# Meet Up: Vancouver Fermi Estimation

20 October 2012 08:17PM

WHEN: 25 October 2012 06:00:00PM (-0700)

WHERE: 885 west georgia vancouver,bc

Hi LW, you should come to our meetup.

As usual, we meet downtown in the HSBC building lobby at 18:00, on Thursday. Try to be there before 18:30.

This time, I have something for you. I've determined that Fermi estimation and quantitative intuition in general are super valuable rationality skills, so I'm going to attempt to learn us some fermi estimation. It will be hopefully a well prepared lesson full of excercises, concepts and optional homework. Can't find the right word for it. Is it a lesson? An excercise? Activity? Drill? We'll see. I promise fun and math. Come on out.

Everyone is welcome and encouraged to come. If I can and it makes sense, I'll post some optional prepatory material before the meetup and all my notes after.

There will probably be other exciting things happening too. Good stuff happens when you get a bunch of aspiring rationalists together.

Our mailing list is vancouver-rationalists on google groups.

Be there.

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Comment author: 22 October 2012 01:08:15AM 1 point [-]

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.