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If $8.8 \times 10^{16} m^3$ of water volume is added to the surface of Venus, how would I calculate how much of Venus would be covered in water?

The land of Venus has its topographical features with plains, valleys, high lands, etc, therefore water would not be distributed evenly across the surface. Water will obviously find its way to the lowest ground available.

Therefore, knowing only the water volume wouldn't help unless you know how the volume is distributed over the surface of Venus. Given topographical features of Venus, how would we calculate the distribution of water volume over the surface?

Other things to consider

  1. In answering the question, you could safely ignore the atmosphere, and if it matters, assume the atmosphere is earth-like.
  2. The topographic map of venus is available on the net, but frankly I have no idea what to do on them. I have no experience nor expertise on working with topographical map, which probably needs some kind of geological degree or perhaps computer degree to make sense of the map. So it would be very helpful to have an answer that provides a more understandable interpretation of the map, in a form usable for calculations knowing only the extra volume of the water. This is not an absolute requirements.

This question was on Sandbox.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ I suspect you'd define a "standard" spherical representation of Venus, the way we have done with Earth (where it's called "mean sea level"). You would then figure out how the actual topography relates to this standard level, calculate the total displacement volume of the actual topography, and from there can adjust the simple "all liquid evenly distributed over the surface" model to better fit reality. However, I don't know how you'd go about doing it, and certainly don't have any scientific papers to show for it. $\endgroup$ – a CVn Dec 7 '17 at 12:10
  • $\begingroup$ @MichaelKjörling That's why I asked how the water be distributed, because I don't know how to calculate it. The water worked out in my paper as ~192 meters of water, if the surface of Venus is flat, and perfectly spherical, but more than that I don't know how to do it, or even work with topographic map. $\endgroup$ – Hendrik Lie Dec 7 '17 at 12:21
  • $\begingroup$ A factor that should be considered is Venus' temperature - some of the added water will boil. $\endgroup$ – Alexander Dec 7 '17 at 18:26
  • $\begingroup$ @Alexander It's all going to boil, as the surface temperature is at around 465°C. How long it would take is a different question. However, I think the original question could be improved by clarifying if we mean actual Venus, or a planet like Venus in a different orbit. $\endgroup$ – richardb Dec 7 '17 at 23:32
  • $\begingroup$ @richardb it's mentioned in the question, you could safely ignore its atmosphere, and if required, assume the atmosphere is earth-like $\endgroup$ – Hendrik Lie Dec 8 '17 at 2:10
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Numerical Solution Required

Numerical solutions are used when there isn't any way to perform a "simple" calculation and find the answer. Your question demands that we account for the actual topography of Venus. This can't be done by hand and will involve a lot of calculations in MatLab (or Octave). The below approach requires no idealized altitude datum since we only care about the absolute difference between max and min altitudes.

Also, this solution assumes that somehow liquid water can be deposited on the surface of Venus which is currently cannot.

How to Calculate It the Digital Way

  1. Download a topographic map of Venus as a resolution that suits you. Likely these will be in the form of bitmaps. For this answer, I'm going to assume that the height data comes as bitmaps.
  2. Use Photoshop to massage the height map/image to be just the data (no frames, scales, legends, etc).
    1. MatLab/Octave have robust facilities for translating pixel values into n-degree matrixes. This is how you'll translate the color values of the bitmap to actual heights.
  3. Calculate the volume of space that each pixel would take up. For example, at the equator, a pixel on this bitmap represents 0.1$^\circ$ by 0.1$^\circ$. From the bitmap, you'll know how many levels of elevation gradation the map offers, say each color value represents 500m of altitude. Given Venus' diameter, find the volume of each pixel by multiplying the width, height and depth of that pixel.

    1. You'll have to be careful as you approach the poles since the normal maps you find about any spherical object will tend to stretch the poles. Stereographic projects are a real pain to deal with. This stretching prevents you from just assuming that each pixel is 0.1$^\circ$ because towards the poles it might be 0.1$^\circ$ by 10$^\circ$ (I just made that number up. You'll have to figure it out yourself.)
    2. There is a projection called a cubed-sphere that handles a lot of these projection problems pretty well but I don't know enough about the conversion between normal spheres and cubed-spheres. It's a well understood topic but one that I only have the briefest of introductions to.
  4. You'll want to build up this height model in MatLab as a 3D grid of some sort. Conceptually, at this point the solution is pretty simple. Find the lowest pixel, fill it then subtract that volume from the $8.8 \times 10^{16} m^3$. Repeat this process for each of the lower pixels till you run out of liquid to deposit.
  5. At the end, you'll know which pixels have water in them and which ones don't.

