I’m traveling until Thursday night. I should be able to post consistently again once I get back. In the mean time, try calculating this: How fast would you half to flap your arms to fly?
The physics of this problem is essentially the same as the “Beat It, Mothra” problem, so I’ll use the same formulas. A person’s wingspan and arm width are about 2.0 m (~6.6 ft) and 10 cm, respectively. This means the “wings” have a total area of about 0.2 m2. I’ll assume the arms/wings move 0.30 m (~1.0 ft) down with each thrust, so that each push propels 6.0×10-2 m3 of air downward. As before, we can compute the total mass of air moved with each thrust,
air mass = (density) · (volume)
= (1.2 kg/m3) · (6.0×10-2 m3)
= 72 g.
For a 75 kg person, the force of gravity is
gravitational force = (mass) · (gravitational acceleration)
= (75 kg) · (9.8 m/s2)
= 740 N.
We can solve for the frequency using the equation from the Mothra problem,
frequency = {(gravitational force) / [(air mass) · (range)]}1/2
= {(740 N) / [(72 g) · (0.3 m)]}1/2
= 185 Hz
This mean you’d have to flap your arms up and down at least 185 times each second in order to just get off the ground. I’ve simplified both this and the Mothra problem considerably. As Rich pointed out in the Mothra comments,
“…wouldn't there be a force downward when Mothra lifts her wings up? I haven't really studied how moths wings work but I would assume that the upswing at least creates some downward force, requiring a faster wing beat in order to compensate for those downward forces.”
It’s a very good point that we neglected in both problems. As I understand it, the down stroke (i.e. the one that pushes the air down and the animal up) is done with the wings spread out as much as possible so that the maximum amount of air is pushed down. The up stroke occurs closer to the body so that less air is propelled. There are reasonably good videos of these dynamics here and here. If the two strokes were the exactly same, they would cancel each other and there would be no net up force. In these problems, I’ve assumed the upstroke is negligible. This assumption may be good for an order of magnitude estimate, but for something more precise you definitely need to include the effect the up stroke.
It’s my wife’s birthday tomorrow. I won’t say her age here, but if she lives to 100, how many birthday candles will she have blown out?
It seems two readers have beat me to the punch. Well played. They’ve both assumed she has one candle on her first birthday, two on her second, three on her third, etc. By making two columns of numbers as A has done, you can see that each pair of numbers on a line sums to 101:
1 +100 = 101
2 + 99 = 101
[...]
49 + 52 = 101
50 + 51 = 101
We can then multiply by 50 rows to get the answer,
There’s a lot of Star Wars physics one could quibble with. Take the Death Star. How, exactly, is a giant laser supposed to blow up a planet? How much energy would it take to blow up Alderaan?
To calculate how much energy you’d need to blow up Alderaan, you first need to know what’s holding it together. Unless the Star Wars universe is vastly different from our own, Alderaan is likely being held together by gravitational forces, i.e. all the different chunks of matter that make up the planet are attracted to each other by Newton’s law of gravity. In order to blow it up, you need to impart at least enough energy to overcome this attraction. Fortunately, physics has a well-known solution for this problem. Most introductory courses on electricity and magnetism teach you to calculate the energy required to assemble a uniformly charged sphere with the result1
E = (3 / 5) k Q2 / r,
where E is the energy, k is a constant, Q is the total charge on the sphere, and r is the radius of the sphere. This can be derived from Coulomb’s law for the force acting on charged particles. Since Newton’s law has the same inverse square dependence on distance as Coulomb’s law, the formula can be modified to calculate the gravitational energy holding together a sphere of uniform mass2,
E = - (3 / 5) G M2 / r,
where G = 6.67×10-11 N·m2/kg2 is a fundamental constant and M is the mass of the sphere. According to Wookieepedia, the radius of Alderaan is 6,250 km and its gravity is “standard,” making it slightly smaller than Earth with a mass of about 5.7×1024 kg. Using these numbers, we can estimate the amount of energy needed to blow up Alderaan,
As Marvin the Martian once said, “Where’s the kaboom?” If you harvested all the solar power that falls on the Earth from the Sun, it would take about 40 million years to collect enough energy to blow up Alderaan.
[1] See, for example, The Feynman Lectures on Physics, Vol. II, Chapter 8.
[2] Strictly speaking, if Alderaan is like Earth, it will not have a uniform mass distribution, but this assumption makes the math a lot easier.
I’ll be traveling a bit for the next two weeks, so my posts may be a little sporadic. When I get back, I’ll try to address all the questions people have left in the comments and I’ll post a few of the Fermi questions readers have sent me. In the mean time, I’m going to try to answer one of Thomas’s questions from a while back:
If we took a cubic meter of the sun and dropped it into the ocean, how much water would boil away?
As a disclaimer, I know very little about astrophysics, plasma physics, and nuclear physics. That said, I think I can still come up with a decent estimate even with my limited knowledge. I’m assuming Thomas means a cubic meter of the Sun’s core, which is both denser and hotter than its surface. According to Wikipedia, the Sun’s core has a temperature of 1.57×107 K and a density of 1.5×105 kg/m3. The core is too hot for atoms to be stable, so it’s basically a bunch of free protons, neutrons, and electrons swishing about. Physicists call this phase of matter “plasma.” When we dump a cubic meter of this plasma into the ocean, it will cool as it gives off heat to the surrounding water. As it cools, the protons will bind with electrons to form atoms—I’ll assume they all form hydrogen atoms—and these hydrogen atoms will then form hydrogen molecules. Each of these phase changes releases some heat1 into the water. To compute how much energy is given to the water, we need to know both the energy gained by cooling and the energy gained by phase changes.
