Where are Voyager and Pioneer heading?

[Cerenkov]

You've just reminded of something I read a long time ago about the chips used in NASA space probes, Alex.

Can't quite recall the details, but it went something like this.

A NASA spokesman commented on the kind of chips put aboard long-duration probes like Pioneer and Voyager, citing the power of the chip and its date of manufacture. This raised a mini media storm because the chip was quite an old one with much less power than was available at the time of their launch.

Questions were asked. Why doesn't NASA use the very latest and most powerful chips rather than these outmoded ones? Can't they afford them? Is this some kind of administrative oversight? And so on.

It turned out that NASA wanted chips with proven reliability rather those with greater power.

The reason being that reliability over years and decades far outweighs any potential gains made in computing power. If it takes over a decade to get to planets like Uranus and Neptune, but you only have two years performance data for the chip - because its so new - then it makes good strategic, long-term sense to go with a chip you know will last for a very long time.

Seeing as the Voyager probes are now in their 49th year of continuous operation NASA made the right choice.

Thanks,

Cerenkov.

 

[AlexB23]

You've just reminded of something I read a long time ago about the chips used in NASA space probes, Alex.

Can't quite recall the details, but it went something like this.

A NASA spokesman commented on the kind of chips put aboard long-duration probes like Pioneer and Voyager, citing the power of the chip and its date of manufacture. This raised a mini media storm because the chip was quite an old one with much less power than was available at the time of their launch.

Questions were asked. Why doesn't NASA use the very latest and most powerful chips rather than these outmoded ones? Can't they afford them? Is this some kind of administrative oversight? And so on.

It turned out that NASA wanted chips with proven reliability rather those with greater power.

The reason being that reliability over years and decades far outweighs any potential gains made in computing power. If it takes over a decade to get to planets like Uranus and Neptune, but you only have two years performance data for the chip - because its so new - then it makes good strategic, long-term sense to go with a chip you know will last for a very long time.

Seeing as the Voyager probes are now in their 49th year of continuous operation NASA made the right choice.

Thanks,

Cerenkov.

Yeah, kinda like an old 1990s Toyota that never breaks vs a high tech Tesla that breaks after a year. I see why NASA has been using older chips. Better to rely on an old tech that has been tested for years vs a new tech. Same reason why some folks wait for more established car companies to make EVs. Luckily, that is happening with Toyota. Finally, full EVs from them are being released.

 

[DaveC426913]

Just stumbled across this:

 

[AlexB23]

Just stumbled across this:
View attachment 375212

I am excited for November 18, 2026 cos of that. One light day, while not as far as sci-fi ships go in movies or books, is still an accomplishment. If only Voyager had working cameras, so people can see what the sun looks like 1/365th of a light year away. Since it is real cold out there and has background radiation for the probe, the fact that the computers still work on any of the Voyagers is impressive.

 

[DaveC426913]

If only Voyager had working cameras, so people can see what the sun looks like 1/365th of a light year away.

^{Artist's impression}

 

[AlexB23]

View attachment 375220
^{Artist's impression}

Wow, that is faint. There would not be any diffused light from the Kuiper belt or dust in the solar system? Plus, would the other planets be too far away to see or too close in perspective to the sun from Voyager's POV? I know about the solar system family portrait that was taken in 1990, but that was when Voyager 1 was at 6 billion km away (~0.25 light-days)

 

[DaveC426913]

Wow, that is faint.

It was a joke.


Seriously though, bit of cheating suggests it would appear as a point (~11 arc-seconds) With a -1.5 apparent brightness. It would rival Sirius - the brightest star in our sky.

 

[AlexB23]

It was a joke.


Seriously though, bit of cheating suggests it would appear as a -1.5 mag star. That would rivel Sirius - the brightest star in our sky.

I don't get jokes all the time, you know, and I thought you ran some computer code to make that dot. You meant -15, not -1.5, right? There is no reference point to -15, as the moon is hundreds of times fainter cos magnitude is not linear.

 

[DaveC426913]

I don't get jokes all the time, you know, and I thought you ran some computer code to make that dot.

I made it in Photoshop.

You meant -15, not -1.5, right?

No. -1.5.

There is no reference point to -15, as the moon is hundreds of times fainter cos magnitude is not linear.

Brighter the object the more negative the mag.

Moon is as much as -13 mag. Sun is -27. Jupiter is about -2.5.

List of sky objects by apparent magnitude.

Many people can see stars as dim as +6. I am not one of those people.

 

[AlexB23]

I made it in Photoshop.


No. -1.5.



Brighter the object the more negative the mag.

Moon is as much as -13 mag. Sun is -27. Jupiter is about -2.

Run the formula, please. It would be -15.8 or so. 1 light day is 173 AU.

 

[AlexB23]

I know how magnitude works, it is like golf scoring where bigger is negative. Brighter objects go really negative, faint ones really dim, as you stated. The sun would not be as faint as Sirius from 1/365th of a light-year away.

By no reference point to -15, I meant there is no object that is readily visible in the sky that has the same brightness as the sun at 1 light-day away, as that would involve an object brighter than a full moon, yet dimmer than the sun at 1 AU away. That would be a way for me to visualize the apparent magnitude of the sun.

