ApolloHoax.net
Apollo Discussions => The Hoax Theory => Topic started by: onebigmonkey on October 03, 2014, 12:32:38 PM
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Over at DIF, user oz93666 is trying to establish that the astronauts should be able to jump much higher than they are observed to do in the lunar footage.
He has taken his idea to a physicsforum:
https://www.physicsforums.com/threads/why-a-man-on-the-moon-can-jump-21-times-higher-than-on-the-earth.774140/ (https://www.physicsforums.com/threads/why-a-man-on-the-moon-can-jump-21-times-higher-than-on-the-earth.774140/)
I am no physicist so find it difficult to prove the maths for either side of a debate, but there are many here who are.
So...erm...discuss :D
Is his logic sound? Are his assumptions reasonable? Here is his post in full:
Two men , one on the Earth, one on the moon, both bend their knees the same amount ready to do a 'standing' jump.
They both start , and accelerate upwards.
The earth man is pushing against the inertia of his mass and the downward force of his earth weight.
The moon man is pushing against the inertia of his mass and the downward force of his moon weight.
Clearly the moon man will have a higher velocity when his feet leave the ground than the earth man, and so will travel OVER six times higher.
How much higher will depend on how long the acceleration zone is.
When I do this I squat right down to the floor and can jump 30cm high. the acceleration zone is 90cm.
The energy the earth man expends against the extra gravity iis 0.9 x( 1 - 1/6)g
Equivalent to 0.9 x5/6 =75cm in earth gravity total jump =30 +75 =105cm
This means he can jump 6.3meters on the moon!! 21 times higher
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I'm no physicist either, but has he taken into account all that mass the 'moon man' would necessarily be carrying? Or the fact that it might not be a particularly great idea for him to jump as high as he can in an environment where landing badly could be life-threatening?
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He's throwing out a hodge-podge of physics-related terms, and he's also missing several important factors in the comparison.
The factors he's missing are the much greater inertia of a space-suited astronaut (ca. 360 pounds-mass) and the frictional assistive devices in the space suit that work to keep joints bent once the astronaut has bent them. The former limits the effective velocity he can attain and the latter limits how much force is applied to the underlying substrate and therefore the reactive force resulting in upward propulsion.
As for the math, I don't even know where to start telling him what's wrong with it. His formulation is wrong literally (and no pun intended) at every step.
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Over at DIF, user oz93666 is trying to establish that the astronauts should be able to jump much higher than they are observed to do in the lunar footage.
He has taken his idea to a physicsforum:
https://www.physicsforums.com/threads/why-a-man-on-the-moon-can-jump-21-times-higher-than-on-the-earth.774140/ (https://www.physicsforums.com/threads/why-a-man-on-the-moon-can-jump-21-times-higher-than-on-the-earth.774140/)
I am no physicist so find it difficult to prove the maths for either side of a debate, but there are many here who are.
So...erm...discuss :D
Is his logic sound? Are his assumptions reasonable? Here is his post in full:
Two men , one on the Earth, one on the moon, both bend their knees the same amount ready to do a 'standing' jump.
They both start , and accelerate upwards.
The earth man is pushing against the inertia of his mass and the downward force of his earth weight.
The moon man is pushing against the inertia of his mass and the downward force of his moon weight.
Clearly the moon man will have a higher velocity when his feet leave the ground than the earth man, and so will travel OVER six times higher.
How much higher will depend on how long the acceleration zone is.
When I do this I squat right down to the floor and can jump 30cm high. the acceleration zone is 90cm.
The energy the earth man expends against the extra gravity iis 0.9 x( 1 - 1/6)g
Equivalent to 0.9 x5/6 =75cm in earth gravity total jump =30 +75 =105cm
This means he can jump 6.3meters on the moon!! 21 times higher
I know this claim in many variations. My answer is as follows:
An astronaut has to accalerate his mass and the mass of his suit.
On Earth an astronaut (estimated weight about 90 kg) an his suit (let's take a round number: another 90 kg) have a mass of 180 kg. A jump with bended knees would be about 10 cm.
On Moon the weight is 1/6 (30 kg) due to the lower gravitation, but the mass (180 kg) is still the same. Youngs salute jump was about 50 to 60 cm high, which is almost 6 times higher.
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By the way, this question comes up a lot and it's based fundamentally on a begged question. We know by watching the Apollo 11 EVA that Armstrong achieved a vertical leap much taller (although certainly not at the theoretical upper limit) than would have been possible on Earth. Most hoax claimants are unaware of this. But the begged question is that the astronauts should have generally leaped higher on the Moon as a demonstration, and that the supposed lack of video evidence for this means it was not actually shot in diminished gravity.
