I posted this on my blog a couple years ago, and I suspect I’ve done a few things wrong. I welcome a critique by physicists:
How much energy, in terms of hours or days of solar output, would it take to strip the atmosphere off of Uranus and get at its valuable rocky core, which I call Terra-Uranus?
Uranus has an escape velocity of 21.3 X 10^3 m s^-1. It has a mass of 8.68 X 10^25 kg, equivalent to 14.5 Earths. The rocky core has a mass of 0.55 Earth’s mass, so the mass of everything else on Uranus you’d want to remove suggests the calculation [(14.5 − 0.55)/14.5] X 8.68 X 10^25 kg = 8.35 X 10^25 kg.
The kinetic energy E of a mass m moving at a subrelativistic velocity v has the formula E = 1⁄2 mv^2.
Therefore I can naively calculate the kinetic energy of moving that much mass away from Uranus at its escape velocity:
E = 1⁄2 X 8.35 X 10^25 kg x (21.3 X 10^3 m s^-1)^2 = 1.89 X 10^34 J
The sun has a power output of 4 X 10^26 J s^-1, so to calculate how long it would take the sun’s power to release that much energy, divide 1.89 X 10^34 J/(4 X 10^26 J s^-1) = 47.3 X 10^6 seconds, or about 548 days of solar output, assuming you could capture it at 100 percent efficiency.
That’s a decent estimate. For some more precision you can take into account that each bit of mass you remove decreases the gravity of the remaining planet. Assuming constant density and dissociating the whole planet, this is a correction factor of 3⁄5.
Alternately, if you can mine and export the core without disassembling the mantle (or equivalently, if you recapture the energy released when disassembled bits of mantle recollapse into a second planet), then the net energy requirement is approximately
I posted this on my blog a couple years ago, and I suspect I’ve done a few things wrong. I welcome a critique by physicists:
How much energy, in terms of hours or days of solar output, would it take to strip the atmosphere off of Uranus and get at its valuable rocky core, which I call Terra-Uranus?
Uranus has an escape velocity of 21.3 X 10^3 m s^-1. It has a mass of 8.68 X 10^25 kg, equivalent to 14.5 Earths. The rocky core has a mass of 0.55 Earth’s mass, so the mass of everything else on Uranus you’d want to remove suggests the calculation [(14.5 − 0.55)/14.5] X 8.68 X 10^25 kg = 8.35 X 10^25 kg.
The kinetic energy E of a mass m moving at a subrelativistic velocity v has the formula E = 1⁄2 mv^2.
Therefore I can naively calculate the kinetic energy of moving that much mass away from Uranus at its escape velocity:
E = 1⁄2 X 8.35 X 10^25 kg x (21.3 X 10^3 m s^-1)^2 = 1.89 X 10^34 J
The sun has a power output of 4 X 10^26 J s^-1, so to calculate how long it would take the sun’s power to release that much energy, divide 1.89 X 10^34 J/(4 X 10^26 J s^-1) = 47.3 X 10^6 seconds, or about 548 days of solar output, assuming you could capture it at 100 percent efficiency.
That’s a decent estimate. For some more precision you can take into account that each bit of mass you remove decreases the gravity of the remaining planet. Assuming constant density and dissociating the whole planet, this is a correction factor of 3⁄5.
Alternately, if you can mine and export the core without disassembling the mantle (or equivalently, if you recapture the energy released when disassembled bits of mantle recollapse into a second planet), then the net energy requirement is approximately