Thought experiment: How would ocean levels change?

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Homework Statement:
Hello, I was just reading some hypothetical questions online when I came across something I was unsure how to approach.
If everyone in the world drank a glass of water from the world’s oceans how would the water level change?
I am more specifically interested in how the water level would change if only one glass of water was removed?
Relevant Equations:
4*π*r^2
I am not too sure how to approach this, initially I thought it may be more of a calculus and related rates of change problem, i.e. finding an expression how the volume and height of water change with respect fo time. I do not know whether this is the right idea or how to progress any further along this train of though.
You are not given any information and it is more of a thinking off the top of your head kind of problem. I have attempted a rather abstract and muddled approach to solving this problem but I would be very grateful for any alternative advice or avenues to explore. It is just a thought experiment for fun really to stretch myself so I welcome any guidance.
Moreover, firstly I thought that one should consider finding the surface area of the Earth, which since I know the radius of the Earth is approximatly 6400km this would be 4*π*r^2=4*π*6400^2=5.15*10^8km^2
Then I thought about finding the volume of the earth being 4/3*π*r^3=4/3*π*6400^3~1.1*10^12 km^3

Then since I know that the Earth is approximately 71% covered by water this means that 71% of the surface area is water, which is about 362,100,000 km^2 of water.

A glass of water could roughly be said to contain 200ml.

I think one should find the total volume of water on Earth, I am not sure how to do this with the information I have so far. Then considering 1km^3 is equal to 1*10^12 litres one could convert this into volume into litres.
Should one then find what percentage 200ml would be of this volume, or subtract 200ml from this total volume of water on Earth?
Should one also consider the volume of water being evenly spread over the surface area of the Earth, finding how many million litres per km^2. From this one could convert from litres per km^2 to m^2 to approximate spreading this over a km^2 of Earth to find how much the water level would fall by.

Thank you to anyone who replies 😄
 

Answers and Replies

  • #2
phinds
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I think one should find the total volume of water on Earth, I am not sure how to do this
If you are only talking about oceans, then to a REALLY good approximation you can just use the assumptions that they all have vertical sides and all have the same uniform depth.

OOPS ... that's only useful if you KNOW the total volume, so yeah, getting it IS the problem. Try Google.
 
  • #3
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If you are only talking about oceans, then to a REALLY good approximation you can just use the assumptions that they all have vertical sides and all have the same uniform depth.

OOPS ... that's only useful if you KNOW the total volume, so yeah, getting it IS the problem. Try Google.
Thank you for your reply. I have looked online but found relatively little on the topic. Any suggestions, however abstract or bizarre are by all means welcome 😀!
 
  • #4
Lnewqban
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I believe that the differential height is independent from the total volume of water that is contained in the oceans.
I would only use the estimated oceanic surface and the calculated total volume of water to be removed.

Imagine a square-shaped pool of sides measuring 19029 km (which is the square root of the calculated wet area).
How deep that pool needs to be to contain the volume of those many glasses of water?
 
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  • #5
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I believe that the differential height is independent from the total volume of water that is contained in the oceans.
I would only use the estimated oceanic surface and the calculated total volume of water to be removed.

Imagine a square-shaped pool of sides measuring 19029 km (which is the square root of the calculated wet area).
How deep that pool needs to be to contain the volume of those many glasses of water?
Thank you for your reply. Ok, I did not realise that the height was independent of the total volume of water. Why would we use the square root of the calculated wet area?
Well the area of the square swimming pool would be the area of a cube which is 6a^2 where a = 19029. Thus, the area would be 2.17 *10^9 km^2.

To determine the depth would we use volume = length *depth *breadth
Therefore, depth=volume/(length*breadth)
So, depth = volume / area

I am still quite unsteady of my approach or of the above jottings
 
  • #6
Lnewqban
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Thank you for your reply. Ok, I did not realise that the height was independent of the total volume of water. Why would we use the square root of the calculated wet area?
Well the area of the square swimming pool would be the area of a cube which is 6a^2 where a = 19029. Thus, the area would be 2.17 *10^9 km^2.

