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I think one thing that really causes people to worry about Kessler Syndrome unnecessarily are satellite visualizations. Out of necessity, they have to display

I think one thing that really causes people to worry about Kessler Syndrome unnecessarily are satellite visualizations. Out of necessity, they have to display every object as at least 1 pixel. But that typically dramatically overstates the size of most objects in relation to the size or earth. Most visualizations make it look like Earth is absolutely covered with satellites. Whereas in reality, those satellites are the size of a car or smaller. 8000 cars might not even fill up a good sized stadium parking lot, much less the entire surface of the Earth. And many of those satellites are orbiting at different altitudes, so they're even more spread out than just the surface of the Earth.

RationPhantoms 2y ago
Genuinely asking, do you think that could be solved by having better orbital visualizations?
- lazide 2y ago
  Not the poster, but the issue is one of scale. Any visualization that showed a full orbit in a visual field that a human can see at once, necessarily has to exaggerate the size of the satellite or it would be completely impossible to see the satellite at that same scale. By millions or even billions of times. A GPS satellite in medium orbit has an orbital diameter of around 45, 000 km but is only a little over 5 meters in size. A ratio of roughly 10,000,000 to 1. If you had a visualization a meter wide of its orbit, the satellite to scale would be less than a micron in size. Completely invisible. The same problem comes up with those ‘topology of the earth’ maps that try to show how things like depth of the oceans, height of the Himalayas, etc. on a globe. They have to dramatically exaggerate the actual altitude differences for them to be perceptible. [https://www.quora.com/If-Mount-Everest-was-on-a-globe-and-was-made-vertically-proportional-to-the-globe-avg-size-how-tall-would-it-be https://www.quora.com/If-Mount-Everest-was-on-a-globe-and-wa...] The actual to scale difference in height of the Himalayan mountains vs sea level on a regular sized globe would be less than .001 inches. Completely imperceptible without precision measuring devices. Even producing such a globe accurately would be a feat of precision manufacturing. Even on a giant (1 meter) globe, we’re talking less than 1mm. Literally smoother than a billiard ball. But they are the highest mountains in the world, and absolutely breaktaking if near them. And space is even bigger, and satellites even smaller.
  - WillAdams 2y ago
    An accurate steel replica of earth the size of a ball-bearing would be the most precise ball bearing ever. Great book on this: https://goodreads.com/book/show/35068671-the-perfectionists https://goodreads.com/book/show/35068671-the-perfectionists
    - coolspot 2y ago
      Doubt it. Earth is not even a sphere.
      - lazide 2y ago
        Looks like you’re right! Earth is within .3% of being a sphere, where ball bearings are within .01%. [https://en.m.wikipedia.org/wiki/Earth%27s_circumference https://en.m.wikipedia.org/wiki/Earth%27s_circumference]
      - WillAdams 2y ago
        Correct --- I should have said "smoothest".
    - sandworm101 2y ago
      I've heard this a couple times and never believed it. Earth is perceptibly not round. Scott Manley did a quick video showing that in some satellite images the difference equates to more than a couple pixels. Some quick googling shows the earth being about 0.3% not-round, which would match a 1000-pixel image being not-round by a few pixels. But a 0.3% deviation is exponentially greater than is allowed by the standards for ball bearings. >> the Earth deviates from spherical by only a third of a percent https://en.wikipedia.org/wiki/Figure_of_the_Earth https://en.wikipedia.org/wiki/Figure_of_the_Earth >> For instance, the Anti-Friction Bearing Manufacturers Association (AFBMA) has a set of grades for bearing balls. A grade three ball has to be spherical within 3 millionths of an inch and the diameter must be accurate within 30 millionths of an inch. This means that for a grade three quarter-inch ball, the diameter would have to be between 0.24997 and 0.25003 of an inch and the smallest diameter measured on the ball has to be within 3 millionths of the largest diameter. https://science.howstuffworks.com/transport/engines-equipment/question513.htm https://science.howstuffworks.com/transport/engines-equipmen...
- aaronharnly 2y ago
  I suppose you could generate an alternate view (maybe someone has?) that doesn't show the individual satellites, but instead only shows little green/yellow/red flashes whenever satellites cross within some radius of one another.
- cruffle_duffle 2y ago
  That is a tricky problem to solve. The scales are so far apart from each other it’s a bit like solving “how do you scale the distance between the sun and the planets in a way that keeps some detail of each while giving a sense of the true amount of nothing between them”. According to ChatGPT, if the earth was the size of a spherical cow (approx 1.5m in diameter) then the space station would be about 8.5 micrometers which is roughly the size of some bacteria. It would be 46 millimeters away from the cow. If earth was a spherical cow then a starlink satellite would be 0.31 micrometers, or 310nm which is about the wavelength of violet light. It would be about a large thumbtack away. The orbital periods for these would be in the order of microseconds.
  - wizzwizz4 2y ago
