Diving on the moon: A brainteaser
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glugo357
Contributor
would the body of water be on the moon or suspended? Would standard diving equipment be used or would we need specialized equipment to protect us from the sun? 
Maybe we need a 
Maybe we need a 
shakeybrainsurgeon
Contributor
The same weight --- all weight is reduced on the moon, including the weight of water displaced and the weight on your belt. Ignoring the fact that there is no air on the moon, the reduction of gravity should not affect buoyancy
Sharky1948
Contributor
Good one! I think you would use the same weight. My logic (though freshman physics is a distant memory) is that the relative volume and mass of the diver, the weights, and the water would be the same as on earth. Therefore it would be the same.
EDIT: By same weight I mean "marked the same." I.e., one would use the same pieces of lead.
Since I'm the first to venture a guess, I'm hoping people will be kind if I'm all wet!
EDIT: By same weight I mean "marked the same." I.e., one would use the same pieces of lead.
Since I'm the first to venture a guess, I'm hoping people will be kind if I'm all wet!
Less weight.magnifica:Theoretically, if you dived in a body of water on the moon how much weight would you need compared to the same dive on earth.
a) More weight
b) Less weight
c) The same weight
Please explain your answer.
If you displaced 1cu of water (64pds displacement on earth) on the moon it was have a bouyant force of about 15pds. But if you weigh 100pds on earth, youd only weigh 25pds on the moon. So on earth you would need 36pds lead, on the moon, you would only need 10pds.
The correct answer to the question you asked is B. The water you displace will weigh less, as the moon's gravity is less, therefore you would need less *weight*.
If, on the other hand, you were intending to ask how much *mass* you'd need, the answer would have been "c) The same mass". Assuming you were in the same gear as you use on your terrestrial dives, you would displace the same volume of water. That water may weigh less, but so will your weights. No changes are necessary.
Now, on the third hand, if you'd intended to ask how much *mass* you'd need, *and* you were trying to make it a pedantic trick question, the answer would be B, but only by the slightest of margins. As water is *not* perfectly incompressible, the reduced weight of the water would mean that the water is slightly less dense, therefore you would displace less mass at a given depth, measured with a line or what-have-you, as your terrestrial depth gauge would not be calibrated for lunar diving.
(As the weight of a given mass on the moon is approximately 1/6 of the weight on the surface of the earth of that same mass, the pressure will increase with depth at 1/6 the rate it does on earth. Assuming 1 ata ambient pressure at the surface, your "15 foot" safety stop would actually be at 90 feet, and the "130 foot" customary recreational limit would be 780 feet down. Your "30 fpm" ascent rate limit would be 180 fpm. It would, therefore, be a total *hoot* to dive the moon.)
If, on the other hand, you were intending to ask how much *mass* you'd need, the answer would have been "c) The same mass". Assuming you were in the same gear as you use on your terrestrial dives, you would displace the same volume of water. That water may weigh less, but so will your weights. No changes are necessary.
Now, on the third hand, if you'd intended to ask how much *mass* you'd need, *and* you were trying to make it a pedantic trick question, the answer would be B, but only by the slightest of margins. As water is *not* perfectly incompressible, the reduced weight of the water would mean that the water is slightly less dense, therefore you would displace less mass at a given depth, measured with a line or what-have-you, as your terrestrial depth gauge would not be calibrated for lunar diving.
(As the weight of a given mass on the moon is approximately 1/6 of the weight on the surface of the earth of that same mass, the pressure will increase with depth at 1/6 the rate it does on earth. Assuming 1 ata ambient pressure at the surface, your "15 foot" safety stop would actually be at 90 feet, and the "130 foot" customary recreational limit would be 780 feet down. Your "30 fpm" ascent rate limit would be 180 fpm. It would, therefore, be a total *hoot* to dive the moon.)
OK... it's not a trick question in the sense that I meant mass rather than weight. I did mean weight (as in you take the same "lead weights" back to earth).
Love the answers so far!
How about the more probable situation of diving on Mars?
Love the answers so far!
How about the more probable situation of diving on Mars?
shakeybrainsurgeon
Contributor
Here's a question for the advanced scientists, since everyone has made a distinction between weight and mass.
A body's mass m enters into two arenas: inertia, the force it takes to accelerate an object (F = ma) and gravity, the m that couples to the gravitational field (aka weight)... the space shuttle is weightless in space but I still couldn't move it.
It's assumed, but not a given, that inertial mass and gravitational mass are identical... question: what is the name given to the assumption that inertial m and gravitational m are the same and, as a bonus, what famous theory is based on that assumption?
A body's mass m enters into two arenas: inertia, the force it takes to accelerate an object (F = ma) and gravity, the m that couples to the gravitational field (aka weight)... the space shuttle is weightless in space but I still couldn't move it.
It's assumed, but not a given, that inertial mass and gravitational mass are identical... question: what is the name given to the assumption that inertial m and gravitational m are the same and, as a bonus, what famous theory is based on that assumption?
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