More Order-of-Magnitude Physics

A very busy day today 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 without getting too bogged down in or turned off by complicated mathematics. 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:

  1. What is the tension in a violin string?
  2. By how much would the temperature of the Earth increase if all its rotational energy were converted to heat?
  3. What fraction of the Earth’s water is in its atmosphere?
  4. How much brighter is sunlight than moonlight?
  5. What is the mass of water in a soap bubble?

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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22 Responses to “More Order-of-Magnitude Physics”

  1. Anton Garrett Says:

    Another nice one: If the electrical energy stored in a fully charged alkaline torch battery were used to raise that battery against the force of gravity, how high would it go?

    • Steve Jones Says:

      D cell battery has capacity of about 10 amp hours

      energy = 1.5 volts x 10 amps x 60 mins x 60 secs = 54,000 Joules

      Gravitational Energy = Mass x g x height

      so, Height = Gravitational Energy / g x mass

      Height = 54,000/ 10 x 0.2kg = 27,000m = 27km

      about 3 times the height of Mount Everest !

      (it would lift a person to the top of a skyscraper!)

  2. For number 3, do you mean free water or total water? The two numbers would be very different, and one is easier to estimate than the other.

    Adrian

    • telescoper Says:

      All the water on Earth…

      • Well, the simple answer first:

        Atmospheric pressure: 10^5 Pa
        Surface area of the Earth: 4*pi*(6400 e3)^2 or about 5e14 m^2.
        Acceleration due to gravity: about 10 N/kg

        So mass of the atmosphere is about 5e18 kg.

        Relative humidity at the surface varies from 20% (in deserts) to 100%. Pick the geometric mean, 45%.

        Average temperature on Earth is 15C. Saturated vapor pressure (calculated by one of any number of ways – I used a standard semi-empirical equation we use a lot in ocean physics): 1.7 kPa
        So saturated specific humidity is 0.622 x 1.7/10^5 (the 0.622 is the ratio of gas constants for dry air to that of water vapor) = 0.01 g(vapor)/g(air)

        So, the specific humidity = 0.45*0.01 = 0.0045 g/g

        So, the mass of water in the atmosphere is 2.25 e16 kg.

        But this is an over-estimate because it assumes a uniform atmosphere with height. This is not so, and the air at high altitude has very little water vapor in it. So take the mean of the above number and 0 to get about 1e16 kg (note that this gives a volume of about about 10^4 km^3 of water vapor in the atmosphere).

        Now the oceans. The average depth is 4 km and the oceans cover about 0.7 of the surface of the planet. This gives a volume of about 14e17 m^3, or a mass of about 10^21 kg.

        We could stop here since most of the free water is in the oceans. But, we will continue and look at the ice sheets.

        Antarctica is about 14e6 km^2 and the mean thickness of the ice sheet is about 2 km. Only a minuscule fraction of the continent is ice free, so the volume of ice is 28e15 m^3. Using 0.9 g/cm^3 as the density of ice, we get a mass of 25e18 kg.

        The other big ice sheet is Greenland. This has an area of about 10% that of the Antarctic ice sheet and a similar thickness, so add 10% to the above ice sheet mass to get the combined mass of 27e18 kg.

        The two ice sheets combined pretty much all the fresh water around, so we’ll ignore the rivers and lakes.

        So the fraction of water in the atmosphere is

        1e16/(1e16 + 1e21 + 27e18) , about 1e-5 or 0.001%

        Now the tough bit. There is a heck of a lot of water in the mantle, which is subducted with crustal material. The problem is that the uncertainties in the amount of water stored in these minerals under these conditions are quite large. As a result, estimates of the about of water stored in the mantle range from about 1 ocean equivalent to 10 equivalents or more. Consequently, the above estimate may be up to an order of magnitude lower.

      • telescoper Says:

        I just counted the oceans – wasn’t aware of the mantle water – but and just assumed an average ocean depth of 4km and standard value for the average vapour pressure of water, giving about 10^-5

  3. My go at 4:
    Moonlight will be primarily due to the sunlight reflected of the moon – and moon radiation will be out of visible, so forget about it.
    The incident solar flux on the moon is approximately the same as on earth, say s.
    The fraction of reflected light will depend on the surface, and I don’t know the moon’s albedo, so I’ll take 0.5
    Light is scattered in all directions, but we don’t know where the Earth is, so we’ll just take the average.
    So, incident intensity I=s \pi r_moon ^2 and what reflects is I/2 is scattered into approximately half of a hemisphere surrounding the moon (it will be scattered further, but my intuition says the intensity of that will be much lower), so the “solar constant of reflected light on earth”:
    s’ = I/(2 (2 \pi r_me^2)), where r_me is the distance from moon to earth
    Hence the estimate will be
    moon brighness / sun brightness = s’ / s = 1/4 (r_m/r_me)^2 = 1/4 (0,27/60)^2 = 200 000

  4. Here’s 2. The rotational energy is about .2 M R^2 w^2. Set equal to M c Delta T. The temperature rise is

    Delta T ~ .2 R^2 w^2 / c.

    R is about 6e6 m, w =2pi/(86400 s). Specific heats of solids are probably of order 1000 in SI units. (Water’s about 4000, right?) Works out to

    Delta T ~ 50 K or so.

  5. 5: Soap bubbles show visible-light interference fringes, so they’re not that much thicker than a wavelength of light. Call it 10^4 nm (20 wavelengths), or 10^-3 cm. If the bubble’s a few centimeters in radius, then its surface area is about 100 square centimeters. So the volume of the spherical shell is 0.1 cc, and the mass is 0.1 g.

  6. telescoper Says:

    Nobody’s done 1 yet, although it’s the easiest one. I assumed the string is tuned to 440Hz (A above middle C), and is about 30cm long. The tricky bit then is to guesstimate the mass per unit length \mu since the frequency is related to \sqrt(T/\mu). Using 1 gram per metre, the answer turns out to be 100 N.

    • I’d go for the value of mu by estimating the cross-section area of the string and its density. mu = pi r^2 rho. With a radius of about .5 mm and a density of about 1 g/cc, you get something close to your value of 1 g/m.

  7. My favourite: Forget gravity and compress the entire visible universe (remember the Hubble Deep Field is about the size of a grain of rice; remember all of Tony Tyson’s images of the sky covered with galaxies) into a ball. What is the radius?

  8. Steve Jones Says:

    How much energy does it take to remove a fridge magnet from the fridge?

    (I don’t know the answer to this one but would like to)

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