Ocean water at depth is higher in salinity so that may impact buoyancy as well. Perhaps not significantly but we're a bunch of nit pickers anyway.

Bottom line, there is no technology to actually measure the difference in mass or weight. It should indeed weigh just a little less as it is closer to the center of the Earth. It is just a mental exercise. No doubt if one knew the precise weight of the ship in dry dock, the weight could be calculated with some fancy math at whatever depth you pick. No one really cares enough to do it, and it is way beyond my math skills.

Actually, it weighs more when closer to the center of the earth.

FiFY

Oh, and mass will not change unless it's converted to energy. Weight is what can/will change, but not by much. Like I said before, at 200,000 kilometers (120,000 miles) above the Earth the weight will still be 94% of what it is at sea level. The weight at the top of Mt Everest will be 0.25% less than the weight at sea level.

displacement of a vessel is an indirect method for assessing the weight of the ship by calculating the weight of the water which is replaced or pushed aside by the ships hull when it is floating, To laymen and for most purposes the weight and mass are rarely considered to be the separate metrics that they really are

a floating 20k ton displacement means that 20k tons of water is pushed aside to support the vessel

weight and mass are 2 different but related quantities

in the British Engineering System mass is measured in slugs

on the earths surface at nominal sea-level
20,000T/32 = 632k slugs

when the ship sinks no matter where it is, as long as it doesn't break up, it always has 632k slugs of mass


when it sinks the weight varies according to Newton F = GMM/r^2 so yes the gravitational pull on the solid structure decreases proportional to r^2
but
on the surface the hull is filled w/ relatively light weight air

at depth the air is replaced by much denser and heavier water so the next q is do you count the the mass of the water inside the sunken vessel in your estimation, for some purposes the answer is, yes, for instance suppose you wanted to know the force necessary to move the boat laterally or to lift it before hitting bottom. Here the interior volume of water is being moved along w/ the ship structure so it is necessary to account for it.

but

the answer most of the time is, no, as usually what you really want to know is the mass of the solid structure + tankage of the vessel whether it is afloat or not

in the vernacular weight and mass are often bandied about w/ no true concern for their actual definition

Best way to understand it is to consider how a submarine works

floating on the surface it might displace ψ tons, but a bathroom scale type weighing device would read 0


to submerge it replaces air in tanks surrounding the pressure hull w/ denser/ heavier sea water until it attains the buoyancy state that it wants, usually close to or at neutral buoyancy, here the mass of the enclosed volume of the sub equals the mass of the equivalent volume of sea water, the mass of the sub is still the same as when it was on the surface but x amount of air has been replaced w/ y amount of water and when the weight of the extra ballast water is added in the sub is far more massive but the mass only affects propulsion in that the entire mass has to be accelerated, If a bathroom scale were placed under the sub to weigh it it would read 0, same as the scale would read on the surface where the neutral bouncy state also existed

Good call, relative density/specific gravity of steel at 7.8

RD = weight of the object in air / weight of object in air - weight of object in water

Of course, the specific gravity of salt water is different from freshwater. Also I believe temperature makes a difference, but that's probably a good start.

If the difference between sea level vs top of Mt Everest is only 0.25%, then I suspect the difference at the bottom of the Mariana Trench would probably be similar, so yes, the boat would weigh less at the bottom of the trench based on specific gravity of steel, and probably even less due to the salinity of the sea water.

https://www.britannica.com/science/seawater/Density-of-seawater-and-pressure
It looks like the density of sea water @ 10km is 1.07, so the steel would be about 87% of the weight in fresh water, but possibly another 7% lighter due to the salinity farther down, maybe more if it was even deeper.

You guys are getting sidetracked. Weight is a measure related to two things, mass and strength of the gravitational field where the mass is. The mass of the ship doesn’t change, but the gravity constant does.

Buoyancy, density, salinity of the water, all that is just irrelevant nonsense.

Right, mass wouldn't change. Gravitational constant changes with altitude. Since we are talking about weight, if you could balance the ship on a scale at the bottom of the Mariana Trench, due to the relative densities of sea water and the materials that the ship is made of would see a change due to bouyancy. If the ocean was Mercury, the ship would never sink.

One point of view, no mass change, so weight would only change due to change in altitude.

Second point of view, assuming no bouyancy due to air pockets, then the material of the ship itself would have some bouyancy (even if it didn't float) and hence the weight would be decreased due to that bouyancy, despite the weight increasing due to change in altitude.

That's how the specific gravity of solids with a specific gravity (relative density) >1.0 is measured. A sample is submerged in fresh water and weighed. The weight is decreased by a certain amount, and that amount gives you them relative density.

