Posted: Fri Jul 24, 2015 9:27 am
Prove: Earth has enough materiel to create a dyson sphere.
Let the radius of the sphere be 8000 million km. That's 8*10^3*10^6^10^3 m, or 8*10^12 m.
Let the thickness of the dyson sphere be 1 km.
Volume of the dyson sphere: ((8*10^12m)^3 - (8*10^12m + 1)^3)*4/3pi. Or, 2*10^20 m^3 * 4/3pi. Yes that's an approximation, and it's rounded up.
The earth is 6.3*10^3km thick, or 6.3*10^3*10^3m thick, or 6.3*10^6m in radius. The volume of the earth, then, is 2.16*10^20 m^3 * 4/3 pi.
(more really, I rounded down to 6.)
Now that's assuming a dyson sphere that's uniformly thick. My room growing up was 10ft^3 and the walls were about .5ft thick. Probably less but shhhh we're going for easy math. That's a ratio to (much, much less than) (10.5ft^3)/(10ft^3) or about 16% of the volume being taken up by materiels. So the real volume of required material for the dyson sphere is less than 10% of the earth's volume.
Energy to get the @#(! to the point in space:
Thank god for basic newtonian mechanics, eh? P = GH, and H = well, we'll call it the average. Because the amount of material being moved is proportionate to the cubed distance it travels, the average is somwhere lower than what we need, which is an overestimate (sorry, drunk, and therefore disinterested in going too deep on this). Let's assume that all the material needs to be moved up to H where H = 8000 million km. Or 8*10^12m. We can also assume that the average position of Earth is 0. If you need me to walk through the logic of the Earth's average position being where the sun is, then shut the $#@! up and kiss my pissflaps.
(if we were clever monkeys, like we seem to think we are, it'd actually be possible to make the delta H in this equation simply equal to the difference of 8000 million km and the earth's orbit, but I like it this way because it's easier on me)
(yes I know Earth's orbit is elliptical, but if you haven't figured out the game of my overestimate-underestimate here, you should consider jumping in a lake)
(well I mean if you're not jumping in a lake right now I don't know what you're doing with your life but it must suck because swimming is one of life's joys. But I meant drowning earlier.)
Anyways the equation gets a little complicated here, but nothing we can't deal with thanks to Sir Isaac Newton. Basically what we need is the sum of the amount of energy needed to move the mass from H = 0 to H = 8000 million km. I don't have a numpad on this so sorry.
The integral of: m * G * H with respect to H, except that G is actually F/m. Which means that it's the integral of F*m*H/m, or FH from 0-8000 million km. F = G (m1)(m2)/(h^2) so our actual equation ends up being the integral from 0-8000m km of G(m1)(m2)(h)/(h^2)dh. This means we need the mass of the materials we're moving. Basically I'm going to argue that the mass is close to 1. Steel, etc, has a mass of 8 or so (in the units we're using), but most people talk about carbon nanotubes and carbon has a density of about .93. Being the nice person I am, let's move our assumed density to 10 because we're an idiot and trying to build our dyson sphere out of more dense steel.
Oh fun fact, there's a helluvalot of constants here, so our integral is really G(m1)(m2) * integral from 0-8000 million km of 1/h dh. Long story short, this is insolvable as written. Oddly, it's NOT insolvable if we just assume it's from 1 m to 8000 million km. Incidentally, this introduces almost 0 error. (ln of 0 is undefined, ln of 1 is 0). So let's move that to 1m - 8000 million km. Virtually no error but we also don't have to deal with some serious bull@#(!. So now we have g(m1)(m2)*ln(8*10^18), or g(m1)(m2)*44. m1 is, according to our guestimation, 2*10^20 m^3 * 4/3 * 1000 kg/m^3 * pi. Or, 2*10^23 * 4/3 * pi. m2 is 2*10^30. Yes, the sun is big. Really big. I mean like, ginormously insanely colossally big. It's pretty much six orders of magnitude larger than our 1km thick dyson sphere that's been carefully constructed outside of Pluto's orbit... at least in terms of mass.
Anyways, let's run the numbers. It takes 4*10^53 * 4/3 pi * 6.6*10^-11, or 6*10^41 J, to get this stuff from point A to point B. The current luminosity of the sun is only about 4*10^26 watts (joules per second). In order to grab the amount of energy we're looking for, it'll take 1.5 quadrillion seconds or only about 50 million years.
