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From the last issue of Orbital Index (https://orbitalindex.com https://orbitalindex.com): > Today’s mind-blowing stat: Voyager’s transmitters use just 23 watts

From the last issue of Orbital Index (https://orbitalindex.com https://orbitalindex.com): > Today’s mind-blowing stat: Voyager’s transmitters use just 23 watts, roughly the same as an incandescent refrigerator bulb, yet we are able to interpret the 0.1 billion-billionth of a Watt that makes it to our 70 m dish from 11 billion kilometers away. Both spacecraft are expected to last another 5 years, until their plutonium batteries decay beyond usefulness and they drift with our golden record more or less forever among the stars.

monocasa 7y ago
I heard a rumour that by the time they've gotten to another solar system, they will have evaporated into a metallic cloud; that's how long this next part of their journey is.
- andrekandre 7y ago
  do you mean the plutonium batteries or the probe itself?
  - monocasa 7y ago
    The whole probe itself.
    - namirez 7y ago
      Technically you're talking about sublimation rather than evaporation. And no, it won't be very likely for the Voyagers to disappear. Most likely they will be damaged by impact or get absorbed by massive object, but sublimation at 10K will take a very very long time. I don't have the data, but if you want to calculate the rate of sublimation, you need to look at the vapor pressure of various metals at interstellar temperatures.
      - Filligree 7y ago
        Some very low numbers indeed. But they're in interstellar space. I imagine high-energy collisions (with hydrogen atoms or loose protons) would be a more likely cause of evaporation, and I don't know the numbers for that.
        jcims 7y ago
        If there’s 1 atom per cubic cm in interstellar space, I’m finding it takes about half a million years to collide with a mole of hydrogen (at 17km/s)
        Jeff_Brown 7y ago
        Google says that's about a gram of hydrogen. Is that enough to destroy a craft? I have little intuition, but I imagine touching a gram of 55,000-degree (Farenheit) anything would be destructive. Will the craft be radiating heat away from itself?
- bufferoverflow 7y ago
  Why would they? They will be very cold. Occasional cosmic rays aren't enough to evaporate them. They can last hundreds of thousands of years, if not millions.
  - monocasa 7y ago
    And how long until they reach the next star? They're not pointed anywhere in particular. We are talking about millions of years here.
    - Koshkin 7y ago
      And even if they were.
      - samstave 7y ago
        Think we can point the James webb at one and get a farewell Pic?
    - mekoka 7y ago
      Millions of years? Fat chance. A few days ago this post https://news.ycombinator.com/item?id=21327269 https://news.ycombinator.com/item?id=21327269 explained that during the galactic collision of the Milky Way and Andromeda in 4B years, no two stars will likely collide because of the sheer amount of empty space. Keeping that in perspective, I highly doubt that Voyager would find its way anywhere in some mere millions of years.
  - njharman 7y ago
    Space is mind boggling large and sparse. Millions of years is not a long time to be traveling through it.
    - vmasto 7y ago
      * even at light speed
ac29 7y ago
23W with extremely high gain though. It was difficult to find a number, but this source [0] gives the gain at 47dB (dBi or dBd isn't specified). With that high of gain, the effective radiated power is over 1 million Watts (because it is concentrated into a very narrow beam). Also using phrases like "0.1 billion-billionth of a Watt" is misleading, and not how radio signals are actually described. For example 0.1 million-billionth of a Watt sounds pretty small too, but describes the power level of a usable, and not particularly uncommon LTE signal level. [0] https://www.quora.com/How-can-Voyager-send-a-signal-strong-enough-for-us-to-receive-in-spite-of-its-enormous-distance-from-us-And-how-can-it-have-the-power-to-do-so-more-than-20-years-after-its-launch https://www.quora.com/How-can-Voyager-send-a-signal-strong-e...
- credit_guy 7y ago
  Great quora link, thank you for posting. Here's a related thought: if we ever manage to send a probe to the closest star, what a tough problem communication with home will be. The distance is about 2000 times higher than where Voyager is now. So the signal will be about 4 million times weaker due to the quadratic attenuation alone, and maybe 100 million if we consider the absorption by the interstellar medium. On top of that, when the probe will phone home, it's going to be a few seconds of arc away from a phenomenally bright source of electromagnetic radiation: the star itself. You can fix this by using a bigger transmitting antenna (but Voyager's is already 3.7m, not small at all), or a bigger receiving antenna (the current ones are 20m). Or a stronger signal. But it appears we'd need to go from 23W to GigaWatts . One thing is sure: we won't be able to send a pound-size probe.
- walrus01 7y ago
  The gain of a 3.5m parabolic antenna on the tx end, and they're using from 45 to 70 meter size dishes on the earth end to communicate with them. The NASA goldstone 70m dish has some serious gain.
m0xte 7y ago
Makes a mockery of the amateur radio “miles per watt” awards :)

monocasa 7y ago

I heard a rumour that by the time they've gotten to another solar system, they will have evaporated into a metallic cloud; that's how long this next part of their journey is.

