<p>I am assuming you are asking for the case of sea level being 200m higher <strong>and</strong> in isostatic equilibrium. In that case we can make use of Airy's isostasy model: <a href="https://en.wikipedia.org/wiki/Isostasy#Airy" rel="noreferrer">https://en.wikipedia.org/wiki/Isostasy#Airy</a></p>

<p>Applied to a water column over the mantle, you have to replace $\rho_c$ with your $\rho_w$. The total increase in ocean depth is $x = b_1 + h_1$, where $h_1$ are the 200 m extra sea level and $b_1$ is the depth increase to isostasy. Using the equation given in the link: 
$$
x =  \frac{h_1 \rho_w}{\rho_m - \rho_w} + h_1,
$$</p>

<p>giving $x =$ 287 m</p>

CC BY-SA 3.0
2017-11-02T12:51:29.417Z
2017-11-02T12:51:29.417Z
ye-ti-800
<p>I'm trying to solve the following problem.
The sea level in the past was 200 m higher than today. The seawater became in isostatic equilibrium with the ocean basin. what is the increase in the depth $x$ of the ocean basins? Water density is $\rho_w = 1000\ kg\ m^{-3}$, and mantle density is $3300\ kg\ m^{-3}$.</p>

<p>Using the compensation column, I reach:</p>

<p>$$
x = (\rho_w * 200\ m)/ 3300 = 60.60\ m
$$</p>

<p>but normally I expected to find 290 m.</p>

<p>Can someone explain to me what's wrong?</p>

CC BY-SA 3.0
2017-11-02T12:35:08.153Z
@user1166251 to clarify: you are asking for *x* for the case where sea level is 200m higher AND in isostatic equilibrium?
ye-ti-800
CC BY-SA 3.0
2015-07-14T00:24:19.780Z
This won't solve all your problems, but the density of seawater is higher than that of fresh water. It's partly due to salinity. On this site seawater density varies from 1020 to 1035 kg/m3. (Seawater density)[http://hypertextbook.com/facts/2002/EdwardLaValley.shtml]
Fred
CC BY-SA 3.0
2015-07-14T00:40:14.293Z
Thank you for your comment. The exercise is based on Archimedes's principle rho_water*height water = rho_mantle*height mantle. I'm not sure the density of water has anything to do here. When I compare the two columns with the compensation depth I don't find the result expected. I'm missing something but don't know what.
user1166251
CC BY-SA 3.0
2015-07-14T13:11:05.393Z
This isn't my field of expertise but if you look at the picture [here] (https://en.wikipedia.org/wiki/Isostasy#/media/File:Backstripping_and_eustasy_correction.jpg) I'm thinking your case is like the middle picture. In the eqn for correct depth there's a term *delta-SL(rho_/(rho_m - rho_w))*. This will give you 286.86, which is near the 290 m you expected. If I could explain it, I'd give you a detailed answer.
Fred
CC BY-SA 3.0
2015-07-14T15:15:29.743Z
@user1166251 *rho_water* **is** the density of water.
casey
CC BY-SA 3.0
2015-07-14T19:01:02.957Z
@Fred Thank you for the link, that is actually very useful :)
user1166251
<geodynamics><homework><isostasy>
How to calculate hydrostatic equilibrium?
user1166251
# How to calculate hydrostatic equilibrium?

(Asked by: username_0 on 2015-07-13)

I'm trying to solve the following problem.
The sea level in the past was 200 m higher than today. The seawater became in isostatic equilibrium with the ocean basin. what is the increase in the depth $x$ of the ocean basins? Water density is $\rho\_w = 1000\ kg\ m^{-3}$, and mantle density is $3300\ kg\ m^{-3}$.

Using the compensation column, I reach:

$$
x = (\rho\_w \* 200\ m)/ 3300 = 60.60\ m
$$

but normally I expected to find 290 m.

Can someone explain to me what's wrong?

**Tags:** geodynamics, homework, isostasy

> **Comments:**
> **username_2**: This won't solve all your problems, but the density of seawater is higher than that of fresh water. It's partly due to salinity. On this site seawater density varies from 1020 to 1035 kg/m3. (Seawater density)[http://hypertextbook.com/facts/2002/EdwardLaValley.shtml]  
> **username_0**: Thank you for your comment. The exercise is based on Archimedes's principle rho_water*height water = rho_mantle*height mantle. I'm not sure the density of water has anything to do here. When I compare the two columns with the compensation depth I don't find the result expected. I'm missing something but don't know what.  
> **username_2**: This isn't my field of expertise but if you look at the picture [here] (https://en.wikipedia.org/wiki/Isostasy#/media/File:Backstripping_and_eustasy_correction.jpg) I'm thinking your case is like the middle picture. In the eqn for correct depth there's a term *delta-SL(rho_/(rho_m - rho_w))*. This will give you 286.86, which is near the 290 m you expected. If I could explain it, I'd give you a detailed answer.  
> **username_3**: @username_0 *rho_water* **is** the density of water.  
> **username_0**: @username_2 Thank you for the link, that is actually very useful :)  
> **username_1**: @username_0 to clarify: you are asking for *x* for the case where sea level is 200m higher AND in isostatic equilibrium?  

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## 1 answers

### Answer by username_1 on 2017-11-02. Score: 5 | Accepted answer

I am assuming you are asking for the case of sea level being 200m higher **and** in isostatic equilibrium. In that case we can make use of Airy's isostasy model: <https://en.wikipedia.org/wiki/Isostasy#Airy>

Applied to a water column over the mantle, you have to replace $\rho\_c$ with your $\rho\_w$. The total increase in ocean depth is $x = b\_1 + h\_1$, where $h\_1$ are the 200 m extra sea level and $b\_1$ is the depth increase to isostasy. Using the equation given in the link:
$$
x = \frac{h\_1 \rho\_w}{\rho\_m - \rho\_w} + h\_1,
$$

giving $x =$ 287 m


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