2 ms·

It's roughly the gravitational acceleration at the surface of an Olympic-size swimming pool of electron-degenerate matter from a white dwarf star. Hope this h

It's *roughly* the gravitational acceleration at the surface of an Olympic-size swimming pool of electron-degenerate matter from a white dwarf star. Hope this helps! (Note: the swimming pool has collapsed under its own weight and is now a sphere).

lucgommans 3y ago
A white dwarf's density is apparently around 1e9 kg/m³. Olympic swimming pools at 50m*25m*2m * 1e9kg/m³ would have a weight of 2.5e12 kg. As a sphere, the radius is (solving: volume_of_sphere = 4/3×π×radius³) $ qalc 50m*25m*2m = 4/3*pi*x^3 (((50 * meter) * (25 * meter) * (2 * meter)) = ((4 / 3) * pi * (x^3))) = approx. (x = 8.4194515 m) Gravitational acceleration (for a point mass) is: the gravitational constant, multiplied by the mass, divided by the radius squared, and we want the result in Earth gravities: $ qalc G * 2.5e12kg / 8.4194515² m² to gee (newtonian_constant * ((2.5 * (10^12)) * kilogram)) / ((8.4194515^2) * (meter^2)) = approx. 0.24 gee You're only one order of magnitude off — if this is correct, which I am really not sure about! Sources: https://en.wikipedia.org/wiki/White_dwarf https://en.wikipedia.org/wiki/White_dwarf · https://en.wikipedia.org/wiki/Olympic-size_swimming_pool https://en.wikipedia.org/wiki/Olympic-size_swimming_pool · https://duckduckgo.com/?q=volume+of+a+sphere&ia=answer https://duckduckgo.com/?q=volume+of+a+sphere&ia=answer · https://www.wolframalpha.com/input?i=surface+gravity+calculator https://www.wolframalpha.com/input?i=surface+gravity+calcula... What I'm even less sure how to calculate is whether an 8-meter sphere can be that heavy. Uranium is ~2e4 kg/m³, but under its own gravity, things shrink until they reach black hole status (infinite smallness; perceived size coming from its event horizon). Basically I'd want to turn the above 1e9kg/m³ into an unknown, but what's the formula for mass given your specified radius and gravitational acceleration? TBD I'm really curious if this was just a few words strung together and it happened to come out to within one order of magnitude by pure coincidence, or if (how) you calculated this!
- perihelions 3y ago
  - " You're only one order of magnitude off" Oh shit you're right. Root cause: I saw "3 g" and wrote "3 m/s^2".
  - lucgommans 3y ago
    :D What calculation did you use though? Like some formula or is there a web utility for this, for example?
    - perihelions 3y ago
      https://www.google.com/search?q=3+m%2Fs%5E2+%2F+%28gravitational+constant+*+%282500+m%5E3+%2F+%28%284%2F3%29+*+pi%29%29%5E%281%2F3%29%29 https://www.google.com/search?q=3+m%2Fs%5E2+%2F+%28gravitati... I think
      - lucgommans 3y ago
        I don't understand, is that what you used originally to come up with your original comment (this one: https://news.ycombinator.com/item?id=37483892 https://news.ycombinator.com/item?id=37483892)? If so, that is surprisingly similar to what I ended up with when trying to validate it!

It's *roughly* the gravitational acceleration at the surface of an Olympic-size swimming pool of electron-degenerate matter from a white dwarf star. Hope this h

lucgommans 3y ago

A white dwarf's density is apparently around 1e9 kg/m³. Olympic swimming pools at 50m*25m*2m * 1e9kg/m³ would have a weight of 2.5e12 kg. As a sphere, the radius is (solving: volume_of_sphere = 4/3×π×radius³) $ qalc 50m*25m*2m = 4/3*pi*x^3 (((50 * meter) * (25 * meter) * (2 * meter)) = ((4 / 3) * pi * (x^3))) = approx. (x = 8.4194515 m) Gravitational acceleration (for a point mass) is: the gravitational constant, multiplied by the mass, divided by the radius squared, and we want the result in Earth gravities: $ qalc G * 2.5e12kg / 8.4194515² m² to gee (newtonian_constant * ((2.5 * (10^12)) * kilogram)) / ((8.4194515^2) * (meter^2)) = approx. 0.24 gee You're only one order of magnitude off — if this is correct, which I am really not sure about! Sources: https://en.wikipedia.org/wiki/White_dwarf https://en.wikipedia.org/wiki/White_dwarf · https://en.wikipedia.org/wiki/Olympic-size_swimming_pool https://en.wikipedia.org/wiki/Olympic-size_swimming_pool · https://duckduckgo.com/?q=volume+of+a+sphere&ia=answer https://duckduckgo.com/?q=volume+of+a+sphere&ia=answer · https://www.wolframalpha.com/input?i=surface+gravity+calculator https://www.wolframalpha.com/input?i=surface+gravity+calcula... What I'm even less sure how to calculate is whether an 8-meter sphere can be that heavy. Uranium is ~2e4 kg/m³, but under its own gravity, things shrink until they reach black hole status (infinite smallness; perceived size coming from its event horizon). Basically I'd want to turn the above 1e9kg/m³ into an unknown, but what's the formula for mass given your specified radius and gravitational acceleration? TBD I'm really curious if this was just a few words strung together and it happened to come out to within one order of magnitude by pure coincidence, or if (how) you calculated this!

perihelions 3y ago
- " You're only one order of magnitude off" Oh shit you're right. Root cause: I saw "3 g" and wrote "3 m/s^2".
- lucgommans 3y ago
  :D What calculation did you use though? Like some formula or is there a web utility for this, for example?
  - perihelions 3y ago
    https://www.google.com/search?q=3+m%2Fs%5E2+%2F+%28gravitational+constant+*+%282500+m%5E3+%2F+%28%284%2F3%29+*+pi%29%29%5E%281%2F3%29%29 https://www.google.com/search?q=3+m%2Fs%5E2+%2F+%28gravitati... I think
    - lucgommans 3y ago
      I don't understand, is that what you used originally to come up with your original comment (this one: https://news.ycombinator.com/item?id=37483892 https://news.ycombinator.com/item?id=37483892)? If so, that is surprisingly similar to what I ended up with when trying to validate it!