To determine the depth would we use volume = length *depth *breadth
Therefore, depth=volume/(length*breadth)
So, depth = volume / area

I am still quite unsteady of my approach or of the above jottings
Think of calculating the thickness of the wall of a basket ball.
Now spread all that surface on a flat surface.

The square root use is just to give you the view of a square-shaped pool of equivalent surface area.
If that pool is shallow and only contains the volume of water in question, how deep would it be?
That depth equals your variation of level.
 
  • #7
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Think of calculating the thickness of the wall of a basket ball.
Now spread all that surface on a flat surface.

The square root use is just to give you the view of a square-shaped pool of equivalent surface area.
If that pool is shallow and only contains the volume of water in question, how deep would it be?
That depth equals your variation of level.
Thank you for your reply. Imagining the thickness of the wall of a basketball and spreading that onto a flat surface has been very helpful to visualise. Oh right, thank you for explaining why the square root is used. When you refer to the volume of water what would this be, since I was uncertain how to calculate this as I stated previously. Would I consider the volume of the Earth and the fact that the Earth is ~71% covered by water?
 
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  • #8
caz
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You have
1) the volume of the water removed, V
2) the amount the ocean level drops, h
3) the radius of the earth, r
4) the % of the earth‘s surface covered by water, 71
You need to come up with a relationship between these quantities. You stated above that depth=volume/area. Try applying it.
 
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  • #9
Lnewqban
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... When you refer to the volume of water what would this be, since I was uncertain how to calculate this as I stated previously....
That would be the volume of water to be removed from the oceans.
One glass or 20 ml times the world population, which, according to the following article, was estimated to have reached 7,800,000,000 persons as of March 2020.

https://en.m.wikipedia.org/wiki/World_population
 
  • #10
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You have
1) the volume of the water removed, V
2) the amount the ocean level drops, h
3) the radius of the earth, r
4) the % of the earth‘s surface covered by water, 71
You need to come up with a relationship between these quantities. You stated above that depth=volume/area. Try applying it.
Thank you for your reply. By relating these quantities with the formula given would this mean that the amount the ocean level drops=the volume of water removed/the 71 % surface area covered by water I.e h=V/A?
Then this could be rewritten in terms of r;
h=4/3πr^3/(0.71*4πr^2)
h= 4/3πr^3/2.84πr^2)
Applying the fraction rule;
h=4πr^3/3* 2.84πr^2
h= 4πr^3/8.52 πr^2

Simplify by cancelling π and applying the exponent rule to r^3/r^2=r^3-2;
h= 4r/8.52

Since r~6400km

h=4*6400/8.52
h=3004.69483 ~ 3005km

This is pure rambling but I would appreciate your thoughts.
 
  • #11
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That would be the volume of water to be removed from the oceans.
One glass or 20 ml times the world population, which, according to the following article, was estimated to have reached 7,800,000,000 persons as of March 2020.

https://en.m.wikipedia.org/wiki/World_population

Thank you for your reply. Ok, so it is the volume of water removed I.e 200ml. I was considering the second problem where only one glass of water is removed so I was disregarding the first scenario involving the world population, although thank you for the population figures. I thought that the second question was more challenging since removing such a small volume of water would have such a slight effect.
 
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  • #12
caz
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You should always make sure that you can visualize the situation. I've attached a drawing.

The top is why you can assume that the earth is flat. Once you gain experience, you will be able to recognize these situations. Ooops. I accidentally left the 4 out of the surface area equations. Since I am taking ratios it cancels out. Sorry.

The bottom is the image of the problem to be solved under this approximation. It is independent of the shape of S.

If I assume that one glass = 200ml = 200 (cm)^3 = 2*10^-13 km^3
I get a height drop of 5.5*10^-16 mm
For 7.8 billion people it is 4.3*10^-6 mm

Fun fact: a water molecule is about .27*10^-6 mm across.

As a side note, a glass of water is roughly 100mm tall. Given that is has to be spread over the entire earth, you would expect the ocean drop to be significantly smaller than this.
Ocean.png
 
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  • #14
jim mcnamara
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https://en.wikipedia.org/wiki/Water_distribution_on_Earth
Try 1.386 billion cubic kilometers as the estimated total volume of water on Earth. Note the word estimate.

