Simple orbital collision problem  need help!

10082023, 06:03 PM
Post: #1




Simple orbital collision problem  need help!
I am giving my "Anime Blinded Me With Science" talk again at this year's Anime Weekend Atlanta. (The talk is on Saturday, 28 Oct, at 3:15. The convention runs Thursday 26 through Sunday 29.)
I'd like to go the extra mile on one particular bit of the talk, and I need some help with the math. For this part, I show an opening scene from the anime "Planetes" that shows a loose aviation screw, apparently in an equatorial orbit, hitting the window of a suborbital commercial passenger spacecraft that is apparently on a polar trajectory. So, roughly traveling at right angles to each other. In the past, I've assumed 8 km/sec for the screw, and ignored the velocity of the spacecraft. Simple high school physics and math (KE = 0.5mv^2) gives 21.7 kilojoules for a 0.68 gram screw, 35 times more energy than a 9mm parabellum fired at point blank. So, plenty powerful enough to punch a hole in the side of the spacecraft. Can someone help me improve the fidelity of my model? 1. Let's assume the likely apogee of a transcontinental suborbital passenger flight is 100 km. (Can we use 110 km to get more impressive numbers, or is that stretching things?) What's a simple model for the horizontal speed at apogee? (And what's the value?) 2. 100 km and V=sqrt(GM/r) gives 7.771 km/sec for the screw's horizontal velocity. 3. My calculus is superrusty! What's the equation that will describe the relative speed at which the screw impacts the side of the spacecraft, assuming their paths are perpendicular? I plan to use this to show kids (potentially interested in STEM careers) that: (a) simple backofthe envelope calculations are possible and often good enough; (b) but there are some nifty math tools that engineers and scientists can use to really predict closely what the real values will be. Thanks! Daily drivers: 15c, 32sII, 35s, 41cx, 48g, WP 34s/31s. Favorite: 16c. Latest: 15ce, 48s, 50g. Gateway drug: 28s found in yard sale ~2009. 

10082023, 06:21 PM
Post: #2




RE: Simple orbital collision problem  need help!
(10082023 06:03 PM)johnb Wrote: 1. Let's assume the likely apogee of a transcontinental suborbital passenger flight is 100 km. (Can we use 110 km to get more impressive numbers, or is that stretching things?) What's a simple model for the horizontal speed at apogee? (And what's the value?) Am I being superdense here in not assuming that the flight is a parabola (which would probably mean the horizontal speed at apogee is negligible)? Daily drivers: 15c, 32sII, 35s, 41cx, 48g, WP 34s/31s. Favorite: 16c. Latest: 15ce, 48s, 50g. Gateway drug: 28s found in yard sale ~2009. 

10132023, 01:00 PM
Post: #3




RE: Simple orbital collision problem  need help!
I'm probably among the least qualified people to answer your question, but since nobody else has, and I think it's interesting, figured I'd chime in.
Why would being a parabola make horizontal speed at apogee negligible? The velocity vector with respect to the "vertical" component (distance from earth's center), would have zero magnitude, just before it changed sign and began descent. The horizontal velocity vector would remain the same, otherwise it's never going to go very far, in which case, why expend all that energy? Presumably it didn't just launch straight up and return the same way. At some point it "tipped" along the azimuth of its intended direction of flight and went really fast above the atmosphere where drag isn't as much of a factor. I'm assuming this isn't a "constant boost" magical rocket, and at some point it's in ballistic flight until such time as it must begin a powered descent to bleed energy and make a controlled landing, presumably to the delight of all the passengers. Any maneuvers to "flip" the spacecraft are likely to make the flight path not be a "perfect" parabola, but I think it's still largely parabolic. Altitude is irrelevant to whether or not the flight is "suborbital," velocity is what matters, so 110KM doesn't make a big difference either way. Most of your V is along the flight path, i.e. "horizontal." For that, you'll have to use the rocket equation applied to your particular spacecraft. Yeah, you're going to get some acceleration due to gravity, but I think your velocity along the direction of flight is the largest component in terms of added energy in the collision. But I'm just guessing here. I'm no rocket scientist. 

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