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Hello I was reading another question asked by zach466920, and when he was trying to calculate the total GPE of a water 'tower', he used this explanation:

enter image description here

He basically used integration to calculate the total GPE of a cylinder by dividing it into disks of infinitesimal thickness, and summing those GPEs. However, another comment said:

Well the integration part is not necessary you know ... You only have to consider the center of gravity of the water pillar when calculating the GPE as the volume is uniform

I do not get this explanation, rather I think the integration method makes sense. Which one is correct?

The post is from How does a ball cause a splash? (With the relevant math)

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  • $\begingroup$ "Well the integration part is not necessary you know ... You only have to consider the center of gravity of the water pillar when calculating the GPE as the volume is uniform". I believe this statement is correct, but rather than say "volume" I would say "density" $\endgroup$
    – Bob D
    Commented Mar 31, 2020 at 14:24
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    $\begingroup$ On Physics SE we actively discourage the use of images for equations and text. You should use text and Mathjax. $\endgroup$
    – StephenG - Help Ukraine
    Commented Mar 31, 2020 at 14:25
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    $\begingroup$ If the height of the water tower is much less than the radius of the Earth, so that the gravitational field strength is approximately constant over the height of the tower, then integration gives the same result as using the centre of mass of the tower. However if the height of the tower is comparable with the radius of the Earth, so that the gravitational field strength varies over the height of the tower, then integration must be used - the centre-of-mass method will give the wrong result. $\endgroup$
    – sammy gerbil
    Commented Mar 31, 2020 at 14:45
  • $\begingroup$ Bob D Sorry can you please elaborate on how I would calculate that? Moreover, does that mean the integration method is wrong $\endgroup$
    – Andrew Norfield
    Commented Mar 31, 2020 at 14:45
  • $\begingroup$ You may be able to find the centre of mass by a symmetry argument. Otherwise you you will have to do an integral, which is essentially the same integral as for the PE. BTW, you can copy/paste the equation from the other Question by clicking on "Edit" there. $$PE=\int_0^{h_t} g \cdot l \cdot \rho \cdot A \ dl={{g \cdot \rho \cdot A \cdot {h_t}^2} \over 2}$$ $\endgroup$
    – Keith McClary
    Commented Mar 31, 2020 at 18:26

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