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Question about the weight of things deep in the ocean

If a 20,000 ton displacement ship sinks to xx,000 feet to the sea floor, will it still weigh out at 20,000 tons at that depth? Assume no buoyant compartments or similar conditions.

Thanks,
Bob

I'm smelling a trick question the likes that even Supreme Court Justice Ketanji Brown Jackson would answer as "I can't answer that as I'm not a naval architect."

gravity is pretty much the same, varies slightly with altitude, but yeah, mas o menos, lo mismo

I got your naval architect right here, and raise you a rocket surgeon.

Quote:

Originally Posted by VenezianBlau 87 (Post 11735692)

If a 20,000 ton displacement ship sinks to xx,000 feet to the sea floor, will it still weigh out at 20,000 tons at that depth? Assume no buoyant compartments or similar conditions.

Thanks,
Bob

So you're asking if the weight of the ship would increase/decrease as it sank? Presumably because of the immense pressure?

I have to assume that 1000# block of iron is going to weigh practically 1000# whether it's in orbit in the ISS, sitting in your driveway or at the bottom of the Mariana trench. Obviously, the weight of the water above the ship would be pushing down, but that water (much like air) is pushing on the ship from all directions. Gravity is still pulling on the metal that the ship is made of with roughly the same force.

A quick google has provided that the difference in gravity from sea level to the top of Everest is 0.25%, and even at 200,000km, it's still 94% of sea level.

Now hang on just a doggone minute..... a piece of a 2x4 floats because it displaces the weight of more water than it weighs.... am I right?

<iframe width="560" height="315" src="https://www.youtube.com/embed/IdBkVRjEr4Y" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture" allowfullscreen></iframe>

Yeah, the pressure down there has nothing to do with it. Only the distance to the center of the earth.

It is closer, a thousand feet under the sea.

Thanks for your responses...:)

I read that salt water is more buoyant than fresh...and that water pressure is "more or less" applied evenly at any depth (thanks java and Steve)
No one's weighing things on the surface and the deep ocean, right?
I suspect the salt concentration reduces weight on the sea floor.
Not sure where the difference is offset if water can't be compressed.

Quote:

Originally Posted by VenezianBlau 87 (Post 11735692)

If a 20,000 ton displacement ship sinks to xx,000 feet to the sea floor, will it still weigh out at 20,000 tons at that depth? Assume no buoyant compartments or similar conditions.

Thanks,
Bob

What weighs more, a ton of lead or a ton of feathers?

If you ignore the buoyancy of the ship and consider pure gravity, the ship will weigh less down there because some of the earth's gravity is in the water which is now above the ship, the gravity of the water thereby now acting upwards upon it.

Sent from my SM-G988B using Tapatalk

The mass doesn't change, obviously, but the weight (force) would be less due to buoyancy.

Quote:

Originally Posted by RobFrost (Post 11735736)

I don't think there's going to be enough water to cause much, if any impact to the weight. The diameter of the earth is ~7900 miles. The deepest part of the ocean is almost 36,000 feet, so roughly 7 miles. I don't think 7 miles of water is going to be enough to counteract 3950 miles of "earth".

Quote:

Originally Posted by hbueno (Post 11735738)

The mass doesn't change, obviously, but the weight (force) would be less due to buoyancy.

I don't think there's going to be much effect due to bouyancy on an iron/steel ship at the bottom of the ocean (assuming all hollow spaces have filled with water).

https://physics.stackexchange.com/questions/132935/gravity-beneath-sea-level

this sounds pretty good.

Quote:

I'm assuming the question is "How does gravity vary inside the Earth?"

I'll look at this three ways:

From a rather empirical set of corrections used by geologists for work near the surface of the Earth,
From a physicist's spherical cow perspective, and finally,
From empirical data collected by geologists and geophysicists that give a better picture of gravitation inside the Earth.

Free air and Bouguer corrections

Linearizing Newton's law of gravitation ��=��2
about height at sea level yields ��(ℎ)=��0−2��0��earthℎ, or Δ��(ℎ)=−3.084×10−6s−2ℎ, where ℎ

is in meters. This is the free air correction. It yields an estimate for gravitational acceleration for a balloon flying a small distance above the Earth.