Please look at the article and the graphics, then decide what water volume to use.

The answer you can get is not likely sensible, in that we cannot measure it. You cannot measure a billion cubic kilometers volume with subatomic dimension accuracy. Even if you can calculate it.
 
  • #15
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You should always make sure that you can visualize the situation. I've attached a drawing.

The top is why you can assume that the earth is flat. Once you gain experience, you will be able to recognize these situations. Ooops. I accidentally left the 4 out of the surface area equations. Since I am taking ratios it cancels out. Sorry.

The bottom is the image of the problem to be solved under this approximation. It is independent of the shape of S.

If I assume that one glass = 200ml = 200 (cm)^3 = 2*10^-13 km^3
I get a height drop of 5.5*10^-16 mm
For 7.8 billion people it is 4.3*10^-6 mm

Fun fact: a water molecule is about .27*10^-6 mm across.

As a side note, a glass of water is roughly 100mm tall. Given that is has to be spread over the entire earth, you would expect the ocean drop to be significantly smaller than this.
View attachment 274211
Thank you for your reply and for your brilliant diagram. How did you find the maximum depth of the ocean and the average depth of the ocean? Could you do this without researching ie. just using the information given? Also why is the surface area at the top of the ocean different to the bottom? Is the segment you have drawn representative of a wedge of the Earth?
 
  • #17
caz
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Thank you for your reply and for your brilliant diagram. How did you find the maximum depth of the ocean and the average depth of the ocean? Could you do this without researching ie. just using the information given? Also why is the surface area at the top of the ocean different to the bottom? Is the segment you have drawn representative of a wedge of the Earth?

I googled ocean depth and found:
https://oceanexplorer.noaa.gov/facts/ocean-depth.html

These cannot be calculated from the original info. They are “physical properties” and it is completely legitimate to look them up.

I assumed that the oceans went radially down, i.e. were 71% of the earths surface and that they were 71% of the surface of a sphere whose radius was determined by the bottom of the ocean. Since these two spheres have different radii, they have different surface areas. The wedge drawing of the Earth was to illustrate this.
 
  • #18
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I googled ocean depth and found:
https://oceanexplorer.noaa.gov/facts/ocean-depth.html

These cannot be calculated from the original info. They are “physical properties” and it is completely legitimate to look them up.

I assumed that the oceans went radially down, i.e. were 71% of the earths surface and that they were 71% of the surface of a sphere whose radius was determined by the bottom of the ocean. Since these two spheres have different radii, they have different surface areas. The wedge drawing of the Earth was to illustrate this.

Thank you for your reply. Is there any way to substitute the ocean depth and average ocean depth algebraically then if this cannot be calculated from the information given?

Oh right, so you are referring to one sphere with a radius of the Earth and another with a radius to the ocean floor?
 
  • #19
caz
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Thank you for your reply. Is there any way to substitute the ocean depth and average ocean depth algebraically then if this cannot be calculated from the information given?

Oh right, so you are referring to one sphere with a radius of the Earth and another with a radius to the ocean floor?

The ocean depth does not matter other than as a sanity check.

The first figure allows me to justify the second figure.

Look at my second figure. You are interested in calculating h. You know the volume removed V and you know the surface area S=0.71*4*pi*r^2. What is the relationship between h, V and S?

Yes.

You could calculate the initial volume of the ocean, subtract the amount removed and then calculate the change in height. The problem with doing this is that you are subtracting a small number from a big number. This can result in rounding errors and errors with significant digits. Coming up with a relationship between h,V and S avoids these difficulties.
 
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  • #21
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The ocean depth does not matter other than as a sanity check.

The first figure allows me to justify the second figure.

Look at my second figure. You are interested in calculating h. You know the volume removed V and you know the surface area S=0.71*4*pi*r^2. What is the relationship between h, V and S?

Yes.

You could calculate the initial volume of the ocean, subtract the amount removed and then calculate the change in height. The problem with doing this is that you are subtracting a small number from a big number. This can result in rounding errors and errors with significant digits. Coming up with a relationship between h,V and S avoids these difficulties.

