Water dissolving and water removing. There is water at the bottom of the ocean.
May 10, 2011 2:00 PM Subscribe
I have a perfectly spherical lead weight that weighs 1 kilogram. I drop it from the surface of the ocean directly into the Mariana Trench. How long does it take to hit the bottom, and how fast is it traveling when it hits?
For the sake of the question, let's say I drop the sphere directly from sea level, with no additional downward push, and it's going to land 35,994 feet at the bottom of the trench, known as the Challenger Deep. Let's assume there's a straight, unobstructed shot from the surface to this spot. Let's also assume there are no currents, waves or interference. Just the weight, the water and gravity.
Go, go, MetaFilter physics.
For the sake of the question, let's say I drop the sphere directly from sea level, with no additional downward push, and it's going to land 35,994 feet at the bottom of the trench, known as the Challenger Deep. Let's assume there's a straight, unobstructed shot from the surface to this spot. Let's also assume there are no currents, waves or interference. Just the weight, the water and gravity.
Go, go, MetaFilter physics.
Too busy to crunch the numbers right now, but here's some background reading to get you started.
posted by schmod at 2:10 PM on May 10, 2011
posted by schmod at 2:10 PM on May 10, 2011
The Titanic sank in about 12,000 feet, and various web sites claim it took 2 hours to sink, so that's equivalent to 6 hours for 36,000 feet, as Meta-4 said.
posted by lukemeister at 2:12 PM on May 10, 2011
posted by lukemeister at 2:12 PM on May 10, 2011
Play with Wolfram Alpha for a decent approximation. I started with these values and got just over an hour, 3m/s at impact.
posted by t_dubs at 2:22 PM on May 10, 2011 [1 favorite]
posted by t_dubs at 2:22 PM on May 10, 2011 [1 favorite]
t_dubs' answer seems to be a better answer, especially as the Trieste took four hours and 48 minutes in a controlled descent, making my first post seem way off.
posted by Meta-4 at 2:35 PM on May 10, 2011
posted by Meta-4 at 2:35 PM on May 10, 2011
1 kg of lead forms a sphere 2.76 cm in radius, giving a projected surface area of 95.7 cm^2 / 2 (only one half is projected) = 47.9 cm^2.
Using t_dubs' Wolfram link, I get a final answer of 44 minutes.
Assuming pure water, no currents, etc.
posted by supercres at 2:42 PM on May 10, 2011
Using t_dubs' Wolfram link, I get a final answer of 44 minutes.
Assuming pure water, no currents, etc.
posted by supercres at 2:42 PM on May 10, 2011
I'm sorry; the projected surface area is just the 2-d area of the largest cross-section of the sphere. So that's 23.93 cm^2. This is the final answer, then: 31 minutes.
posted by supercres at 2:46 PM on May 10, 2011
posted by supercres at 2:46 PM on May 10, 2011
That's going to be really difficult to calculate, because the density of the water changes on the way down. Part of that is due to pressure, but there are also temperature changes and changes in salinity, and those are not linear. They're also not consistent. They change from place to place, and from time to time. Seasons and weather can affect them, as well as special events like tsunamis.
Submarines carry sophisticated sensors for that kind of thing, because it was learned a long time ago that sonar would bounce off of the thermocline (and to a lesser extent, the halocline), and it was safer for the sub to hide lower than that.
But the density due to pressure is also an issue. Your lead ball won't compress as well as water does, so its relative weight (and thus the force pushing it downward) will be reduced as it sinks lower. (Its density will rise, but the water density will rise faster.)
posted by Chocolate Pickle at 3:13 PM on May 10, 2011 [5 favorites]
Submarines carry sophisticated sensors for that kind of thing, because it was learned a long time ago that sonar would bounce off of the thermocline (and to a lesser extent, the halocline), and it was safer for the sub to hide lower than that.
But the density due to pressure is also an issue. Your lead ball won't compress as well as water does, so its relative weight (and thus the force pushing it downward) will be reduced as it sinks lower. (Its density will rise, but the water density will rise faster.)
posted by Chocolate Pickle at 3:13 PM on May 10, 2011 [5 favorites]
Actually, I don't know if lead is better at resisting pressure than water. It could be the other way, with the lead compressing more, and thus the relative weight increasing.
posted by Chocolate Pickle at 3:15 PM on May 10, 2011
posted by Chocolate Pickle at 3:15 PM on May 10, 2011
Note that salt water is slightly more dense than freshwater, 1027 kg/m3 vs 1000 kg/m3, resp. It doesn't affect the end result by more than a minute, but it would be a factor for dropping something into the Mariana Trench.
posted by Blazecock Pileon at 3:15 PM on May 10, 2011
posted by Blazecock Pileon at 3:15 PM on May 10, 2011
Pressure at the bottom of Marianas Trench is over 1000 bars, so the lead sphere must compress somewhat, as Chocolate Pickle said. Does anyone have an idea how much?
posted by lukemeister at 3:23 PM on May 10, 2011
posted by lukemeister at 3:23 PM on May 10, 2011
Pressure at the bottom of Marianas Trench is over 1000 bars, so the lead sphere must compress somewhat, as Chocolate Pickle said. Does anyone have an idea how much?