    Please don't trust ChatGPT. It's got the right orders of magnitude, which is lucky, but it's still giving you the wrong numbers. If the earth were 1.5m in diameter, the space station would be proportionally 12.83µm long[0], orbiting at a distance of 50.66mm (2 inches)[1]. [0]: https://www.wolframalpha.com/input?i=iss+length+%2F+earth+diameter+*+1.5m https://www.wolframalpha.com/input?i=iss+length+%2F+earth+di... [1]: https://www.wolframalpha.com/input?i=iss+height+%2F+earth+diameter+*+1.5m https://www.wolframalpha.com/input?i=iss+height+%2F+earth+di...
- LeifCarrotson 2y ago
  I don't. The human sensorium (whether actually looking at an image, or visualizing something based on some numbers) has limited dynamic range. Sometimes, it can help a little to convert to something vaguely human-scale, which probably doesn't fit on a screen. The surface area of the planet is ~500 million km^2, there's a large park near me with long, clear sight lines where I go for runs play disc golf that's about 5 square km. If I shrink the ~3x2 meter satellite dimensions by an equivalent factor of about 100M, it's an object with a size of 30x20 micrometers (unless I've screwed up a square-cube ratio somewhere), on the order of a grain of silt, the thickness of a Sharpie ink mark, or something a little smaller than a cross-section of a hair. My intuition approximates those dimensions and probabilities of two of those infinitesimal specs all the way to zero. On the other hand, tens of thousands of satellites, those barely-palpable specs crossing that region of space once every 90 minutes, for hundreds of years? Then my intuition says it's a certainty! Facing the multiplication of an impossibility with a certainty, intuition says they cancel out and it's a 50/50 chance, which is just not how it works. Sometimes you have to just do the math and trust the numbers.
schiffern 2y ago
It's popular to point out how satellite visualizations exaggerate the size, but we never seem to notice how satellite visualizations shrink the duration of time. You only look at a visualization for maybe a minute, whereas these objects are actually out there whizzing around 24/7 for decades or even millennia. Both effects are misleading dataviz artifacts, but at least they're in opposite directions! A great ESA video, linked from Don Kessler's website, shows a time-lapse simulation that better captures the seriousness of space junk. In the simulation @5:25, realize that each of those flashes represents an Iridium-Cosmos magnitude collision event: http://www.youtube.com/watch?v=RvZ3Lr-Tj6A http://www.youtube.com/watch?v=RvZ3Lr-Tj6A
- bryanlarsen 2y ago
  The third & fourth factor are whether the satellites are actively avoiding collisions and whether the satellites are in a self-clearing orbit. Depending on the assumptions you make, we could have billions or even trillions of Starlink satellites without triggering Kessler simply because - Starlink satellites are actively controlled and steer to avoid collisions. Yes there are sometimes a couple of failed Starlinks in orbit, but since the active ones avoid the dead ones, you have to calculate the risk of one dead satellite hitting another dead satellite. And that's less than random because they quickly decay out of their crowded orbital plane to a different less crowded one. - dead Starlinks are in a self-clearing orbit, deorbiting in under 5 years. As the article indicates Kessler is a decades-long process. Which can't happen if the satellites don't stay in orbit that long. - most dead Starlinks aren't completely dead, or give alerts before they die, allowing SpaceX to actively deorbit them safely and quickly - the vast majority of particles from a collision in a self-clearing orbit result in immediate clearing. If the original dead Starlink that would take 5 years to clear suddenly fragments, most fragments de-orbit within a single orbit (90 minutes). Elastic deformation reduces their velocity, lowering the orbit. Half of fragments are toward a lower orbit. The other half only boost apogee and either lower perigee or leave it unchanged. Being smaller, their surface area / mass ratio increases, reducing deorbit times. Basically, Kessler is not a concern at VLEO altitudes. Unfortunately most proposed Starlink competitors cannot afford to launch the thousands of satellites necessary for a viable VLEO constellation and instead are launching dozens or hundreds into LEO orbits which aren't self-clearing and do pose a Kessler concern.
inglor_cz 2y ago
For some related ideas... Let us imagine that there are 100 000 Starlink-sized satellites on the orbit. At a given time, approximately 29 000 of them are over dry land and the rest over the sea. 29 000 car-sized object randomly moving across all 6 continents, including terrains such as the Himalayas. It would be pretty infrequent for them to come close to one another.
- inhumantsar 2y ago
  that's being generous too, since the spherical area those satellites cover is considerably larger than it is at ground level. Douglas Adams nailed it. "Space is big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space." Very few people, even us nerds who are aware of the scale, and actually hold that scale in their minds properly. And that's just low earth orbit. Geosynchronous, the Lagrange points, an Astronomical Unit, a Light Minute, a Light Year. It's one thing to understand what these things mean, but it's entirely another to put them into a human-scale context. It's like the difference between knowing that there are 20 million people in the New York area and trying to name each one of them individually.

