How much solar power could the Earth produce? (Part 1)

I just filled up the car, paying $3.87 for midgrade.  Almost back to the four dollar gallon of gas!  A website I'm newly a fan of is the Energy Information Agency (www.eia.gov).  They have so much information on energy, use, efficiency, etc, and it's all available in easy to download Excel files so you can make cool graphs like this on your own!

Anyway, with rising petroleum costs, (please note the sharp drop in gas prices which corresponded to a drop in the price of crude oil when the United States declined to elect another hawkish president in November 2008.  The war drums are being beaten again over Iran, and look!  Crude oil just peaked at $114 / barrel), it gets one thinking about what it would take to completely get off natural gas and depend only on solar power.

This is going to be a bit more complicated than some of the previous ones, so I'm going to do this in parts.  In this one, I'm going to tackle how much solar energy reaches the Earth.

Okay, we'll start with finding out how much energy the sun emits.
Solar luminosity (the amount of energy emitted by the sun) is about 3.8e26 Watts.
If the Earth is, on average, 1.5e11 meters (one astronomical unit) from the sun, we can use the equation for the surface area of a sphere to find out how much area that energy spreads out at our radius from the sun:

A = 4 * π * r²

  = 4 * (3.14) * (1.5e11 m)²

A = 2.8e23 m²

Now for how much of that the Earth takes up.  The diameter of Earth is, on average (Earth is not a perfect sphere, but we'll cheat and say that it is) 12,742 km.  So the radius would then be about 6,371,000 m, and with the equation for area of a circle (A = π*r²), the area it would present to the sun would be about 1.3e14m².  Divide the two areas and you get that Earth gets about

Amount of solar energy that reaches Earth:

                     Cross-sectional area of Earth

Energy of the Sun * --------------------------------

                     Area of Earth's Orbital Sphere

              1.3e14 m²

(3.8e26 W) * ----------- ≈ 1.8e17 W

              2.8e23 m²

So the Earth receives about 1.8e17 W of heat from Earth.

Deriving the equation for centripetal acceleration

I'm going to be completely honest about my last post in which I looked at how fast you would have to spin a spaceship to make artificial gravity.  I didn't remember how to calculate centripetal force, and I didn't have my phone or computer or physics books to look up the equation.  I was able to do it on one bar napkin though.

Assuming an object orbits a central point and has its position defined by the functions:

x(t) = r * cos(ωt)

y(t) = r * sin(ωt)

Take the first derivative to get velocity and you have:

vx(t) = x'(t) = -ω * r * sin(ωt)

vy(t) = y'(t) = ω * r * cos(ωt)

Take another derivative for acceleration:

ax(t) = vx'(t) = x''(t) = -ω² * r * cos(ωt)

ay(t) = vy'(t) = y''(t) = -ω² * r * sin(ωt)

So now we can just use the Pythagorean Theorem (x² + y² = r²) to find the magnitude of centripetal acceleration.  Rearranging:

a = √(ax² + ay²) 

It looks like it's going to get hairy with all the squaring and square rooting, but a quick substitution will simplify it. Substitute and multiply out:

a = √( (-ω² * r² * cos(ωt) )² + (-ω² * r² * sin(ωt) )² ) 

a = √(   ω⁴ * r² * cos²(ωt)   +   ω⁴ * r² * sin²(ωt)   )

Divide out ω⁴ * r²:

a = √(  ω⁴ * r² * ( cos²(ωt) + sin²(ωt)  )

a = ω² * r * √( cos²(ωt) + sin²(ωt) )

Remember the property of sin²ϑ + cos²ϑ = 1?  If we just apply that, we are left with the equation of a = ω² * r.  Derivation complete!!

Making fake gravity on the way to Mars

The New York Times had an article this week "Squashed Eyeballs Are a Danger for Astronauts" discussing how eyeballs deform in zero gravity and the complications it poses for long-term space missions, such as a two year trip to Mars.  It discusses rotating spaceships to produce artificial gravity.

My question is this: how fast would a spaceship have to rotate to simulate normal Earth gravity?  I grew up reading and watching a lot of sci-fi where this was done, such as in Babylon 5:

2001: A Space Odyssey:

I always accepted that rotation would work to provide artificial gravity, but never actually looked at the numbers involved with it.  Next time I watch some B5 I'll certainly be taking a stopwatch to the rotation of the space station.