Let's get started boys!
Let the radius of the sphere be 8000 million km. That's 8*10^3*10^6^10^3 m, or 8*10^12 m.
Let the thickness of the dyson sphere be 1 km.
Volume of the dyson sphere: ((8*10^12m)^3 - (8*10^12m + 1)^3)*4/3pi. Or, 2*10^20 m^3 * 4/3pi. Yes that's an approximation, and it's rounded up.
The earth is 6.3*10^3km thick, or 6.3*10^3*10^3m thick, or 6.3*10^6m in radius. The volume of the earth, then, is 2.16*10^20 m^3 * 4/3 pi.
(more really, I rounded down to 6.)
Now that's assuming a dyson sphere that's uniformly thick. My room growing up was 10ft^3 and the walls were about .5ft thick. Probably less but shhhh we're going for easy math. That's a ratio to (much, much less than) (10.5ft^3)/(10ft^3) or about 16% of the volume being taken up by materiels. So the real volume of required material for the dyson sphere is less than 10% of the earth's volume.
Energy to get the @#(! to the point in space:
Thank god for basic newtonian mechanics, eh? P = GH, and H = well, we'll call it the average. Because the amount of material being moved is proportionate to the cubed distance it travels, the average is somwhere lower than what we need, which is an overestimate (sorry, drunk, and therefore disinterested in going too deep on this). Let's assume that all the material needs to be moved up to H where H = 8000 million km. Or 8*10^12m. We can also assume that the average position of Earth is 0. If you need me to walk through the logic of the Earth's average position being where the sun is, then shut the $#@! up and kiss my pissflaps.
(if we were clever monkeys, like we seem to think we are, it'd actually be possible to make the delta H in this equation simply equal to the difference of 8000 million km and the earth's orbit, but I like it this way because it's easier on me)
(yes I know Earth's orbit is elliptical, but if you haven't figured out the game of my overestimate-underestimate here, you should consider jumping in a lake)
(well I mean if you're not jumping in a lake right now I don't know what you're doing with your life but it must suck because swimming is one of life's joys. But I meant drowning earlier.)
Anyways the equation gets a little complicated here, but nothing we can't deal with thanks to Sir Isaac Newton. Basically what we need is the sum of the amount of energy needed to move the mass from H = 0 to H = 8000 million km. I don't have a numpad on this so sorry.
The integral of: m * G * H with respect to H, except that G is actually F/m. Which means that it's the integral of F*m*H/m, or FH from 0-8000 million km. F = G (m1)(m2)/(h^2) so our actual equation ends up being the integral from 0-8000m km of G(m1)(m2)(h)/(h^2)dh. This means we need the mass of the materials we're moving. Basically I'm going to argue that the mass is close to 1. Steel, etc, has a mass of 8 or so (in the units we're using), but most people talk about carbon nanotubes and carbon has a density of about .93. Being the nice person I am, let's move our assumed density to 10 because we're an idiot and trying to build our dyson sphere out of more dense steel.
Oh fun fact, there's a helluvalot of constants here, so our integral is really G(m1)(m2) * integral from 0-8000 million km of 1/h dh. Long story short, this is insolvable as written. Oddly, it's NOT insolvable if we just assume it's from 1 m to 8000 million km. Incidentally, this introduces almost 0 error. (ln of 0 is undefined, ln of 1 is 0). So let's move that to 1m - 8000 million km. Virtually no error but we also don't have to deal with some serious bull@#(!. So now we have g(m1)(m2)*ln(8*10^18), or g(m1)(m2)*44. m1 is, according to our guestimation, 2*10^20 m^3 * 4/3 * 1000 kg/m^3 * pi. Or, 2*10^23 * 4/3 * pi. m2 is 2*10^30. Yes, the sun is big. Really big. I mean like, ginormously insanely colossally big. It's pretty much six orders of magnitude larger than our 1km thick dyson sphere that's been carefully constructed outside of Pluto's orbit... at least in terms of mass.
Anyways, let's run the numbers. It takes 4*10^53 * 4/3 pi * 6.6*10^-11, or 6*10^41 J, to get this stuff from point A to point B. The current luminosity of the sun is only about 4*10^26 watts (joules per second). In order to grab the amount of energy we're looking for, it'll take 1.5 quadrillion seconds or only about 50 million years.
Let's get started boys!