RationPhantoms 2y ago

Genuinely asking, do you think that could be solved by having better orbital visualizations?

lazide 2y ago
Not the poster, but the issue is one of scale. Any visualization that showed a full orbit in a visual field that a human can see at once, necessarily has to exaggerate the size of the satellite or it would be completely impossible to see the satellite at that same scale. By millions or even billions of times. A GPS satellite in medium orbit has an orbital diameter of around 45, 000 km but is only a little over 5 meters in size. A ratio of roughly 10,000,000 to 1. If you had a visualization a meter wide of its orbit, the satellite to scale would be less than a micron in size. Completely invisible. The same problem comes up with those ‘topology of the earth’ maps that try to show how things like depth of the oceans, height of the Himalayas, etc. on a globe. They have to dramatically exaggerate the actual altitude differences for them to be perceptible. [https://www.quora.com/If-Mount-Everest-was-on-a-globe-and-was-made-vertically-proportional-to-the-globe-avg-size-how-tall-would-it-be https://www.quora.com/If-Mount-Everest-was-on-a-globe-and-wa...] The actual to scale difference in height of the Himalayan mountains vs sea level on a regular sized globe would be less than .001 inches. Completely imperceptible without precision measuring devices. Even producing such a globe accurately would be a feat of precision manufacturing. Even on a giant (1 meter) globe, we’re talking less than 1mm. Literally smoother than a billiard ball. But they are the highest mountains in the world, and absolutely breaktaking if near them. And space is even bigger, and satellites even smaller.
- WillAdams 2y ago
  An accurate steel replica of earth the size of a ball-bearing would be the most precise ball bearing ever. Great book on this: https://goodreads.com/book/show/35068671-the-perfectionists https://goodreads.com/book/show/35068671-the-perfectionists
  - coolspot 2y ago
    Doubt it. Earth is not even a sphere.
    - lazide 2y ago
      Looks like you’re right! Earth is within .3% of being a sphere, where ball bearings are within .01%. [https://en.m.wikipedia.org/wiki/Earth%27s_circumference https://en.m.wikipedia.org/wiki/Earth%27s_circumference]
    - WillAdams 2y ago
      Correct --- I should have said "smoothest".
  - sandworm101 2y ago
    I've heard this a couple times and never believed it. Earth is perceptibly not round. Scott Manley did a quick video showing that in some satellite images the difference equates to more than a couple pixels. Some quick googling shows the earth being about 0.3% not-round, which would match a 1000-pixel image being not-round by a few pixels. But a 0.3% deviation is exponentially greater than is allowed by the standards for ball bearings. >> the Earth deviates from spherical by only a third of a percent https://en.wikipedia.org/wiki/Figure_of_the_Earth https://en.wikipedia.org/wiki/Figure_of_the_Earth >> For instance, the Anti-Friction Bearing Manufacturers Association (AFBMA) has a set of grades for bearing balls. A grade three ball has to be spherical within 3 millionths of an inch and the diameter must be accurate within 30 millionths of an inch. This means that for a grade three quarter-inch ball, the diameter would have to be between 0.24997 and 0.25003 of an inch and the smallest diameter measured on the ball has to be within 3 millionths of the largest diameter. https://science.howstuffworks.com/transport/engines-equipment/question513.htm https://science.howstuffworks.com/transport/engines-equipmen...
aaronharnly 2y ago
I suppose you could generate an alternate view (maybe someone has?) that doesn't show the individual satellites, but instead only shows little green/yellow/red flashes whenever satellites cross within some radius of one another.
cruffle_duffle 2y ago
That is a tricky problem to solve. The scales are so far apart from each other it’s a bit like solving “how do you scale the distance between the sun and the planets in a way that keeps some detail of each while giving a sense of the true amount of nothing between them”. According to ChatGPT, if the earth was the size of a spherical cow (approx 1.5m in diameter) then the space station would be about 8.5 micrometers which is roughly the size of some bacteria. It would be 46 millimeters away from the cow. If earth was a spherical cow then a starlink satellite would be 0.31 micrometers, or 310nm which is about the wavelength of violet light. It would be about a large thumbtack away. The orbital periods for these would be in the order of microseconds.
- wizzwizz4 2y ago
  Please don't trust ChatGPT. It's got the right orders of magnitude, which is lucky, but it's still giving you the wrong numbers. If the earth were 1.5m in diameter, the space station would be proportionally 12.83µm long[0], orbiting at a distance of 50.66mm (2 inches)[1]. [0]: https://www.wolframalpha.com/input?i=iss+length+%2F+earth+diameter+*+1.5m https://www.wolframalpha.com/input?i=iss+length+%2F+earth+di... [1]: https://www.wolframalpha.com/input?i=iss+height+%2F+earth+diameter+*+1.5m https://www.wolframalpha.com/input?i=iss+height+%2F+earth+di...
LeifCarrotson 2y ago
I don't. The human sensorium (whether actually looking at an image, or visualizing something based on some numbers) has limited dynamic range. Sometimes, it can help a little to convert to something vaguely human-scale, which probably doesn't fit on a screen. The surface area of the planet is ~500 million km^2, there's a large park near me with long, clear sight lines where I go for runs play disc golf that's about 5 square km. If I shrink the ~3x2 meter satellite dimensions by an equivalent factor of about 100M, it's an object with a size of 30x20 micrometers (unless I've screwed up a square-cube ratio somewhere), on the order of a grain of silt, the thickness of a Sharpie ink mark, or something a little smaller than a cross-section of a hair. My intuition approximates those dimensions and probabilities of two of those infinitesimal specs all the way to zero. On the other hand, tens of thousands of satellites, those barely-palpable specs crossing that region of space once every 90 minutes, for hundreds of years? Then my intuition says it's a certainty! Facing the multiplication of an impossibility with a certainty, intuition says they cancel out and it's a 50/50 chance, which is just not how it works. Sometimes you have to just do the math and trust the numbers.

