## Saturday, 4 August 2007

### Shiveluch - measuring height from shadow

What on Earth is that? This is a satellite image of an erupting volcano. What you can see is the plume from exactly above, before it got smeared by the winds.

Shiveluch volcano is on the Kamchatka peninsula, here are the details of this eruption in last winter, and a larger image.

Can we guess the height of the plume?

Well, we have to find out how high was the Sun (that casts this shadow) on the sky at that moment. The page says Aqua satellite took this picture at 02:00 UTC on March 29. Kamchatka is only about 20 degrees west from the date line (which is about opposite from the 0 meridian, just to appreciate how vast Russia is). The exact latitude and longitude of Shiveluch is 56°63' N, 161°32' E. That means local noon is about 161°/15° = 10.75 hours = 10 hours 45 minutes before noon in Greenwich (the Earth turns 360 degrees in 24 hours, or 15° in one hour). Or to put it another way, it's 10:45 later than the Universal Time. So the local time was 02:00 + 10:45 = 12:45, or 3/4 after noon, when the picture was taken. (Of course this is a perfectly "local time" defined by the apparent movement of the sun - we don't care for time zones that can be arbitrarily defined by authorities). However, we need a correction to this, because Earth's orbit is not a perfect ring, and that results in a slight change of the apparent speed of the Sun's movement on the sky (as for the effects of the eccentricity of Earth's orbit, see the explanation). This correction is never more than 15 minutes though, and on March 29 it is only about 5 minutes (see the explanation). Now let's make a small cheat here. Let's just say it was 12:00 local time, to make things easier. Local noon means the Sun is at its highest elevation (on a given day). The highest elevation depends on the part of the year - it climbs much lower in winter and higher in the summer. But note that March 19 was only 8 days after the spring equinox. So let's just say it was approximately spring equinox (our 2nd cheat). This we prefer because it's easier to tell, how high the Sun climbs. On equinox day, if you stand on the equator (latitude 0°), the Sun passes exactly overhead. If you stand on the North or the South pole (latitude 90°), the Sun circles exactly on the horizon. And if you stand on latitude 56° N, the Sun will appear to move around along a circle in a plane that is tilted by 56 degrees from vertical - or 90°-56°=34° above the horizon. So it reaches its highest elevation on the sky at 34°. So we know that at the time the above photo was taken, the Sun was 34° high on the sky. (actually i think the two cheats i made (it's a bit later in the day, but it's also a few days after equinox) somewhat cancel out each other, so this might be a fairly good approximation, if i didn't do any mistakes of course.) The photo appears to be oriented north (check an atlas or google maps), so the fact that shadows fall almost exactly up/north reassures us that it indeed was pretty close to noon. (Making the above calculations of the local time somewhat unnecessary...)

I measured the length of the shadow (the shadow of the whole big plume) to be 53.4 pixels on average, which equals to a length of 13 km. In the center of this ash cloud is an other shadow, which probably indicates a secondary, smaller plume (if it's possible at all) that rises above the large one, and casts a shadow maybe 6-12 pixels long or 1.5-3 km. Now knowing the sun was 34 degrees high, a simple trigonometry gives us tan(34°)*13 km = 7.8 km (4.9 miles) for the height of the main plume, or probably 8.7-9.6 km for the whole thing.

As you can see on the drawing above, this is the height of the plume compared to the plains around. The height of the volcano itself is about 3 km, so the height of the ash cloud may be a little less. From this image it seems that it's probably surrounded by low lying plains (I tried to check it on google maps but the surroundings of the volcano are clouded there by an other ash cloud!). So maybe it really rises 2-3 km above its surroundings.

Unfortunately i don't know much about eruptions. Is it realistic that 10 minutes after the eruption the plume is at 8-9 km? Probably yes.

This way, one could teach a little physics (or astronomy), armed with only this beautiful photograph. Actually you can go further. Like start thinking about why is the landscape so snowy in late March, when Kamchatka is about as far north as Denmark or Britain?