## Order-of-magnitude Physics

A very busy day today so I thought I’d wind down by giving you a chance to test your brains with some order-of-magnitude physics problems. I like using these in classes because they get people thinking about the physics behind problems without getting too bogged down in or turned off by complicated mathematics. I’ve also kept some of these in archaic units just to annoy people who can only do things in the SI system. I think it’s good to practice swapping between systems, especially for us astro-types who use all kinds of bizarre units, so if you don’t know the units, look them up! And if there’s any information missing that you need to solve the problem, make an *order-of-magnitude* estimate!

Give order of magnitude answers to the following questions:

- What is the mass of a body whose weight is equivalent to the total force exerted by a 40 mph gale on the side of a house 40 ft long and 20 ft high?
*Express your answer in tons.* - What is the power required to keep in the air a helicopter of mass 500 kg whose blades are 3m long?
*Express your answer in kilowatts.* - The base of the Great Pyramid is 750 ft square and its height is 500ft. How much work was done building it?
*Express your answer in Joules.* - How high would the jet of a fountain reach if it were aimed vertically up and supplied by a water main in which the pressure is 3 atmospheres?
*Express your answer in feet*.

There’s no prize involved, but feel free to post answers through the comments box. It would be helpful if you explained a bit about how you arrived at your answer!

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November 14, 2011 at 7:26 pm

okay, I am attempting #3 – Work done in constructing a pyramid.

this one is especially entrancing because it made me visualize a square-bottomed pyramid inside a cube, and trying to determine what percentage of the cube the pyramid occupies. i drew out a couple of sketches and just kindof intuited that it was around 33%. but then i drew a top-front-side view and since it occupies 100% x 50% x 50% i believe it is 25%. and then i sketched out an isometric and i think that from the mating piece that would complete the cube you could construct 3 other identical pyramids, which confirms the 25% idea. this is fun and i haven’t even started!!

the volume of the cube with the 500’x750’x750′ dimensions is 281.25E6 ft^3, which corresponds to 8E6 m^3. we are only talking about 25% of this volume, so the volume of the pyramid is 2E6 m^3.

the pyramids are made out of limestone (mostly) which has an approximate density of 2500 kg/m^3 (actually slightly higher, but i am rounding down, ever so slightly – probably not enough – to account for the packing factor) (granite and mortar – also used in much smaller amounts – have comparable densities). so the mass of the pyramid is an estimated 5E9 kg.

so at this point i am taxed with doing calculus for the first time in 8 or so years. must relate W=Fd to a triple integral. um, i will break them up. first, triple integral to determine the height of the center of mass of the pyramid. then i can just use that height and it should all even out (i think?)

(calculator’s note: i got partway through and got frazzled. confirmed where i was error-ing online, and let’s just say i successfully figured out that the height of the CM of a perfect pyramid (which this is not,) is h/4. so the CMsubZ is 500/4 = 125 ft = 38 meters.

W = Fd = Delta-PE = mgh = (5E9 kg)(10 m/s^2)(38 m) = 1.9E12 J

November 14, 2011 at 8:43 pm

40 mph is about 20 m/s which we’ll approximate as 10 m/s. The density of air is about 1 kg/m^3. The dimensions of the building can be estimated at 10 m x 10 m. In one second, the total volume of air that hits the building is 10^3 m^3 which is about 1000 kg. The momentum of that mass is (1000 kg)(10 m/s). Assuming the air slows to v=0 upon hitting the wall, the rate of change of momentum is (10 000)/1 which is a force of 10 000 N. This corresponds to a mass of about 1000 kg. 1 kg is approximately 2 pounds. Thus, 1000 kg is approximately 2000 pounds which gives our answer on the order of 1 ton.

November 15, 2011 at 9:55 am

It took 100,000 people 20 years to build the Great Pyramid. Assuming that they worked an 8 hour day, and each produced

100 W of continuous power used as work, then the total work done is about 10^15 J.

November 15, 2011 at 11:22 am

Most of the 100W produced by humans goes to maintaining their body temperature rather than doing mechanical work, so your answer is an overestimate.