It begs the question of propriety. Just because they theoretically could have done a certain thing doesn't mean it was wise to do it. Apollo 11 astronauts, for example, operated under severe limits for mobility, flexibility, and locomotion because engineers advocated conservative operations until they could get the EVA suits back for a performance analysis. By Apollo 17 the astronauts were happily leaping all over the place. (The joke passed up to Harrison Schmidt was that the Houston Ballet wanted him to audition.) Not 5-7 meters above the surface, as the ignorant claims here suggest, but clearly not in Earth gravity.
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Do that on the Earth.. ;)
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This (http://www.youtube.com/watch?v=16D0hmLt-S0) is why you really don't want to try a really high jump in a spacesuit.
'That ain't very smart', indeed!
It also looks nothing like jumping on Earth, or wire work for that matter.
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This (http://www.youtube.com/watch?v=16D0hmLt-S0) is why you really don't want to try a really high jump in a spacesuit.
'That ain't very smart', indeed!
It also looks nothing like jumping on Earth, or wire work for that matter.
Game, Set and Match raven ;D
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Clearly the moon man will have a higher velocity when his feet leave the ground than the earth man, and so will travel OVER six times higher.
I'm just not quite grasping this concept. He's under 1/6 gravity, so he should leap *higher* than 6 times an earth jump?
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I am a physicist, and his physics is wrong. Utterly and totally wrong. So wrong in fact, I have no idea where he started with his formulation. He's just throwing numbers around.
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Agreed.
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I am a physicist, and his physics is wrong. Utterly and totally wrong. So wrong in fact, I have no idea where he started with his formulation. He's just throwing numbers around.
I know, right? It's like the phrase goes: "That's not right. That's not even wrong." I don't know how else to describe but it as a tossed salad of words from physics along with arithmetic that bears no relationship to any concept in Newtonian dynamics.
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This (http://www.youtube.com/watch?v=16D0hmLt-S0) is why you really don't want to try a really high jump in a spacesuit.
'That ain't very smart', indeed!
It also looks nothing like jumping on Earth, or wire work for that matter.
Game, Set and Match raven ;D
Thank you. I admit, I don't know the math of physics, but I do know that trying to jump straight up with a heavy load on your back without forward support means whoops! And when that heavy load is keeping you alive, quintuple whoops, at least!
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This (http://www.youtube.com/watch?v=16D0hmLt-S0) is why you really don't want to try a really high jump in a spacesuit.
'That ain't very smart', indeed!
It also looks nothing like jumping on Earth, or wire work for that matter.
Game, Set and Match raven ;D
Thank you. I admit, I don't know the math of physics, but I do know that trying to jump straight up with a heavy load on your back without forward support means whoops! And when that heavy load is keeping you alive, quintuple whoops, at least!
Similar to you I'm not a physicist. But I do know how some important details work, and I do know how to do the maths behind it (most of the times ;D) And I think the most important difference between us (you, me and all the others) and the hoax believer is: We know, what to do, when there's something we don't understand. And this is simply asking experts in the special field.
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This (http://www.youtube.com/watch?v=16D0hmLt-S0) is why you really don't want to try a really high jump in a spacesuit.
'That ain't very smart', indeed!
If you asked Charlie Duke, he would probably agree with you.
:o
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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!
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Well done! Seems like you have accounted for most of the variables and matched the observational values. :) The CT's won't budge either.
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You can simplify the analysis even further so you can do it in your head. Gravity changes weight but not mass. As you leave the ground, all your energy is kinetic, and that depends only on mass. Assuming that your legs work equally well in both places, you should have equal amounts of kinetic energy and therefore equal velocities as you leave the ground.
Yes, that's a first order estimate that ignores the distance you have to lift your weight as you straighten your legs. But we're assuming a quick jump where most of the energy goes into accelerating your mass upward.
After your feet leave the ground, your kinetic energy is converted to potential gravitational energy and back into kinetic energy as you fall. Potential energy is proportional to gravity and distance, so a given amount of kinetic energy will therefore take you six times higher on the moon, no more.
On top of that, on the moon you're wearing a heavy suit/PLSS that weighs about as much as you do, and it's so stiff from inflation that your legs can't possibly work as well as they do on the earth without a suit.