Nope. Just increases density of the water - has no effect on the mass of an object or the gravitational pull on it. 33 foot depth of saltwater is about 34 foot of depth of freshwater is about 1atm pressure (14.7psi)

I see what you mean. My last two lines above are where I went wrong.
I hadn't even considered changes in gravitational force!
Thanks all! This is, probably not surprisingly, only the tip of things I don't understand.

http://forums.pelicanparts.com/uploa...1657133135.jpg

Questions are the beginning of wisdom.

Based on the bolded portion, I assume you mean, the ship is COMPLETELY flooded with no air pockets, and no materials that would normally float (wood, foam, etc... for instance) You want to know if the weight of the materials is the same.
Actually, if it's possible to weigh a material at the bottom of the ocean, then because of the relative densities, it would absolutely have a positive (upward) force applied due to bouyancy when submerged in salt water. THere would also be an increased force applied downward (gravity) that is greater than the force applied to the same object at sea level.

It's that ability to weigh an object under water which enables us to measure the specific gravity (relative density) of substances that can be submerged in water.

Based on the previous information that I found that says that the effect of gravity on an object at the top of Mt Everest is 0.25% less. I'm going to say that the effect of gravity on an object at the bottom of the deepest part of the ocean may be similar but opposite. If you then look at the specific gravity (relative density) of steel, 7.8, then steel submerged in fresh water will only weigh about 87% of what it weighs in air. If submerged in salt water, the weight would be even less. The mass hasn't changed, but there is a force exerted on the object by water which causes it to weigh (based on a scale of some sort) less.

If you were able to put a ship in dry dock and weigh it, and then you were able to sink it in the ocean, so that it was just barely covered in sea water, and you had a way (no pun intended) to measure the weight of the submerged boat, it would weigh measurably less than the boat in dry dock (assuming completely flooded with no air pockets. That's exactly how specific gravity is measured, by weighing objects before and after being submerged in water. Changing from distilled water to salt water would then make a difference that could be as little as 2.5% or as high at 7% when 10km down.

http://rogermarjoribanks.info/wp-con...y-1024x575.jpg

https://scientificknowledge.in/wp-co...eriment-04.png

https://media.cheggcdn.com/media/982.../phprIfwbq.png

https://i1.wp.com/www.cbsetuts.com/w...12%2C514&ssl=1

That weight doesn’t change, the buoyant force just opposes it.

F=ma

Weight is force F, which is the product of mass x acceleration.

The weight is less underwater because the buoyant force is pushing up while gravity is pulling down.

I agree and disagree.
The mass doesn't change. Yes, the bouyant force is another force acting on the object in the opposite direction to the force of gravity. The weight will change by some (very small) amount based on "altitude" (whether it's up or down).

But certainly if you try to measure the weight of the object, it will measure differently at the bottom of the ocean than it would in a hole in the ground that was as deep as the bottom of the ocean due to the bouyancy.

The first definition of weight on the Merriam Webster website is, "the amount that a thing weighs." If you then check the definition of "weighs," the first definition is "to ascertain the heaviness of by or as if by a balance."

Using those definitions, the bouyant effect of a liquid on an object would affect the weight of the object. If you consider weight to be "mg" (mass x gravitational constant) and you know the mass (no need to measure it) then yes, the weight using that formula would not change unless either the gravitational constant changes or the mass changes.

Technically, our weight probably changes whether it's midnight on a full moon or midday on a new moon because the gravity of the moon does impart some (albeit REALLY small) force on us. Clearly there is some impact from the gravity of the moon or we wouldn't have tides. Just another funny to consider datapoint due to forces.

It's all about definitions.

Is an item's weight what you can measure with a scale?
Is an item's weight what you calculate using a known mass and a known gravitational constant, then the bouyant effect from water doesn't matter. (net weight vs gross).
If we are going to use W=mg, then are we considering ONLY g (like above) or are we looking at a free body diagram of forces.

It really depends on how you define weight. The classical layman's context is "force normal to ground" than buoyancy matters, if you use rigorous engineering context of force exerted by gravity than you wouldn't include the buoyancy.

A bit of a semantics thing. I have designed instruments that measure weight on bouyant systems (drill collars) for 20 years, and the one thing that is consistent is people mean different things when they say weight. It is better to more precisely define what one means by weight than have a semantic argument over what the term means in isolation.

exactly

We were talking about physics here. In physics, weight is defined using two inputs, mass and gravity. That’s it.

Buoyancy is another force, altogether.

Whatever’s in the dictionary that is in customary usage by the masses is somewhat irrelevant because very few of them can pass a basics physics exam. I would say weight is one of the most misunderstood terms, as most people equate it with mass.

I pulled into Nazareth, feeling ‘bout half past dead…