The best way to do that problem is to work out the energy needed to raise the centre of mass of the pyramid…

November 15, 2011 at 12:26 pm

My answer was slightly tongue-in-cheek.I think we produce a lot more power than 100W, but agree that much of this power goes into heating the body. Certainly one can power a dynamo to light a lightbulb, so I can’t be that far off. My solution takes into account all the work done against friction in sawing stones, dragging them along and up ramps, heaving them in to place etc. You asked “How much work was done building it?” not “What is its gravitational potential.” – the answers will differ by more than an order of magnitude I suspect.

November 15, 2011 at 1:34 pm

I’m not sure exactly how it is calibrated, but there are stationary cycles which are used to test the response of the heart. Rather than a speedometer, there is a gauge showing watts. It’s quite a bit of effort to get over 100 watts, and only someone very well trained could keep this up for more than a few minutes. So, my guess is that someone doing sustained manual labour could manage a few dozen watts at most (in addition to maintaining body temperature etc).

November 15, 2011 at 1:39 pm

I have an exercise bike in my bathroom that monitors the energy used. It’s quite depressing, actually. My standard workout is 30 minutes at 30 km/h, which is quite good going for an oldie like me. It burns up about 240 Calories, about the equivalent of a Mars bar.

November 15, 2011 at 3:20 pm

Factor of 10 too high? That looks like about half a kW!

Could you raise the temperature of 240 litres of water by one degree in half an hour?

Sadly, it probably takes several hours to get rid of the calories in a Mars bar.

Or have

made a factor-of-10 mistake?November 15, 2011 at 3:21 pm

Or have

Imade a factor-of-10 mistake?November 15, 2011 at 7:02 pm

Phillip: your metabolic rate is faster for several hours after doing sustained aerobic activity. So you continue to burn off that Mars bar even sitting in an office afterwards, compared to a no-exercise regime. It’s fairly obvious to me that I don’t need to wear a sweater for a while afterwards in an environment where I normally would. It’s also why exercise to lose weight is more effective earlier in the day.

November 16, 2011 at 9:16 am

“My standard workout is 30 minutes at 30 km/h, which is quite good going for an oldie like me. It burns up about 240 Calories, about the equivalent of a Mars bar.”

Depressing? That looks like a pretty amazing work rate to me (560 W) – more than I ever achieve in the bathroom anyway.

November 16, 2011 at 4:17 pm

Someone else also thinks Peter claims to be 10 times as powerful as he is! 🙂

Order of magnitude: 100 thousand seconds in a day, 100 W power, that’s 10 MJ per day, or about 2500 kCal. That’s about the recommended number. 10 Mars bars per day would be enough energy to live on. Sounds about right. So, you would need one for each two-an-one-half hours; no way you could burn it off in half an hour. (The difference is a factor of 5 not 10 since your bike only measures the extra effort, apart from maintaining body temperature etc.)

November 16, 2011 at 5:01 pm

Check out the following table, according to which 30 minutes cycling is about 180 Calories. I keep at quite a high speed, so it will be a bit more than this in my case. I’m sure about 250 Calories is OK.

http://www.bupa.co.uk/individuals/health-information/directory/e/exercise-weight-control

November 16, 2011 at 5:03 pm

ps. Cycling is certainly able to power several domestic light bulbs through a dynamo. I know this is true as we have a lecture demonstration about it….

November 16, 2011 at 5:20 pm

OK but plug in the numbers and your power comes out at half a kW!

I remember once getting up to 130 W, but could keep it up for only a few minutes. I just can’t believe 3 or 4 times that for half an hour. Yes, a few 100 W bulbs could be powered by a bike, but half an hour long?

November 16, 2011 at 5:28 pm

Half an hour’s gentle running expends about 400 Calories, which is up near a kW, much higher than my cycling output. And I’ve kept that work rate up for well over 3 hours, running marathons….although alas I can do it no longer.

November 17, 2011 at 10:13 am

I am pretty sure that Peter’s exercise bike is estimating his body’s total energy use rather than his mechanical work turning the pedals. It must assume some efficiency factor (say 25%), which would bring his mechanical power down nearer to 100 W

rather than 500 W. There are some interesting graphs for Tour de France riders here http://www.sportsscientists.com/2011/07/tour-de-france-stage-4-power-output.html

November 17, 2011 at 2:08 pm

That would explain the discrepancy. Now for the faster-than-light neutrinos. 🙂

November 19, 2011 at 1:31 pm

I don’t know if anyone watched the episode of “Bang Goes the Theory” called “Human power station” where they powered the energy needs of a ordinary family entirely by pedal power for a whole day.

