4 ms·

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 exp

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.
- gamegoblin 5y ago
  I totally agree with you and the other commenters that the total area of the spherical shell is only increasing by a few %, but I still think for the purposes of "escaping earth" as the top level comment was talking about, all that matters is how spread out things are directly above you. When talking about the area of the spherical shells, it conflates what is above me as equally relevant to things that are on the other side of the planet from me. That is, satellite X and Y may be over me at 500km and 1000km distance, respectively. Later, they may be directly through the earth from me at distance 13200km and 13700km distance. In the first case, if I shine a laser straight up, my probability of hitting X is 4 times higher than my probability of hitting Y. In the second case, (if I could somehow shine a laser straight through the earth), the probabilities are nearly equal. But my intuition is that for the purpose of escaping earth, this second case does not matter, because we are just dealing with what is above us, not the entire spherical shell.
tgb 5y ago
If your original claim were true, then you'd also be saying that there is twice as much area at 2m above sea level as there is 1m above sea level, and infinitely more than at sea level. If you truly believe this, then I will be willing to make you a trade where I give you one tenth floor condo in exchange for 10 first floor condos.
- deleted 5y ago
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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.

gamegoblin 5y ago
I totally agree with you and the other commenters that the total area of the spherical shell is only increasing by a few %, but I still think for the purposes of "escaping earth" as the top level comment was talking about, all that matters is how spread out things are directly above you. When talking about the area of the spherical shells, it conflates what is above me as equally relevant to things that are on the other side of the planet from me. That is, satellite X and Y may be over me at 500km and 1000km distance, respectively. Later, they may be directly through the earth from me at distance 13200km and 13700km distance. In the first case, if I shine a laser straight up, my probability of hitting X is 4 times higher than my probability of hitting Y. In the second case, (if I could somehow shine a laser straight through the earth), the probabilities are nearly equal. But my intuition is that for the purpose of escaping earth, this second case does not matter, because we are just dealing with what is above us, not the entire spherical shell.

tgb 5y ago

If your original claim were true, then you'd also be saying that there is twice as much area at 2m above sea level as there is 1m above sea level, and infinitely more than at sea level. If you truly believe this, then I will be willing to make you a trade where I give you one tenth floor condo in exchange for 10 first floor condos.

deleted 5y ago
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