4 ms·

> And the satellites, being at 100x higher altitude, are 100x more spread out than the airplanes. This doesn't feel right. Shouldn't the right frame of referen

by jsnell 5y ago

> And the satellites, being at 100x higher altitude, are 100x more spread out than the airplanes. This doesn't feel right. Shouldn't the right frame of reference be the distance to the center of the earth, not the sea level? Then it is not a 100x difference, but more like a 10% one.

karatekidd32v 5y ago
Agreed rough math there doesn't seem right. Also, I wonder if there's an added benefit from the inverse square law here: https://en.wikipedia.org/wiki/Inverse-square_law https://en.wikipedia.org/wiki/Inverse-square_law
montenegrohugo 5y ago
Here's the actual correct math: --- Area of sphere: 4πr^2 Airplane cruising height: ~10km Satellite orbit height: ~550km Radius of earth: 6371km thus, relative increase: (6371+550)/(6371+10) = 1.084 = 8.4% increase Squared (because of first formula) that corresponds with a 1.084^2 = 1.175 = 17.5% increase in area. --- Still, a few caveats: 1. Earth is huge. 510.1 million km² is a lot of space (+20% at 550km altitude). We could have a million sats with each having more than 1km^2 to themselves. 2. Satellites orbit at different heights. Amazon's and SpaceX's satellites will not be on the same orbit. 3. Starlink satellites are at a sufficiently low altitude that Kessler Syndrom is not a problem; even if they all simultaneously turned into millions of pieces of dead debris at the same time, the atmospheric drag would make them lower their orbits and burn up in just a few months.
gamegoblin 5y ago
If we were also launching from the center of the earth I would agree, but as we are launching from sea level, I think the general idea still holds. Thought experiment: You are standing in a field. There is a 10x10 meter plate hovering 10 meters above you. There is a 1 square meter target on it. You fire a gun upwards at a random location on the plate. There is a 1 in 100 probability you hit the target. Now imagine there is a 20x20 meter plate hovering 20 meters above you. It is perfectly occluded by the original 10x10 meter plate. It also has a 1 square meter target on it. When you fire at a random location on the original plate, the bullet passes through it and continues on to the higher plate. What is the probability you hit the target on the higher plate? I believe it is 1 in 400. From this thought experiment it seems that altitude from launch point is what counts.
- shkkmo 5y ago
  Except you can't generalize a square plate to a spherical shell...
  - gamegoblin 5y ago
    Can you explain why? If I am launching a rocket, and there is a 1x1 meter satellite orbiting 1000km above me. What is the probability that my rocket hits that satellite compared to an identical satellite at 2000km above me? The area of the angular sector of the sky that the 2000km altitude satellite is 1/4 of the 1000km altitude satellite.
    - shkkmo 5y ago
      The orbital launch will follow a curved path through succesive spherical shells. The percentage of the spherical shell occupied by the 1000km satellite is not 4x larger than the percentage of the spherical shell occupied by the 2000km satellite.
    - only_as_i_fall 5y ago
      This is the wrong mental model entirely unless your rocket launch is directly vertical and instantaneous.
- froh 5y ago
  (10x10):(20x20) sounds like 1:4 to me, instead of 1:400, no?
  - gamegoblin 5y ago
    It's not 1:400 -- it's 100:400 which is equal to the 1:4 you mention. That is, you 2x the height of the target which results in the probability of hitting being 4x less.
- jsnell 5y ago
  Now expand your thought experiment to 4 people, standing in a 10 meter square from each other. They each have their own 10x10 and 20x20 plates, and their own targets. Clearly the 10x10 plates are lining up (modulo curvature). But the 20x20 plates are not, they're overlapping. So when I shoot through a random location of my own 10x10 plate, there's a chance that I'll hit a target from somebody else's 20x20 plate. Sum up those additional chances, and they'll cancel out the 4x difference you found. This feels like it would make a nice puzzle, your phrasing makes for a great misdirect / sleight of hand.

by jsnell 5y ago

karatekidd32v 5y ago

Agreed rough math there doesn't seem right. Also, I wonder if there's an added benefit from the inverse square law here: https://en.wikipedia.org/wiki/Inverse-square_law https://en.wikipedia.org/wiki/Inverse-square_law

montenegrohugo 5y ago

Here's the actual correct math: --- Area of sphere: 4πr^2 Airplane cruising height: ~10km Satellite orbit height: ~550km Radius of earth: 6371km thus, relative increase: (6371+550)/(6371+10) = 1.084 = 8.4% increase Squared (because of first formula) that corresponds with a 1.084^2 = 1.175 = 17.5% increase in area. --- Still, a few caveats: 1. Earth is huge. 510.1 million km² is a lot of space (+20% at 550km altitude). We could have a million sats with each having more than 1km^2 to themselves. 2. Satellites orbit at different heights. Amazon's and SpaceX's satellites will not be on the same orbit. 3. Starlink satellites are at a sufficiently low altitude that Kessler Syndrom is not a problem; even if they all simultaneously turned into millions of pieces of dead debris at the same time, the atmospheric drag would make them lower their orbits and burn up in just a few months.

gamegoblin 5y ago

If we were also launching from the center of the earth I would agree, but as we are launching from sea level, I think the general idea still holds. Thought experiment: You are standing in a field. There is a 10x10 meter plate hovering 10 meters above you. There is a 1 square meter target on it. You fire a gun upwards at a random location on the plate. There is a 1 in 100 probability you hit the target. Now imagine there is a 20x20 meter plate hovering 20 meters above you. It is perfectly occluded by the original 10x10 meter plate. It also has a 1 square meter target on it. When you fire at a random location on the original plate, the bullet passes through it and continues on to the higher plate. What is the probability you hit the target on the higher plate? I believe it is 1 in 400. From this thought experiment it seems that altitude from launch point is what counts.

shkkmo 5y ago
Except you can't generalize a square plate to a spherical shell...
- gamegoblin 5y ago
  Can you explain why? If I am launching a rocket, and there is a 1x1 meter satellite orbiting 1000km above me. What is the probability that my rocket hits that satellite compared to an identical satellite at 2000km above me? The area of the angular sector of the sky that the 2000km altitude satellite is 1/4 of the 1000km altitude satellite.
  - shkkmo 5y ago
    The orbital launch will follow a curved path through succesive spherical shells. The percentage of the spherical shell occupied by the 1000km satellite is not 4x larger than the percentage of the spherical shell occupied by the 2000km satellite.
  - only_as_i_fall 5y ago
    This is the wrong mental model entirely unless your rocket launch is directly vertical and instantaneous.
froh 5y ago
(10x10):(20x20) sounds like 1:4 to me, instead of 1:400, no?
- gamegoblin 5y ago
  It's not 1:400 -- it's 100:400 which is equal to the 1:4 you mention. That is, you 2x the height of the target which results in the probability of hitting being 4x less.
jsnell 5y ago
Now expand your thought experiment to 4 people, standing in a 10 meter square from each other. They each have their own 10x10 and 20x20 plates, and their own targets. Clearly the 10x10 plates are lining up (modulo curvature). But the 20x20 plates are not, they're overlapping. So when I shoot through a random location of my own 10x10 plate, there's a chance that I'll hit a target from somebody else's 20x20 plate. Sum up those additional chances, and they'll cancel out the 4x difference you found. This feels like it would make a nice puzzle, your phrasing makes for a great misdirect / sleight of hand.