Ah, ok. Also how have you calculated the average depth from the maximum depth?

Moreover, where you state the surface area at the bottom of the ocean is 4*0.71*pi*rb^2 is this calculating the surface area of the Earth minus the area of the ocean?(since the depth of the ocean is subtracted from the radius of the Earth)

Then is the surface area at the top of the ocean, being 4*0.71*pi*(rb +average depth)^3 equal to the entire surface area of the Earth since this includes the depth of the ocean?

So the volume removed is 2*10^-13 km^3 and the surface area of the ocean is S=0.71*4*pi*r^2.

When referring to the relationship between h, S and V would this be that the height of the Earth is equal to 2r, so that SA=4*pi*h/2^2 and V=4/3*pi*h/2^3?

I am not sure how to relate them otherwise but I do not believe I am thinking along the right lines here.

How would I embark upon the method to calculate the initial volume of the ocean, subtract the amount removed and then calculate the change in height? Thank you again for all of your diligent help 👍
 
  • #23
hmmm27
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How is this "Advanced Physics" ? Seems more like "Basic Arithmetic".

Also, if you wanted something useful, how about "How many people would get sick from drinking the nearest water, polluted by their unregulated industrial/agricultural runoff and raw human excrement ?"
 
  • #24
caz
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Ah, ok. Also how have you calculated the average depth from the maximum depth?

Moreover, where you state the surface area at the bottom of the ocean is 4*0.71*pi*rb^2 is this calculating the surface area of the Earth minus the area of the ocean?(since the depth of the ocean is subtracted from the radius of the Earth)

Then is the surface area at the top of the ocean, being 4*0.71*pi*(rb +average depth)^3 equal to the entire surface area of the Earth since this includes the depth of the ocean?

So the volume removed is 2*10^-13 km^3 and the surface area of the ocean is S=0.71*4*pi*r^2.

When referring to the relationship between h, S and V would this be that the height of the Earth is equal to 2r, so that SA=4*pi*h/2^2 and V=4/3*pi*h/2^3?

I am not sure how to relate them otherwise but I do not believe I am thinking along the right lines here.

How would I embark upon the method to calculate the initial volume of the ocean, subtract the amount removed and then calculate the change in height? Thank you again for all of your diligent help 👍
You cannot calculate the average depth from the maximum depth. There is not enough information.

Look at the second picture. You are taking a glass with volume V from the ocean. Since this will lower the ocean by a height h, you can view this volume as originally being distributed as a layer with a thickness h at the top of the ocean. The ocean has a surface area S. So V=Sh. Solving for h, gives h=V/S=200ml/(0.71*4*pi*r^2) where r=6400km.

The non-recommended way to do it. Look at the second picture. The volume of the ocean is the surface area multiplied by it’s average depth. Subtract the glass of water to get the remaining volume. Divide this by the surface area to get the new depth. Subtract this from the original depth to get the change in height. You will find that you will have round off problems if you do this numerically. If you do this analytically you will end up with h=V/S as above.
 
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You cannot calculate the average depth from the maximum depth. There is not enough information.

Look at the second picture. You are taking a glass with volume V from the ocean. Since this will lower the ocean by a height h, you can view this volume as originally being distributed as a layer with a thickness h at the top of the ocean. The ocean has a surface area S. So V=Sh. Solving for h, gives h=V/S.

The non-recommended way to do it. Look at the second picture. The volume of the ocean is the surface area multiplied by it’s average depth. Subtract the glass of water to get the remaining volume. Divide this by the surface area to get the new depth. Subtract this from the original depth to get the change in height. You will find that you will have round off problems.
Thank you for your reply. How did you find the average depth then?

So just to check would the surface area be 0.71*4*pi*6400^2?

Ok, attempting the non-recommended method... what value for the volume would I be subtracting 2*10^-13 km^3 from? i.e. how would I find the ocean volume using the information given. Also, is the original depth taken to be 3.7km?

Would volume of oceans be .0.71*4/3*pi*6400^3~1.22*10^8 ?
 
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