## Raindrops, stars and photons

Here is another one of those occasional posts with mazin facks bout stronomy. (See also Almost Nothing and A Dim Glimmer).

Today’s rambling is inspired by some fascinating tweets emitted by darkskyman aka Steve Owens of Dark Sky Diary fame. First I will give you the gist of Steve’s ‘believe it or not’ sequence. From the average rainfall in the UK last year (1.33m), the area of the UK (2.426×10^11 m^2), and the average volume of a raindrop (10^-9 m^3) he calculated that the total number of raindrops that fell on the UK was 10^20. That is a big big number. However, continues Steve, the number of stars in the Universe is approximately 10^23.

So as Steve told us ‘for every raindrop that hit the UK last year there are 1000 stars in the Universe’.

Very nice, but I found myself thinking – how does the rate of raindrops compare to the rate of photons? The average raindrop flux is 13 raindrops/m^2/s. Lets compare to the Sun. The solar constant is 1360 W/m^2. If we take the typical photon as being at about 500nm with energy hc/lamda = 4×10^-19J, we get roughly 3×10^21 photons/m^2/s.

So we get much much much more sunlight than rain! Woohee!

What about starlight? Well, as Mr Olbers pointed out, a Universe full of stars would make a sky as bright as the Sun in every direction. However, the Milky Way fades out, and the universe runs out too, in time and therefore space. So lets just get empirical. Cosmology types will often plot the ‘cosmic optical background’ at a level of about 10^-8 W/m^2/sr, about a factor of a thousand less than the CMB. However, that is the extragalactic background light;  the summed emission from nearby stars is in fact much more. According to my Trusty Allen, star light from the whole sky is equivalent to 460 V=0 stars, or one star of V=-6.7. The  apparent mag of the Sun is V=-26.7. So the scattered starlight is 20 magnitudes fainter or a factor of 10^8.

So in super-crude terms, starlight is giving us something like 3×10^13 photons/m^2/s. Still lots more than the 13 raindrops/m^2/s.

But what energy? Scaling down from the solar constant, starlight is giving us an energy flux of about 1.4×10^-5 W/m^2. What about raindrops? Each of those raindrops has mass 10^-6 kg. The terminal velocity of a raindrop depends on size, but at 1mm its about 4 m/s. So the KE per raindrop is about 8×10^-6 J and the energy flux is therefore 10^-5 W/m^2, about 6 times as much as starlight.

So… in particle count terms, the Sun wins hands down; starlight is down a factor of hundred million but still huge; and the raindrop count is pitiful, another factor of a trillion down.

In energy terms, the Sun still wins easily, with starlight a hundred millions times down; but the rain carries more energy than the starlight – just.

Please do check the maths… Anyhoo. Better do some real work today.

### 4 Responses to Raindrops, stars and photons

1. What is the total amount of extra-solar-system energy received by all radio telescopes (pre-SKA) in all of history?

• andyxl says:

Didn’t you do this one before? I think the answer is diddly-squat, but please do tell us ole bean

• Yes, I did. Less than the kinetic energy of a snowflake.

2. Will Sutherland says:

On a slightly related note, while on a long stay at Paranal I calculated the solar energy as equivalent depth of oil “rainfall”. Taking day/night average for a desert as 300 W/m^2, that’s 9.5 GJ/m^2/year.

The energy density of oil is ~ 35 MJ/litre (crude, petrol and diesel are quite similar according to Wikipedia), so this gives around 270 mm depth of “oil-fall” per year, i.e. nearly half of London’s average rainfall.
Clearly there is a large problem actually catching this and transporting or storing it; but imagining the Sahara shin-deep in oil each year does show very intuitively just how huge the possible solar power is.

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