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Voyager 1 is currently 22 hours, 32 minutes and 33.1467 seconds away. I want to celebrate when it is a "Light Day" away. https://theskylive.com/how-far-is-voya

Voyager 1 is currently 22 hours, 32 minutes and 33.1467 seconds away. I want to celebrate when it is a "Light Day" away. https://theskylive.com/how-far-is-voyager1 https://theskylive.com/how-far-is-voyager1 A prior thread calculated it to be November 18, 2026. [see here: https://news.ycombinator.com/item?id=31794292 https://news.ycombinator.com/item?id=31794292]

xingped 2y ago
Wait why is the 1 LD from earth sooner than 1 LD from the sun? I'd expect it would be the opposite?
- GenerWork 2y ago
  Because the earth is a bit closer to Voyager than the sun is.
  - xingped 2y ago
    If the earth is closer to voyager than the sun, then the 1 LD from the sun date should be sooner than the 1 LD from the earth date. But in the linked calculations it's later?
- seabass-labrax 2y ago
  Given that the Earth orbits the Sun, I would expect it to swap every half a year. For one half of the year, the Sun should be ever so slightly closer to Voyager 1 than the Earth, and the other half of the year the opposite should be true. However, considering that the Earth and the Sun are no more than ten light-minutes away from each other, I shouldn't think this would make much difference to our festivities :)
- huppeldepup 2y ago
  Earth orbits the Sun at an average distance of 149.60 million km [WP] 45 degree angle (see Quora animation): 150000000 * cos(45) = 79000000 Earth projection sits between Voyager and Sun. "...travelling at around 61,500km/h ": 79000000/61500/24 = 53 days to add the equivalent of that distance in LD time. Edit: huh, since Earth sits between voyager and the Sun, the LD distance is satisfied for the Sun first.
  - xingped 2y ago
    Yeah, that's why I'm consued why the LD from the sun is later than the LD from the earth.
  - dogma1138 2y ago
    The light delay to the sun is 8 min which means that the biggest light delay difference to voyager will be about 16min give or take if the sun is directly between earth and voyager. That’s more or less negligible when you are measuring 1 light day worth of delay.
    - huppeldepup 2y ago
      That's the point: it's not negligible when you have to convert it to the distance this craft has to fly. It's a bit of mental gymnastics, I suggest you do the exercise with paper & pencil to see what happens.
      - dogma1138 2y ago
        At most it would take Voyager about 190 days to travel 16 light minutes. Considering it’s been flying for 47 years and that it will take 3 more years for it to reach 1 light day of delay half a year more on top of that is pretty negligible in my book.
eddieroger 2y ago
The enormity of this statement is really blowing my mind. /If/ we could travel the speed of light, which we can't, it would still take nearly a day to get to Voyager 1, which seems both infinitely far given our current capabilities and remarkably close given how long it's taken to get that far.
- onetimeuse92304 2y ago
  No. If you could travel at the speed of light you would take no time to reach Voyager 1. Actually, you could get anywhere in the reachable Universe. For example, it takes only 29 years to from Earth to Andromeda galaxy assuming constant 1g acceleration for half of the trip and then constant 1g deceleration for the other half.

xingped 2y ago

Wait why is the 1 LD from earth sooner than 1 LD from the sun? I'd expect it would be the opposite?

GenerWork 2y ago
Because the earth is a bit closer to Voyager than the sun is.
- xingped 2y ago
  If the earth is closer to voyager than the sun, then the 1 LD from the sun date should be sooner than the 1 LD from the earth date. But in the linked calculations it's later?
seabass-labrax 2y ago
Given that the Earth orbits the Sun, I would expect it to swap every half a year. For one half of the year, the Sun should be ever so slightly closer to Voyager 1 than the Earth, and the other half of the year the opposite should be true. However, considering that the Earth and the Sun are no more than ten light-minutes away from each other, I shouldn't think this would make much difference to our festivities :)
huppeldepup 2y ago
Earth orbits the Sun at an average distance of 149.60 million km [WP] 45 degree angle (see Quora animation): 150000000 * cos(45) = 79000000 Earth projection sits between Voyager and Sun. "...travelling at around 61,500km/h ": 79000000/61500/24 = 53 days to add the equivalent of that distance in LD time. Edit: huh, since Earth sits between voyager and the Sun, the LD distance is satisfied for the Sun first.
- xingped 2y ago
  Yeah, that's why I'm consued why the LD from the sun is later than the LD from the earth.
- dogma1138 2y ago
  The light delay to the sun is 8 min which means that the biggest light delay difference to voyager will be about 16min give or take if the sun is directly between earth and voyager. That’s more or less negligible when you are measuring 1 light day worth of delay.
  - huppeldepup 2y ago
    That's the point: it's not negligible when you have to convert it to the distance this craft has to fly. It's a bit of mental gymnastics, I suggest you do the exercise with paper & pencil to see what happens.
    - dogma1138 2y ago
      At most it would take Voyager about 190 days to travel 16 light minutes. Considering it’s been flying for 47 years and that it will take 3 more years for it to reach 1 light day of delay half a year more on top of that is pretty negligible in my book.

eddieroger 2y ago

The enormity of this statement is really blowing my mind. /If/ we could travel the speed of light, which we can't, it would still take nearly a day to get to Voyager 1, which seems both infinitely far given our current capabilities and remarkably close given how long it's taken to get that far.

onetimeuse92304 2y ago
No. If you could travel at the speed of light you would take no time to reach Voyager 1. Actually, you could get anywhere in the reachable Universe. For example, it takes only 29 years to from Earth to Andromeda galaxy assuming constant 1g acceleration for half of the trip and then constant 1g deceleration for the other half.