From the density, we know that the total mass of one cubic meter of the Sun’s core is about 1.5×105 kg. Since the atomic weight of a hydrogen atom is 1.008 g/mol, we can compute the total number of hydrogen atoms we’ll get out of the chunk o’ Sun,
number of atoms = (mass) / (atomic weight)
= (1.5×105 kg) / (1.008 g/mol)
= 1.5×105 mol.
This is equivalent to 9.0×1031 hydrogen atoms. The ground state of a hydrogen atom is –1300 kJ/mol (~-13.6 eV). From this we can compute the amount of energy released in the formation of hydrogen atoms,
energy released = (energy per atom) · (# of atoms)
= (1300 kJ/mol) · (1.5×105 mol)
= 2.0×108 kJ.
It takes two hydrogen atoms to form a hydrogen molecule. This means the 9.0×1031 hydrogen atoms in our glob will form 4.5×1031 molecules or 7.5×104 moles of molecules. Each H2 molecule has a binding energy of 436 kJ/mol, so the total heat released by the formation of molecules is
energy released = (energy per molecule) · (# of molecules)
= (436 kJ/mol) · (7.5×104 mol hydrogen molecules)
= 3.3×107 kJ.
There’s also energy released in the cooling process. I’ll assume both the plasma and molecular hydrogen are well approximated by an ideal gas, which has a specific heat of 20.79 J/K·mol at constant pressure2. From this, the temperature change, and the number of moles of hydrogen atoms, we can estimate the heat released by cooling,
energy released = (specific heat) · (# of atoms) · (temperature change)
= (20.79 J/K·mol) · (1.5×105 mol) · (1.57×107 K)
= 4.9×1010 kJ.
As you can see, the heat released by cooling is much larger than either of the phase changes indicating that this is the only significant factor. This means about 4.9×1010 kJ of energy is used to boil ocean water.
To determine how much water boils away, we need to know water’s specific heat (~4.186 kJ/kg·K) and its latent heat of vaporization (~2260 kJ/kg). If the ocean starts at room temperature (~298 K = 25 °C), it will have to increase by about 75 K in order to boil. Using this data, we can estimate the total mass of water that would be boiled away3.
It would boil away about 20 million kilograms of water. That’s about a cube of water 27 m on a side.
[1] Physicists call this “latent heat.”
[2] The specific heat of an ideal gas at constant pressure is equivalent to 5R/2, where R is Boltsmann’s constant.
[3] For simplicity, I’m assuming all the heat gets transferred directly to the water that gets boiled off (i.e. the rest of the ocean stays at room temperature.)
Same rules as before. The winner gets a signed copy of How Many Licks? Email your entries to "aaron at aaronsantos period com" by May 1, 2010 to be eligible.
This is a problem that has vexed me for a while, but I think I finally have a good method to solve it: How far would the oceans sink if we took all the fish out?
I’m told Twitter is a great way to get exposure. Not including family and friends, I get a new follower about once every eight days. To reach Sarah Killen’s follower status, it will take me 622 years.
How many times has our lovable Italian, shall we say, left the platform?
According to Wikipedia, over 210 million Mario games have been sold, but this doesn’t include the number games that have been downloaded, purchased at yard sales, or resold on E-bay. To account for these extra players, I’ll assume 400 million people have owned Mario games. One usually gets bored with a game after a few months at which point the amount of game play decreases. If you average an hour per day for the first few months, then you’ll have about 100 total hours of game play in this time. Occasionally, nostalgia kicks in and you start playing again. I’ll assume 100 hours of nostalgia play giving a total of 200 hours per player. Some players are so good they never die, but most Mario players, or rather their Mario avatars, die fairly regularly especially when starting out. A reasonable estimate is one Mario death per minute. I say “reasonable” because even terrible players don’t die every 6 seconds, and even fairly good players usually can’t last 10 minutes without accidentally falling into a bottomless pit. Using these numbers, we can estimate the total number of Mario deaths:
# of deaths = (# of game owners) · (hours of play per owner) · (deaths per minute)
= (400 million game owners) · (200 hours per owner) · (0.5 deaths per minute)
= 4.8×1012 Mario deaths
Congratulations to our winner, Melgamoose!
A couple of observations about the contest
There really weren’t any bad entries. There were 29 entries in total and the average answer was 8.35×1012 Mario deaths. This makes me feel reasonably confident that my answer isn’t too far off since it’s within a factor of 2 of the mean. Fermi estimates are only expected to be good to within an order-of-magnitude, so it’s interesting to look at the distribution of magnitudes:
The mean magnitude was 11.5 and the standard deviation was about 1.3, i.e. roughly an order of magnitude. This tells me two things. First—forgive the sucking up—I have a really smart readership. Second, just about any of these answers could reasonably have won the contest. So please enter our next contest which will be announced this weekend, because next time the winner could be you!
