Fermi problems You start by making reasonable guesses, like to answer, “How many piano tuners are in Chicago,” you make a guess about how many people have pianos. Then you figure how many pianos a man could tune in day.

How much does a cow weigh? Start with something you already know. Like, Jerry knows how much he weighs, 160 lb, so for back of the envelope he rounds to 150 because its easier to work with. If you crumple him into a ball, then he’d probably be about 1.5 feet. Then you make a guess about how big the cow would be if you crumpled into it into a ball. So how many Jerry-balls could you fit into the Cow-ball?

Keep your units consistent.

Your answer will be no more envelope than your least accurate guess.

We are using the back of the envelope approach to mock up a scale system of the universe in the room.

Using the Sun as our base unit, so the Sun = 1

The Earth is 100 solar distances from the Sun, or 1 AU, and 1/100th the size of the sun.

Jupiter is 500 solar distances from the Sun, or 5 AU, and 1/10th the size of the sun.

Uranus is 2000 solar distances from the Sun, or 20 AU, and 1/30th the size of the sun.

Pluto is 4000 solar distances from the Sun, or 40 AU, and 1/∞ the size of the sun.

If we used a beachball to represent the Sun, the Pluto would be about a mile away.

If the beachball represented the sun, then Earth would be a piece of steel shot.

Mars would be a mustard seed.¹

Jupiter would be a ping pong ball

Uranus and Neptune would be the 1/2″ steel ball

Moving to the sun being a ball that’s 5″ in diameter.

Earth would be two grains of salt.

Jupiter would be the 1/2″ steel ball

But Pluto would still be a 1/3rd a mile away.

So, if the sun is a tennis ball, which is 2 1/2″ then Jupiter would be a pea.

The Earth would be a single grain of salt, 21 feet away.

If the Sun is a pingpong ball, then Jupiter would be a BB 75 feet away.

The Earth would be about 12 feet away.

So, in order to fit a scale model of the solar system in our classroom² the Sun would have to be the size of mustard seed. ³

Now imagine that you have a character on Saturn who needs to get to Mars. It’s a long darn ways off.

Alpha Centuri is 4.2 ly Since 1 ly is 63,000 AU. That means that Alpha Centuri would be would be 250,000 AU to Alpha Centauri. In this scale, our AU is 10 inches. So… that’s 2.5 million inches to Alpha Centauri. Do the math to come up with miles and it would be 31 miles away and Alpha Centauri A is the size of another mustard seed. Go find that!

And this is why you can crash two galaxies together and stars won’t collide.

Next he moved on to talking about orbiting.

Stuff wants to fall toward the center of the planet. If an object is moving fast enough, then by the time you’ve fallen the distance to the planet, the planet has curved away from you. So maintaining an orbit means finding the velocity at which you don’t hit the surface.

v=√rg

V is velocity

R is radius

G is g-force

Then they did math and my head imploded.

v²/r = g will give you the gravity on a rotating space station.

If you get over two rpm you have problems standing up because of the corolious force.

If you double your velocity, you quadruple your radius.

Then we⁴ calculated the gravity on the space station in 2001.

So the space station rotates at 1rpm.

We estimate that the space station is 160 meters radius

The circumference is 1000 meters

v = 16.6 meter per second

v² = 275m²/s²/160m =1.7m/s²

Which is about lunar gravity, but they walk normally on the station.

Mike says that it’s a beautifully shot film that got the physics almost totally right, except for a few minor things.

1. 1/10th of an inch [↩]
2. which is 31 feet [↩]
3. It makes you think that kids shouldn’t do solar system models because it gives them a false sense of how big the solar system is [↩]
4. ”We” meaning our math literate people, while I looked on and oooed [↩]

Tags: Journal, LaunchPad Astronomy Workshop

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