Negligibly. The bulk modulus of Pb is 46 GPa, so a pressure of around 1100 bar will reduce its volume by around 0.2%, its cross section by about two-thirds of that.
posted by Mapes at 4:31 PM on May 10, 2011 [4 favorites]
Negligibly. The bulk modulus of Pb is 46 GPa, so a pressure of around 1100 bar will reduce its volume by around 0.2%, its cross section by about two-thirds of that.
posted by Mapes at 4:31 PM on May 10, 2011 [4 favorites]
Mapes, how much does water compress at that pressure?
In other words, what is the relative weight of the lead (after buoyancy) at the top and at the bottom?
posted by Chocolate Pickle at 6:04 PM on May 10, 2011
In other words, what is the relative weight of the lead (after buoyancy) at the top and at the bottom?
posted by Chocolate Pickle at 6:04 PM on May 10, 2011
Formula's for water density at a particular temp and pressure can be found here. I'm not sure how well these hold up for salt water though and they're a PITA to calculate.
posted by Confess, Fletch at 7:03 PM on May 10, 2011
posted by Confess, Fletch at 7:03 PM on May 10, 2011
The assumption of a drag coefficient=0.47 is dependent on the flow being fully turbulent. My unconfirmed estimate of the Reynolds number (based on numbers I'm finding on the internet and without rigourous checking of the math) is 175000 where the temperature is near 0 deg C, which puts it in an interesting spot on the curve--right in the transition between laminar and turbulent flow.
Like I said, I'm not 100% confident in the number, but the drag estimate maybe significantly off.
posted by cardboard at 7:49 PM on May 10, 2011
Like I said, I'm not 100% confident in the number, but the drag estimate maybe significantly off.
posted by cardboard at 7:49 PM on May 10, 2011
It will take 49.71 something seconds to hit the bottom. Here is the code you can use to do the math yourself.
I apologize for the poor commenting but it is mostly self-explanatory. (I hope.)
posted by hariya at 4:29 AM on May 11, 2011
I apologize for the poor commenting but it is mostly self-explanatory. (I hope.)
posted by hariya at 4:29 AM on May 11, 2011
The assumption of a drag coefficient=0.47 is dependent on the flow being fully turbulent.
That's not what that link is saying. The drag coefficient varies between 0.4-0.6 for Re between 300-300,000, in the regime preceding turbulent flow.
posted by Mapes at 5:33 AM on May 11, 2011
That's not what that link is saying. The drag coefficient varies between 0.4-0.6 for Re between 300-300,000, in the regime preceding turbulent flow.
posted by Mapes at 5:33 AM on May 11, 2011
Doesn't terminal velocity come in to play here? No way it would reach the bottom in 49.71 seconds. Maybe in a vacuum, but definitely not through water.
posted by Grither at 5:35 AM on May 11, 2011
posted by Grither at 5:35 AM on May 11, 2011
Oops, ginormous brain fart happened on the way here. I forgot to notice the Reynolds numbers. I should use the turbulent drag equation, not form drag.
Here is the correct version of the code.
The spehere reaches the bottom in 2007.15 seconds.
posted by hariya at 5:54 AM on May 11, 2011
Here is the correct version of the code.
The spehere reaches the bottom in 2007.15 seconds.
posted by hariya at 5:54 AM on May 11, 2011
2807.145 seconds not 2007.15 seconds.
It is not like I am trying to make first post or anything. Damn!!
posted by hariya at 6:11 AM on May 11, 2011
It is not like I am trying to make first post or anything. Damn!!
posted by hariya at 6:11 AM on May 11, 2011
That's not what that link is saying. The drag coefficient varies between 0.4-0.6 for Re between 300-300,000, in the regime preceding turbulent flow.
Fair enough. 0.47 assumes laminar flow. The problem remains that 0.47 is based on assumptions that have not yet been validated.
posted by cardboard at 6:40 AM on May 11, 2011
Fair enough. 0.47 assumes laminar flow. The problem remains that 0.47 is based on assumptions that have not yet been validated.
posted by cardboard at 6:40 AM on May 11, 2011
This thread is closed to new comments.
posted by Meta-4 at 2:08 PM on May 10, 2011 [3 favorites]