Brief clip here

November 21, 2011 at 5:01 pm

A lot more interesting than the superluminal neutrinos is the claim of cold fusion in the ‘e-cat’ device, but the popular media are being a lot more cautious after the fiasco 20 years ago. Time will settle both claims.

November 15, 2011 at 10:57 am

My favourite is here: http://cosmic-horizons.blogspot.com/2011/09/long-way-to-chemists-rough-guide-to.html

AFAIK, it’s my own invention.

November 15, 2011 at 10:57 am

Yes, the link above is correct, but what I’m referring to is in the comments.

November 15, 2011 at 2:49 pm

For #4, the relevant number is that 1 atm is enough to raise water 30 feet. So if that 3 atm is a gauge pressure (i.e., pressure above atmospheric), the answer is 90 feet, and if it’s absolute pressure (above vacuum), then it’s 60 feet.

That 30-foot value is just the height of a water barometer. If you happen to remember that a mercury barometer is 76 cm tall and that mercury is about 14x that of water, then you can conclude that a water barometer is about 14 times .76 meters high, which is 10 meters, which is 30 feet. Alternatively, if you know that atmospheric pressure is 10^5 pascals, the density of water is 1000 kg/m^3, and g is 10 m/s^2, Then h = P/(rho g) = 10 meters.

#2 seems significantly harder than the others. I’m not surprised that it’s the last one to be tackled.

November 15, 2011 at 2:52 pm

Oh, one more thing: if you know some history of technology you might know the 30 feet for another reason. It’s the maximum height you can raise water with a pump. I believe mining engineers as far back as the 18th century had found this out.

November 16, 2011 at 7:15 pm

It’s the maximum height you raise water by “pulling” it with a pump at the top of the pipe, there is no limit to the height you can raise water by “pushing” it from the bottom.

November 15, 2011 at 3:29 pm

The correct answer to No. 4 is indeed 60 ft.

November 15, 2011 at 3:31 pm

I’m not sure No. 2 is as hard as people seem to think, but for those who are struggling here’s a hint: assume that all the air beneath the circle swept out by the blades is moved uniformly downwards.

November 16, 2011 at 4:31 am

Not sure if you noticed but your post got picked up on the physics subreddit here: http://www.reddit.com/r/Physics/comments/mbz4e/some_backofanenvelope_physics_problems_for_your/

I’ll copy my answer to 2 from there.

The lift force required to keep the helicopter flying is provided by the rate change of momentum of the air passing through the plane of the rotor. The rotor has rotational symmetry so the rate change of momentum can only depend on r, and not theta. Now it turns out any dependence on r doesn’t influence the final answer and it’s not hard to show this, but I would need to write out integrals and I don’t know how to do that on reddit. So we’ll presume the speed of the air through the rotor is independent of r.

Momentum change can be expressed as the product of mass flow rate (mdot) and velocity (v).

F = m g = mdot v

The mass flow rate through the plane of the rotor is given by

mdot = rho A v

Combine the two equations and v2 = m g / (rho A), which gives v ~= 13 m/s

The work done by the helicopter manifests as a change in kinetic energy of the air that passes through the rotor.

P = 1/2 mdot v2

Plug in some numbers and you get the power required is ~30kW, so O(10-100 kW)

To see if this estimate is reasonable-ish let’s see how much fuel is required to fly for 15 minutes. Wikipedia says the energy density of helicopter fuel is ~50 MJ/kg. Let’s say the chopper is 10% efficient at turning stored energy into the lift (absolutely no clue if that’s right). That means we need very roughly a kilo of fuel for every 10 s of flight, so 15 minutes would require 900 kilos of fuel. That seems a little high, but at least it’s not batshit out of whack.

June 27, 2012 at 1:20 pm

[…] today so I thought I’d wind down by giving you a chance to test your brains with some more order-of-magnitude physics problems. I like using these in classes because they get people thinking about the physics behind problems […]

April 25, 2019 at 1:57 pm

[…] so I thought I’d just do a quick post to give you a chance to test your brains with some more order-of-magnitude physics problems. I like using these in classes because they get people thinking about the physics behind problems […]