The Analog Approach

The above approach is daunting to someone who isn't used to Matlab. Here's an easier, craftier approach.

  1. Print out as many copies of the topographic map of Venus as there are divisions of altitude (so if the map has 10 layers of altitude, print off ten copies.)
  2. Label each copy of the map with a number. Sort all the copies by number.
  3. For each copy of the map, blacken out with a marker, all the areas higher in elevation than this map.
  4. For each copy of the map, with as much accuracy as you care to, measure the area of the unblackened portions.
    1. You'll still have the same kind of distortions resulting from projecting a sphere down to a plane but at this resolution level, it may not matter much.
  5. Make a quick spreadsheet with one row for each layer of the map. Record the volume for each layer.
  6. For each row, assign as much of the $8.8 \times 10^{16} m^3$ of liquid that will fit in that volume.
  7. Do this till you run out of liquid to deposit. You'll have a decent idea of which areas are filled and which ones aren't.

This is how you would do it, not the doing of it. ;)

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  • $\begingroup$ This is going to demand expertise in using applications in question. Actually, the way it is answered really answers my question, i should have expected that this kind of answer may appears. This is a compliment :). Unfortunately my Law major wouldn't help much. I'll try to use programs you mentioned and see what i can do with it. $\endgroup$ – Hendrik Lie Dec 7 '17 at 14:30
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    $\begingroup$ @HendrikLie if I may set an expectation for you. Octave and Matlab are very deep products that can be daunting to learn. This will not be quick to figure out and it will be frustrating along the way. Take your time and be patient. I spent considerable time on a similar project. Good luck. $\endgroup$ – Green Dec 7 '17 at 14:38
  • $\begingroup$ The edit is unexpected, that surely looks workable by hand. Only if i could upvote twice $\endgroup$ – Hendrik Lie Dec 8 '17 at 2:15
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Order of magnitude approximation
8.8 x 10^16 cubic metres = 8.8 x 10^7 cubic km The North Atlantic Ocean is 14.6 * 10^7 cubic km so that’s about 60% of the North Atlantic Ocean being added to Venus or about 6% of Earth’s oceans worth.

Average depth of the North Atlantic ocean = 3519m and it covers 41490000 sq km Surface area of Venus = 460230000 sq km So 41490000*100/460230000 =about 9% of the area of Venus so very roughly 9% of 3519 = 317m across the entire surface of the Venus assuming its flat. Obviously there will be some deep depressions soaking a lot of water and lots of raised areas so there's going to be a lot of dry land.

Plenty of water to splash around in, lots of big shallow lakes and seas over low lying areas and plains, but not really a proper ocean. For depth comparison Lake Superior average depth 147m, max depth 406m

Look at this reference https://www.lpi.usra.edu/resources/venus_maps/2444_4/ note the area covered in green. This is 1km below the datum level so will absorb roughly 3 times the overall 317m worth. So the seas going to be based here but will be a bit deeper and wider but not enough to fill up to the 1km contour. As a first estimate take the green areas and join them up and expand them via the deep brown areas only.

Oceans ref: https://www.ngdc.noaa.gov/mgg/global/etopo1_ocean_volumes.html
Venus ref: https://en.wikipedia.org/wiki/Venus

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  • $\begingroup$ I have calculated that assuning Venus surface is flat and perfectly spherical, that volume divided by Venus's surface we get water tables of around 192m deep. And the question asks about how to calculate distribution of water over the surface given certain topology of today's venus, :) $\endgroup$ – Hendrik Lie Dec 7 '17 at 15:03
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    $\begingroup$ @HendrikLie It’s a tough call. You really need to integrate over all of the different depth areas to find the true sea level. I don't think this is solvable with anything more than a crude approximation without a programmatic approach and more detailed listing of the Venusian surface contours with intervals at the very least of 100m (rather than 1km) given our previous calculations. But depending on what you want it for, you may be able to approximate by eye - see my edited answer. $\endgroup$ – Slarty Dec 7 '17 at 16:32
  • $\begingroup$ The link you provided surely is helpful, upvoted for that :) $\endgroup$ – Hendrik Lie Dec 7 '17 at 17:32

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