# How can you usefully determine the dimensions of the perseides meteor shower band earth passes through?

Yesterday the annual perseides meteor shower peaked (and I was lucky to see a handful of wakes). Since the bits from Swift-Tuttle have distributed to form a band of debris over the millenia and earth passes through that band, I wonder how to describe the band's dimensions.

I'd go along these lines:

• "shower period" is from July 17 – August 24, i.e. 39 days
• Earth orbits the sun at $$1.07×10^5$$ km per hour says Cornell

I'd assume:

• Earth moves through the band's full diameter, not e.g. tangentially
• the band is symmetrical, i.e. the asymmetrical pre and post-maximum rates do not matter

So that gives a simplified $$39 × 24 × 1.07×10^5 = 1.00152×10^7$$ km as the plausible diameter.

Is my rule of thumb way to calculate this reasonable? As in, campfire discussion precise? Or am I missing relevant details?

• – uhoh
Aug 14, 2022 at 8:42
• FWIW, here are the orbital elements of 8P/Tuttle. Its inclination is fairly steep: almost 55°. It has SPKID 90000179, which you can use to make a 3D interactive plot of its orbit using my live Python script at the end of this answer. To plot the whole comet orbit, I suggest a step of 30 days. The SPKIDs of the Sun & Earth are 10 & 399. Aug 14, 2022 at 9:01
• @PM2Ring that is an especially useful suggestion. Thanks! Layperson as I am, I'll take my time looking at the script. Aug 14, 2022 at 10:37
• Glad you like it. :) The script is mostly Python, but it uses Sage's vector objects & 3D plotting functions. Of course, the meteoroid orbits aren't exactly the same as the comet orbit, but they're pretty close, and almost in the same plane. Aug 14, 2022 at 10:47