# 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
Commented Aug 14, 2022 at 8:42
• – uhoh
Commented Aug 14, 2022 at 8:43
• 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. Commented 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. Commented 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. Commented Aug 14, 2022 at 10:47