We'll start with the equation for centripetal acceleration.

a = ω² * r 

Where ω is the angular velocity (speed of rotation) and r is the radius of the spaceship.

Centripetal force is that force which pulls on a rotating object to keep it in its curved trajectory.  If you were to swing a yo-yo around in the air over your head, the force you would pull on the string with would be centripetal force.  An equal force, centrifugal force, is felt pushing outward, and that would be perceived as gravity by the crew of the ship.

Since we already know the value of acceleration we want to simulate (9.8 m/sec² or 32 ft/sec²), we're going to rearrange the equation to be in terms of ω:

        a 

ω = √ (---)

        r    

I want my answer in rpm (revolutions per minute), so I'm going to add some conversion factors.  This equation will solve for ω in terms of radians per unit time, so I need to convert radians into revolutions (2π radians per revolution).

        a     60 sec      rev

ω = √ (---)*(--------)*(--------)

        r      min       2π rad

Now to plug some numbers in:

r (ft)	ω (rpm)
100	5.4
300	3.12
250	3.42
500	2.42
2640	1.05
5280	0.74

 or in metric:

r (m)	ω (rpm)
50	4.23
100	2.99
250	1.89
500	1.34
1000	0.95
5000	0.42

So all of these are pretty fast.  If I had a spaceship that was 1 mile across (r = 2640 ft), I would need to spin it around at just under one rpm to simulate normal gravity.

A good read I'm halfway through is Mary Roach's Packing for Mars, which takes a light-hearted look at some of the obstacles to tackle with space travel, specifically those relating to human limitations.  I had previously read her first book Stiff: The Curious Lives of Human Cadavers, which was also a great read.

How thick is the hull on James Cameron's submarine?

Okay, so we figured out that James Cameron's submarine probably can withstand external pressure of about 17 - 18 kpsi.  How thick would the hull have to be to survive?  The NYT article mentioned that the capsule is only 43 inches wide.  I assume that that is on the inside not the outside.

The thing is shaped like a torpedo, so I'll use the model of a hollow cylinder with a half sphere at either end.  Spheres are the strongest solids, so the cylinder will be the limiting factor.  Granted it's going to have ribs that will increase the strength of the cylinder, but this will work for a good ballpark figure.

I got this equation for hoop stresses (the inward or outward stresses on a circular piece of metal) from efunda.com:

      P * r

σθ = -------

        t

Where:

σθ = Hoop stress (in psi)

P = pressure

r = internal radius of the vessel

t = wall thickness

James Cameron's submarine has an internal diameter of 43 inches, the radius is 21.5 inches.

The hoop stress will be the main stress of concern, which we calculated to be about 17kpsi.  So we rearrange and get:

     P * r

t = -------

      σθ

Now for a material.  I have no idea what that thing is made of, but let's assume a titanium alloy (it's pretty strong, right?) with a yield strength of about 120ksi (120,000psi), so...

     P * r     (17 ksi) * (21.5 inches)

t = ------- = -------------------------- ≈ 3 inches

      σθ              (120 ksi)

Bear in mind that this would cause the submarine to start to deform at the bottom of the trench, so give it a safety factor of about 1.5 and you get a thickness of about 4.5 inches.  Add support ribs on the inside and the thickness can be reduced.

So all in all, the thickness of the hull is probably about 3 to 4 inches. 

How much pressure does James Cameron's submarine have to withstand?

I was reading in the NYT the other day about James Cameron's minisub that can (I hope for his sake) safely get to the bottom of the Challenger Deep.  The deepest point is just under 7 miles down.  Having been on a submarine for a couple years, one thinks about sea pressure around the boat.  I wonder what kind of pressure James Cameron's submarine has to endure.  On the boat, our rule of thumb was about 44 psi for every 100 feet of depth.  Normal submarines operate at a mere fraction of the depth that Cameron's does, so he'll actually experience more pressure per hundred feet than we would, but since the Challenger deep is a bit less than 7 miles deep, if I round up the depth, it'll work out.

            44 psi      5280 feet

7 miles * ---------- * -----------  ≈ 16.25 kpsi (16,250 psi)

           100 feet      1 mile

That's really a lot.  Consider that normal sea level pressure is 14.7psia, and that works out to be about 1,100 times greater than atmospheric, or in submarine terms, about 46 times the unclassified depth of 800 feet that U.S. submarines dive to.