schiffern 2y ago

It's popular to point out how satellite visualizations exaggerate the size, but we never seem to notice how satellite visualizations shrink the duration of time. You only look at a visualization for maybe a minute, whereas these objects are actually out there whizzing around 24/7 for decades or even millennia. Both effects are misleading dataviz artifacts, but at least they're in opposite directions! A great ESA video, linked from Don Kessler's website, shows a time-lapse simulation that better captures the seriousness of space junk. In the simulation @5:25, realize that each of those flashes represents an Iridium-Cosmos magnitude collision event: http://www.youtube.com/watch?v=RvZ3Lr-Tj6A http://www.youtube.com/watch?v=RvZ3Lr-Tj6A

bryanlarsen 2y ago
The third & fourth factor are whether the satellites are actively avoiding collisions and whether the satellites are in a self-clearing orbit. Depending on the assumptions you make, we could have billions or even trillions of Starlink satellites without triggering Kessler simply because - Starlink satellites are actively controlled and steer to avoid collisions. Yes there are sometimes a couple of failed Starlinks in orbit, but since the active ones avoid the dead ones, you have to calculate the risk of one dead satellite hitting another dead satellite. And that's less than random because they quickly decay out of their crowded orbital plane to a different less crowded one. - dead Starlinks are in a self-clearing orbit, deorbiting in under 5 years. As the article indicates Kessler is a decades-long process. Which can't happen if the satellites don't stay in orbit that long. - most dead Starlinks aren't completely dead, or give alerts before they die, allowing SpaceX to actively deorbit them safely and quickly - the vast majority of particles from a collision in a self-clearing orbit result in immediate clearing. If the original dead Starlink that would take 5 years to clear suddenly fragments, most fragments de-orbit within a single orbit (90 minutes). Elastic deformation reduces their velocity, lowering the orbit. Half of fragments are toward a lower orbit. The other half only boost apogee and either lower perigee or leave it unchanged. Being smaller, their surface area / mass ratio increases, reducing deorbit times. Basically, Kessler is not a concern at VLEO altitudes. Unfortunately most proposed Starlink competitors cannot afford to launch the thousands of satellites necessary for a viable VLEO constellation and instead are launching dozens or hundreds into LEO orbits which aren't self-clearing and do pose a Kessler concern.

inglor_cz 2y ago

For some related ideas... Let us imagine that there are 100 000 Starlink-sized satellites on the orbit. At a given time, approximately 29 000 of them are over dry land and the rest over the sea. 29 000 car-sized object randomly moving across all 6 continents, including terrains such as the Himalayas. It would be pretty infrequent for them to come close to one another.

inhumantsar 2y ago
that's being generous too, since the spherical area those satellites cover is considerably larger than it is at ground level. Douglas Adams nailed it. "Space is big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space." Very few people, even us nerds who are aware of the scale, and actually hold that scale in their minds properly. And that's just low earth orbit. Geosynchronous, the Lagrange points, an Astronomical Unit, a Light Minute, a Light Year. It's one thing to understand what these things mean, but it's entirely another to put them into a human-scale context. It's like the difference between knowing that there are 20 million people in the New York area and trying to name each one of them individually.