelerate his mass so he would be going faster than his jumping speed on the earth. If his jumping force is 2g, then 1.8333g of that energy will be used for acceleration on the moon, whereas only 1.00g will be used for acceleration on the earth. I haven’t done the math, but 11 (1.8333 * 6) sounds like a good answer if his jumping force is 2g.

My own first swag at these kinds of questions is to narrow it down to three choices: scales linearly, or goes by the square, or by the square root. That’s just because I know the formulas for acceleration and distance have some squares in them.

So I could see wondering about whether it’s 6x, 36x, or 2.45x, but I don’t know about the 11 and 22 factors she came up with.

For the long jump, I think your long jump distance should go up by 36x, if you could reach the same running speed and jump vertical speed. That would be hard to do on the Moon however.
ETA: I see bizerta and zut have pointed out factors I didn’t think of. 11 could be correct.

(Missed the edit window)

bizerta and I are essentially saying the same thing. However, the answer really depends on what assumptions one makes. For example, the assumption that “the energy the jumper can put into the jump is the same” is pretty reasonable, but it might not be perfectly accurate - just how fast can the jumper flex their leg muscles, anyway, and does energy input drop off with muscle flex speed, and at what point is the gravity so low that the jumper can’t accelerate him/herself that much?

Since the original question is slanted toward third-graders who are just beginning to think about gravity, I think the nuances here are a bit much. I would just say, “assuming you jump up at the same speed, you jump six times as high on the Moon as you do on Earth.” That should be both simple and correct.

I knew “x 6” was too simple; but it’s perfect for 8 year olds.

Thank you, all.

the spacesuit should be taken into consideration, because:

how far can an astronaut bend down on the moon?
https://www.youtube.com/watch?v=NiJ54Jj2rck

Actually, the question is entirely interesting, and a good demonstratin of body mechanics vs. gravity.
Get a bunch of grade 3 students, have them stand between a video camera and a wall grid, and throw progressively heavier weights as high as they can. (Maybe aim for a 60-degree angle so they don’t hit their head!)

If as the 11x proponents say, part of the effort of muscle use is to simply fight gravity, then: your muscle can impart X energy. For a 5-pound weight, that means you can give X-(5lbsxG), for a 10 lb weight, X-(10xG) and so on.

So the height you can throw something in the air drops until you reach the point that it’s all you can do to lift it. (Weight = X/G); you should not find that you can throw a 5-lb twice as far as a 10lb and 4 times as far as a 20lb weight. Therfore on the moon you could throw a 10lb weight based on imparting force of X-(10lbxG/6).

Sounds like a fun aftenoon experiment.

I think the point made about the next level of complexity is actually really important, and even for 8 year olds holds a very valuable lesson. The difficulty is that there is no single answer. Whilst 6 times is a defensible answer, with simple assumptions, it is wrong. 11 is also wrong, but for different reasons. The trouble with 11 is that it is a particular solution for a question that actually has an infinite number of solutions. Whilst it is defensible to an extent, it is just an average number for an average human. Anyone who claims 11 is the right answer is in some respects even more wrong than someone who says 6. Even though 11 contains a derivation process that is closer to correct, without stating the assumptions, which are complex, just saying 11 is a very bad answer.

As pointed out above, there are boundary cases. One to think about, if you have some lard arsed kid, so fat that they cannot actually get off the ground with a squatting jump, you will probably find that on the moon, they can jump a reasonable height. The multiplier here is infinity, not 11. You could get a flea, and work out the multiplier. It could be insane.

Perhaps one way of explaining the issue to kids is to look at the real astronauts. We know they could jump on the moon, there is even video, and at least one tried to see how high he could jump, with nearly disastrous consequences. However on the Earth, astronauts needed help to even stand up when they had the full suit and backpack on. There was no chance at all that they would have been able to jump off the ground in them. The entre suit had a mass of nearly 100kg, and exceeded the mass of the astronauts. Thinking about trying a squatting jump whilst carrying someone your own weight would convey the issue. Then imagine trying the same jump, but if you weighed one third of what you normally do.

I think we need to assume an enclosed shirtsleeves habitat on the surface of the Moon, not a bulky spacesuit.