andrekandre 7y ago
do you mean the plutonium batteries or the probe itself?
- monocasa 7y ago
  The whole probe itself.
  - namirez 7y ago
    Technically you're talking about sublimation rather than evaporation. And no, it won't be very likely for the Voyagers to disappear. Most likely they will be damaged by impact or get absorbed by massive object, but sublimation at 10K will take a very very long time. I don't have the data, but if you want to calculate the rate of sublimation, you need to look at the vapor pressure of various metals at interstellar temperatures.
    - Filligree 7y ago
      Some very low numbers indeed. But they're in interstellar space. I imagine high-energy collisions (with hydrogen atoms or loose protons) would be a more likely cause of evaporation, and I don't know the numbers for that.
      - jcims 7y ago
        If there’s 1 atom per cubic cm in interstellar space, I’m finding it takes about half a million years to collide with a mole of hydrogen (at 17km/s)
        Jeff_Brown 7y ago
        Google says that's about a gram of hydrogen. Is that enough to destroy a craft? I have little intuition, but I imagine touching a gram of 55,000-degree (Farenheit) anything would be destructive. Will the craft be radiating heat away from itself?
bufferoverflow 7y ago
Why would they? They will be very cold. Occasional cosmic rays aren't enough to evaporate them. They can last hundreds of thousands of years, if not millions.
- monocasa 7y ago
  And how long until they reach the next star? They're not pointed anywhere in particular. We are talking about millions of years here.
  - Koshkin 7y ago
    And even if they were.
    - samstave 7y ago
      Think we can point the James webb at one and get a farewell Pic?
  - mekoka 7y ago
    Millions of years? Fat chance. A few days ago this post https://news.ycombinator.com/item?id=21327269 https://news.ycombinator.com/item?id=21327269 explained that during the galactic collision of the Milky Way and Andromeda in 4B years, no two stars will likely collide because of the sheer amount of empty space. Keeping that in perspective, I highly doubt that Voyager would find its way anywhere in some mere millions of years.
- njharman 7y ago
  Space is mind boggling large and sparse. Millions of years is not a long time to be traveling through it.
  - vmasto 7y ago
    * even at light speed

ac29 7y ago

23W with extremely high gain though. It was difficult to find a number, but this source [0] gives the gain at 47dB (dBi or dBd isn't specified). With that high of gain, the effective radiated power is over 1 million Watts (because it is concentrated into a very narrow beam). Also using phrases like "0.1 billion-billionth of a Watt" is misleading, and not how radio signals are actually described. For example 0.1 million-billionth of a Watt sounds pretty small too, but describes the power level of a usable, and not particularly uncommon LTE signal level. [0] https://www.quora.com/How-can-Voyager-send-a-signal-strong-enough-for-us-to-receive-in-spite-of-its-enormous-distance-from-us-And-how-can-it-have-the-power-to-do-so-more-than-20-years-after-its-launch https://www.quora.com/How-can-Voyager-send-a-signal-strong-e...

credit_guy 7y ago
Great quora link, thank you for posting. Here's a related thought: if we ever manage to send a probe to the closest star, what a tough problem communication with home will be. The distance is about 2000 times higher than where Voyager is now. So the signal will be about 4 million times weaker due to the quadratic attenuation alone, and maybe 100 million if we consider the absorption by the interstellar medium. On top of that, when the probe will phone home, it's going to be a few seconds of arc away from a phenomenally bright source of electromagnetic radiation: the star itself. You can fix this by using a bigger transmitting antenna (but Voyager's is already 3.7m, not small at all), or a bigger receiving antenna (the current ones are 20m). Or a stronger signal. But it appears we'd need to go from 23W to GigaWatts . One thing is sure: we won't be able to send a pound-size probe.
walrus01 7y ago
The gain of a 3.5m parabolic antenna on the tx end, and they're using from 45 to 70 meter size dishes on the earth end to communicate with them. The NASA goldstone 70m dish has some serious gain.

m0xte 7y ago

Makes a mockery of the amateur radio “miles per watt” awards :)