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Thank you, ka9q. I wanted to mention the mass is still constant, so the better formula is probably something like f = ma minus gravitational loss to arrive at initial velocity. But i'm home sick and can barely handle logic today, much less math.
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...you should have equal amounts of kinetic energy and therefore equal velocities as you leave the ground....
...so a given amount of kinetic energy will therefore take you six times higher on the moon, no more.
On top of that, on the moon you're wearing a heavy suit/PLSS that weighs about as much as you do, and it's so stiff from inflation that your legs can't possibly work as well as they do on the earth without a suit.
You're only accounting for one aspect the reduced gravity changes : acceleration due to gravity. It also changes the net force of your jump. If all other factors are the same, you could jump MUCH higher than six times higher on the moon. However, as you point out (and I as well in my original post), all factors are not the same. The restriction in movement (limiting the amount of time you can apply your force) and the fact that there is an additional 90 kg of mass is the difference.
If you have initial velocity (which you would have to if your jump leaves the ground), then that velocity is a result of a NET force upwards. This is the sum of all forces - including gravity - which is six times stronger on the Earth. On the moon, with the exact same force of your jump, the NET force would be much larger - therefore your initial velocity much higher.
For example, if your mass is 80 kg and you accelerate your body upward at 10 m/s², then the NET force on your body is 800 Newtons. This is not the force of your jump - it is the force of your jump IN EXCESS of Earth's gravity. To calculate the total force of your jump, you have to account for gravity - which on Earth would be 784 Newtons. If the NET force is upward at 800 Newtons, then the total upward force would be the 784 Newtons it takes to overcome gravity plus the 800 Newtons in addition to it - for a total of 1,584 Newtons. On the moon, this force of 1,584 Newtons is only countered by the moon's gravity - so 128 Newtons on your 80kg body. The NET force then is 128 Newtons to overcome the moon's gravity, and the rest - 1,456 Newtons is a NET force upwards, resulting in an acceleration of 18.2 m/s². You must also account for the fact that since your acceleration is higher, you will leave the ground faster (so the force is applied for less time), but the initial velocity is still higher. In my original comment, I have an edit where I do a rough calculation on this - and came to the reasonable conclusion that if all factors are the same with regards to range of motion and mass, a person could potentially jump about 12 times higher on the moon. You must account for the change in gravity both for the rate at which something falls AND the counter force to your jump. I understand that my original post was pretty convoluted, and simplification is better for a lot of people (and I would tend to be one of those people), but if your simplification arrives at the conclusion that six times higher is the absolute highest anyone could jump on the moon, then your simplification is so simple, it's leaving something out.
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Has anyone accounted for the fact that the astronauts may not have wanted to jump as high as they could?
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Has anyone accounted for the fact that the astronauts may not have wanted to jump as high as they could?
Yes, this is another - probably more relevant - reason most of the "jumps" are not as high as hoaxers expect. I'm not using those jumps as a basis for my comparison though. There is at least one clip of astronauts jumping as high as they can on Apollo 16. Charlie Duke and John Young estimate their jumps to be about 4 feet which is "not bad for a 380 pound guy" (addressing his earth weight including suit). Which is consistent with my estimation that one could potentially jump 12 times higher on the moon. This translates to a 4 inch vertical on Earth with all that mass and restricted movement vs 8 inches if you are only able to jump 6 times higher - which I have a hard time believing they could achieve on Earth. The vast majority of "jumps" on the moon aren't jumps, but hops and are just the way astronauts figured put was the best way to move on the moon since gravity was lower, but their inertia was not. Walking normally was too hard, and you're right - those hops are not an example of how high the astronauts could jump on the moon.
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If you have initial velocity (which you would have to if your jump leaves the ground), then that velocity is a result of a NET force upwards. This is the sum of all forces - including gravity - which is six times stronger on the Earth.
You are correct. In the limit of a very sudden, instantaneous jump, the momentary upward force of your muscles swamps your downward weight on both the earth and moon, so you'd rise 6 times as high on the moon. But you are correct that when you consider the upward force to be of finite duration, some of it is required to overcome weight, so the difference would be greater.
It's the same thing as "gravity loss" in a rocket, the reason why lots of liftoff acceleration is much better than just a little more than that required to overcome gravity, even if the total impulse (thrust times time) is the same.
I had to explain this once to my significant other after she put an low thrust engine in one of my rockets at a launch event. Despite having the same total impulse as a correctly sized engine, it valiantly struggled off the pad and made it to only maybe 100' before burning out and nosing over into (into) the earth before the ejection charge could fire...