 

[DaveC426913]

Mea culpa.

When i said: "with a bit of cheating" I meant "i asked a chatbot" what Sol's apparent mag would be from a distance of 1 light day.

I gotta lotta egg on my face. I have clearly not learned my lesson yet. You were right to call me out.

 

[AlexB23]

Mea culpa.

When i said: "with a bit of cheating" I meant "i asked a chatbot" what Sol's apparent mag would be from a distance of 1 light day.

I gotta lotta egg on my face. I have clearly not learned my lesson yet. You were right to call me out.

Thank you. I call people out when things are off. That part of my personality can be good, and bad. You know what, a local LLM could probably do this better. Let me see if it can. Also, some things are better left unsaid or not participated in, and most things are best not to be offloaded to AI or LLMs, and I am saying that as a man who runs local LLMs as a hobby (but never uses them on this site, as per rules).

Whelp, it looks like muh Gemma 4 got the right answer given the prompt: "Please calculate the apparent magnitude of the Sun from a distance of one light day, and show your work with the general formula and final answer. Use your Python tool calling skills in order to do so."

AI Disclosure (The following screenshot is the results of Gemma-4-26B-A4B. No other parts of this reply besides below is from AI):

 

[Hornbein]

Do these probes have enough velocity to escape the Solar System?

 

[AlexB23]

Do these probes have enough velocity to escape the Solar System?

Yes. Both are going ~17 km/s.

Source:
https://www.astronomy.com/space-exploration/will-the-voyager-satellites-ever-return-to-earth/

 

[Filip Larsen]

Do these probes have enough velocity to escape the Solar System?

They are on what is called a hyperbolic trajectory relative to the Sun and will as such drift infinitely far from the Sun with speed to spare, called hyperbolic excess speed or $v_\infty$. From the standard gravitational parameter for the Sun and post-Saturn Voyager 1 orbital data, for example, this speed can be calculated as $v_\infty \approx 16.6 \text{ km/s}$, so only slowed very little compared to its speed now. Voager 2 will end up a tad slower with $v_\infty \approx 14.9 \text{ km/s}$.

 

[Filip Larsen]

If only Voyager had working cameras, so people can see what the sun looks like 1/365th of a light year away.

You can get an idea of this by noting that at Voyager 1's current distance of 173 AU the intensity (power/area) of the Sun will be 173² = ~30'000 times less than at Earth's distance. So while Earth receive around 1360 W/m², Voyager 1 currently only receives around 45 mW/m², roughly corresponding to a 100W incandescent light bulb seen at over 13 m.

The diminishing solar intensity outside the inner planets is also the main reason these probes use an RTG instead of solar panels, allowing decades of operation.

 

[Hornbein]

The escape velocity from Earth is 5km/sec. Gravity assist from a big planet helps.

 

[Filip Larsen]

The escape velocity from Earth is 5km/sec. Gravity assist from a big planet helps.

Not sure what you mean by this. You need around 7.8 km/s for low Earth circular orbit and an additional $\sqrt{2} -1 \approx 40\%$ to escape Earth on a parabolic trajectory, so around 11 km/s in total relative to Earth.

 

[Hornbein]

Not sure what you mean by this. You need around 7.8 km/s for low Earth circular orbit and an additional $\sqrt{2} -1 \approx 40\%$ to escape Earth on a parabolic trajectory, so around 11 km/s in total relative to Earth.

What my source meant was from a distance from the sun equal to that of the Earth. They say 151,300 km/h = 42 km/sec is escape velocity. Quite a disagreement.

 

[Filip Larsen]

What my source meant was from a distance from the sun equal to that of the Earth. They say 151,300 km/h = 42 km/sec is escape velocity. Quite a disagreement.

OK, the 42 km/s is the speed relative to the Sun needed for an object at Earths distance to escape the Sun, that is around 40% more than the Earths orbital speed of 29 km/s around the Sun. However, absolute orbital speeds are rarely "useful" as indicator for how much speed (delta-V) is actually required.

For a realistic estimate of how much delta-V you need to "add" in order to achieve escape in case of multiple bodies (e.g. Earth and Sun) you need to work up a delta-V budget. For example, taking Voyager 2's trans-Jupiter heliocentric orbit, it corresponds to an injection speed of around 9.4 km/s more than Earths orbital speed at its distance from Sun and relative to the Sun. A simple model is then to say we need to inject from LEO into trans-Jupiter orbit such that the geocentric hyperbolic excess speed ends up at those 9.4 km/s, which from 300 km LEO corresponds to around 14.4 km/s relative to Earth or, on top of the 7.8 km/s LEO speed, an additional 6.7 km/s which would need to be provided by the Jupiter insertion stage.

 

[Cerenkov]

Do these probes have enough velocity to escape the Solar System?

Hornbein,


I began this thread by citing this science paper.

https://arxiv.org/pdf/1912.03503

Which details the stars that the probes will be passing by on their journey through the Milky Way galaxy.

This confirms that they do indeed possess velocities sufficient to escape from our solar system.

Thanks,


Cerenkov

Do these probes have enough velocity to escape the Solar System?