Meteor Showers & the Zenith Hourly Rate
Today (6th May 2010) was the maximum of a meteor shower called the eta-Aquarids. This shower lasts many weeks, beginning on 19th April and ending around 28th May, with a peak of activity today. If you look up info for this shower you will find quoted a Zenith Hourly Rate (ZHR) of 85.
That means if you were oberving this shower in perfect conditions – no light pollution, no cloud, no moon, and with the radiant of the shower (the point where the meteors appear to emerge from) directly overhead (at the zenith) you would expect to see around 85 shooting stars per hour.
When people talk about meteor showers they sometimes quote the ZHR as an indication of how many you can expect to see, but this is often misleading, for reasons I’m about to discuss.
Let’s use the eta-Aquarids shower as an example.
If I wanted to view the shower at it’s most advantageous position I’d need to go out around 0200 tomorrow morning, after the radiant had risen in the east, and before the sky starts to brighten as we enter twilight.
The radiant at this point will be very low, about 5° above the horizon, well short of the 90° it needs to be for the radiant to appear directly overhead.
This isn’t the only problem; the Moon is up in last quarter phase, rising before the radiant, and so the Moon’s light will drown out the fainter meteors.
Then there’s light pollution. From my garden I can see stars (and meteors) down to around magnitude 4, and so the fainter ones won’t be visible through the orange city glow. Let’s assume the Moon’s glow won’t be any worse than the man-made light pollution.
And finally there’s the more mundane problem of cloud cover. If my sky isn’t entirely clear of clouds, then I may miss some of the shooting stars.
How do these factors effect the number of meteors I’ll see? We can put them into an equation to find out:
Actual Hourly Rate = =(ZHR x sin(h))/((1/(1-k)) x 2^(6.5-m)) where
h = the height of the radiant above the horizon
k = fraction of the sky covered in cloud
m = limiting magnitude
Tonight for the eta-Aquarids at 0200 ZHR = 85, h = 5°, m = 4 (optimistically), k = 0 (very optimistically)
Therefore the actual hourly rate of eta-Aquarids I can expect is
(85 x sin(5))/((1/(1-0)) x 2^(6.5-4)) = 1.3, or just over one per hour. Not great at all, and that’s assuming a cloudless sky!
But don’t let that put you off. You never know, if you pop out at 0200 (perhaps taking a break from the UK election coverage) look over to the east and maybe you’ll catch that one bright eta-Aquarid meteor…