  # How big is a meteor? (shooting star)

After that brief flash as a meteor (a shooting star) traces a short line across the night sky, have you ever asked your self how big it is?

The meteor burns up before it reaches Earth – otherwise it would be a meteorite, so it seems to be a difficult question. But scientists love questions like this. A car-ful of UNSW physicists (Anna Gribakin, Gleb Gribakin and Joe Wolfe), returning from watching the annual Leonid meteor shower, enjoyed making an estimate based on what we had seen. Traditionally, such calculations are done on the back of an envelope, but it was dark—and this was before mobile phones. So of course we had to rely on data that we could remember.

A quirky game for passing time in the car, you might think, but it's actually typical of 'order-of-magnitude' calculations that are required often in science. 'Is this effect big enough to be important?' 'Is it large enough to be measured?' 'Could it be dangerous?' The answers to such questions often do not need precision: if it turns out to be so small that we cannot measure it, we may not need a better answer. And of course if the rough answer is interesting, we can then take the time to do a more serious calculation.

We reproduce the arguments here as an example of using simple physical arguments and plausible values to estimate unknown quantities. ### The first calculation

We needed to make several rough approximations, of which perhaps the roughest was saying that the meteor looked about as bright as an average star (and about the same colour), so it looks about as bright as the Sun (an ordinary star) would look if it were 10 light years away, ie reasonably close by stellar standards. We then reasoned as follows:

The intensity of sunlight – solar power received per square metre – is 1.4 kW/m2. (This constant is well known because one needs it for solar energy, ecology, architecture etc). The Sun is 8 light minutes away. If it were 10 light years away, then its intensity, I, would be given by the inverse square law:

I = (1.4 kW/m2) * (500 light seconds/300 million light seconds)2 = 4 nW/m2
Suppose the meteor is up where the Earth's atmosphere is rather thin, say 100 km away, so its radiated power, P, is spread over a sphere of radius 100 km, so

P = (area of sphere) * (4 nW/m2) = 2πr2 = 500 W
It burns at its brightest (comparable to brightish stars) for about one second, so its energy is E = Pt = 500 J.

Another rough assumption (justified in an appendix calculation below) is that this energy is roughly equal to the initial kinetic energy of the meteor.

Now Earth is 8 light minutes from the Sun, so it travels 3000 light seconds per year around the sun = 30 km/s. The meteor has 'fallen' from the outer solar system, so it is going faster than the Earth, and we strike it at an angle. We put the collision speed at 60 km/s, which is no rougher than some of the other approximations. So

½mv2 = 500 J   gives

m ~ 0.3 mg

Now make it a sphere with radius r and density ρ = 2000 kg/m3, m ~ 4ρr3

This gives a radius of 0.3 mm – about the size of a grain of sand. On our return, we were happy to see that this was broadly in line with information from other sources, such as satellite measurements.

### Does it really convert most of its kinetic energy into light?

At room temperature of 300 K, molecules travel a bit faster than sound due to their thermal motion. An object travelling at 200 times the speed of sound would therefore have enough kinetic energy in collisions to raise its temperature to ~2002 times room temperature, ie 12 million K. Objects vaporise and ionise at temperatures of thousands of K, so the meteorite needs only about 0.1% of its kinetic energy to vaporise and ionise, and nearly all of the rest is emitted as radiation. Most of this radiation will be emitted when it is vaporised and ionised, ie at temperatures of several thousand K, roughly the surface temperature of the sun, so a lot of that radiation will be in the visible. So the implicit assumption that it converts its energy to light with the same efficiency as does the sun is rough but not bad.

### Alternative calculation: can we work it out directly from the brightness and the colour?

Again we imagine the meteor as a little ball heated by air drag to a sun-like temperature, and again suppose it is about 100 km away from us. How big should it be to appear similar in brightness to a sun-like star 10 light years away from us?

The intensity of radiation it emits is proportional to its radius squared, and inversely proportional to the square of distance to it. So:

diameter of meteor = diameter of the sun * (100 km/10 light years)
The angular size of the sun is about 0.5 degree ~ 1/120 radian, so its diameter is then 150 * 106 km /120, and we get :

diameter of meteor = (1 * 109)(105)/((10 light years) * (3*107 seconds/year) * (3*108 m/s)) metres

= approximately 1 mm

This number is bigger than the previous one: ~3 times the diameter, or ~30 times the volume. But this is the size of the shining meteor, not the cold hard lump flying through space. It is quite reasonable that the shining ball of plasma is larger than the original cold dust grain, after its surface heats up and evaporates. But there are also several crude approximations in both calculations. (The guesses about the temperature are particularly serious, because the radiated energy depends on T to the fourth power.)

An order of magnitude is as good as one can hope for in such calculations: for more precision, one needs better data, more careful calculations – and a calculator!

 Joe Wolfe / J.Wolfe@unsw.edu.au, phone 61- 2-9385 4954 (UT + 10, +11 Oct-Mar). School of Physics, University of New South Wales, Sydney, Australia. 