What if instead the intervening space was filled with some material with a density ��
? One way of accounting for that stuff is to pretend that it's an infinite flat of material. This leads to the Bouguer correction, ��(ℎ,��)=4.194×10−10m3kg−1s−2��ℎ, where �� is in kilograms per cubic meter and ℎ

is in meters.

That's for going above sea level. What about below? The free air correction still applies as-is (with height being negative). The Bouguer correction needs a correction. Since that correction is based on a flat plate model, gravity inside the Earth needs a double Bouguer correction.

Assuming sea water with a density of 1025 kg/m3, the above corrections leads to gravitational acceleration under the sea increasing by 2.224×10-6 m/s2 per meter of depth in sea water.
Earth as a sphere with density that varies radially

Per Newton's shell theorem, gravitation inside a sphere whose density depends only you radial distance from the center is given by ��(��)=��(��)��2
, in other words, it's just Newton's law of gravitation, but that ��(��) represents only the mass that is inside a sphere of radius ��

. The stuff above contributes nothing.

Differentiating with respect to distance ��
yields ��=��3(��−2��(��)). For a spherically symmetric object, ��=4��(��)��2, where ��(��) is the density at a distance �� from the center of the Earth. Denoting the mean density ��¯(��) as ��¯(��)≡��(��)��(��)=��(��)4/3��3 lets us write ��(��)=43��¯��3. Applying these results to the expression for �� yields ��=4��(��(��)−23��¯(��)). We want the effects of increasing depth below the surface rather than increasing distance from the center. That's simple: Just negate the above:
��=4��(23��¯(��)−��(��))
where ��

is depth below the surface.

Applying the above to sea water with a density of 1025 kg/m3 and using 5515 kg/m3 as the mean density of the Earth leads to gravitational acceleration under the sea increasing by 2.224×10-6 m/s2 per meter of depth in sea water, the same result as with the empirical models.
Gravitation inside the Earth

One thing pops right out from the physics-based derivation: Gravitational acceleration increases with increasing depth if the local density ��(��)
is less than 2/3 times the mean density ��¯(��) of all the stuff at a depth ��

or greater. It decreases if the local density is at least 2/3 of the mean density. Surface rock has a density a bit less than 1/2 the mean density of the Earth, so gravitational acceleration initially increases with increasing depth. In fact, gravitational acceleration inside the Earth is above the surface value until just below the core-mantle boundary. The global maximum occurs at the core-mantle boundary. The Earth's core is very, very dense stuff. It's mostly iron and it is highly compressed.

This is exactly what is seen in the Preliminary Reference Earth Model. Below are some links to the PREM.

Wikipedia article: http://en.wikipedia.org/wiki/Preliminary_reference_Earth_model

Model parameters: http://geophysics.ou.edu/solid_earth/prem.html

Paper: Dziewonski, A. M., & Anderson, D. L. (1981). "Preliminary reference Earth model." Physics of the earth and planetary interiors, 25(4), 297-356.

Finally, here's a graph of gravitational acceleration above and inside the Earth.
https://upload.wikimedia.org/wikiped...tyPREM.svg.png

I guess if you want to see the equations, you'll have to follow the link. Interesting stuff.

I need a free body diagram :D

Quote:

Originally Posted by RobFrost (Post 11735736)

Quote:

Originally Posted by hbueno (Post 11735738)

The mass doesn't change, obviously, but the weight (force) would be less due to buoyancy.

These ^

Some of the first gravity experiments were done against a mountain (to Frost's point) to measure the affect of mass.

And relative specific gravity of steel is 7.8 (to hbueno's point) meaning that the weight of the steel will be ~7/8ths of what it is on land.

Just looked up the mountain gravity experiment - https://en.wikipedia.org/wiki/Schiehallion_experiment

Someone wanted a "free body diagram"? This is the best I could do:

http://forums.pelicanparts.com/uploa...1657084465.jpg

A ship on the floor of the sea would be really difficult to lift... You'd have to deal with the suction.
(Depending on how long it's been there)

Quote:

Originally Posted by SCadaddle (Post 11735788)

Someone wanted a "free body diagram"? This is the best I could do:

http://forums.pelicanparts.com/uploa...1657084465.jpg

Her hair isn't parallel!