Fluid mechanics problem: The extent to which an iron deposit can influence water level

what i understand i have to do is find the formula for the potential energy with this iron deposit and substract it from the potential energy before this iron deposit was placed there. I found the solution but i do not undsrstand why the formula for potential gravitation energy was used.

what i did was $U=($\rho$ $\pi$ g r^3 \frac{4}{3} )( \frac{1}{h} - \frac{1}{(h+r)})$
hmmm i dont know why it doesnt work here i tried to write it on discord using LaTex

germoneyswis said:

i do not undsrstand why the formula for potential gravitation energy was used.
what i did was $$U=(\rho\pi g r^3 \frac{4}{3} )( \frac{1}{h} - \frac{1}{(h+r)})$$
hmmm i dont know why it doesnt work here i tried to write it on discord using LaTex

You needed to double up the dollar signs at the ends and remove the embedded ones.

I don’t understand the logic behind your attempt. Please explain your reasoning.

i substracted the gravitational potential energy caused by the iron deposit from the potential energy before it was placed there (mgh) to find H

germoneyswis said:

i substracted the gravitational potential energy caused by the iron deposit from the potential energy before it was placed there (mgh) to find H

I still cannot relate that to the expression you wrote:
$$U=(\rho\pi g r^3 \frac{4}{3} )( \frac{1}{h} - \frac{1}{(h+r)})$$
Both of those terms concern potential energy in the presence of the iron deposit. The second is the change in PE at the (unrisen) sea surface caused by the deposit, while the first is the change in PE at r below the (unrisen) sea surface caused by the deposit, which doesn’t seem to have any possible relevance.

does the solution provided in the pdf make more sense?

germoneyswis said:

does the solution provided in the pdf make more sense?

Yes, except that there seems to be a confusion as to whether H is the extra height (as in the final boxed equation) or the total height, including h, as in the preceding equation.
To be clearer, I would replace H in the boxed equation with ##\Delta h##.

Edit: I inadvertently squared all the distance denominators as though the expression was for force, not work. Corrected below:

Here's the reasoning:
The presence of the iron blob creates a loss of GPE at height h. The extent of that loss is ##\frac {GMm}{h+r}## where M is the extra mass of the blob (compared to the rock it replaces). To offset that, we go up to height ##h+\Delta h##, so by ##\Delta h##. Roughly, that adds ##mg\Delta h## to the GPE. That gives the equation in the solution.

I suppose, strictly speaking, we should also consider how the iron blob affects that. I.e. it really adds ##mg\Delta h+GMm(\frac {1}{h+r}-\frac {1}{h+\Delta h+r})##, so the equation becomes ##mg\Delta h=GMm(\frac {1}{h+\Delta h+r})##, but that gives you a cubic to solve.

I have a different perspective on this. The amount of work required to move a test mass m from radial position r of a sphere to infinite radial position is $$W(r)=\frac{GM_sm}{r}$$ where ##M_S## is the mass of the sphere. So at positions on the earth situated at the top of the ocean far from the iron deposit, $$W=\frac{GM_E \ m}{r_E+h}$$At positions at the top of the ocean immediately above the iron deposit, the work to move a test mass vertically to infinity is $$W=\frac{GM_E\ m}{r_E+H}+\frac{G\Delta M_D\ m}{r_D+H}$$where ##\Delta M_D## in the incremental mass of the deposit over and above the same volume of water, and H is the ocean depth at the deposit. The two work amounts must be equal, so $$\frac{GM_E \ m}{r_E+h}=\frac{GM_E\ m}{r_E+H}+\frac{G\Delta M_D\ m}{r_D+H}$$Since h and H are << r_E we can approximate this equation by $$g(H-h)=\frac{G\Delta M_D}{r_D+H}$$If the change in depth is small compared to the original depth, this becomes: $$g(H-h)=\frac{G\Delta M_D}{r_D+h}$$

Chestermiller said:

I have a different perspective on this. The amount of work required to move a test mass m from radial position r of a sphere to infinite radial position is $$W(r)=\frac{GM_sm}{r}$$ where ##M_S## is the mass of the sphere. So at positions on the earth situated at the top of the ocean far from the iron deposit, $$W=\frac{GM_E \ m}{r_E+h}$$At positions at the top of the ocean immediately above the iron deposit, the work to move a test mass vertically to infinity is $$W=\frac{GM_E\ m}{r_E+H}+\frac{G\Delta M_D\ m}{r_D+H}$$where ##\Delta M_D## in the incremental mass of the deposit over and above the same volume of water, and H is the ocean depth at the deposit. The two work amounts must be equal, so $$\frac{GM_E \ m}{r_E+h}=\frac{GM_E\ m}{r_E+H}+\frac{G\Delta M_D\ m}{r_D+H}$$Since h and H are << r_E we can approximate this equation by $$g(H-h)=\frac{G\Delta M_D}{r_D+H}$$If the change in depth is small compared to the original depth, this becomes: $$g(H-h)=\frac{G\Delta M_D}{r_D+h}$$

i do not understand what $$r_E$$ really means. is it the radius of the iron deposit? if so why did you say its a lot greater than H and h. if its the radius of earth why did you add in the second formula this radius by the height of the ocean?

germoneyswis said:

if its the radius of earth why did you add in the second formula this radius by the height of the ocean?

The work required to move off to infinity a point mass m that is presently at r from
the centre of a uniform sphere of mass M is ##\frac{GMm}r##.
@Chestermiller is considering moving to infinity a test mass that starts at H above the top of the iron deposit. It starts at ##H+r_E## from the centre of the Earth and at ##H+r_D## from the centre of the deposit.

haruspex said:

The work required to move off to infinity a point mass m that is presently at r from
the centre of a uniform sphere of mass M is ##\frac{GMm}r##.
@Chestermiller is considering moving to infinity a test mass that starts at H above the top of the iron deposit. It starts at ##H+r_E## from the centre of the Earth and at ##H+r_D## from the centre of the deposit.

so shouldnt it be $$r_E + (h-H)$$ and $$H+r_D$$

germoneyswis said:

so shouldnt it be $$r_E + (h-H)$$ and $$H+r_D$$

What should be $$r_E + (h-H)$$?
That would be a point inside the Earth.