# For how long will the Voyager records remain playable?

The records attached to the Voyager spacecraft are apparently just like an old-school vinyl disk, except made of gold (sort of like a metal master), in that the information is encoded in mechanically cut grooves. Based on depictions of the spacecraft, it appears that while the record is attached to the outside of the vehicle body, the playing surface is facing in (presumably for both protection and to display some engraved illustrations). In the series "Cosmos: A SpaceTime Odyssey", Neil deGrasse Tyson states rather simply that the record should still be playable in a billion years, but says nothing about the duration as either a minimum or a maximum, or how the recording may be emperiled.

Assuming the Voyager craft don't crash into anything, and encounter nothing larger or more substantial than cosmic ray particles, what would be the limit at which the record would still be in a playable condition? And by what mechanisms might it be damaged to the point its recording could no longer be played (or its content no longer recoverable by any known means)?

Would cosmic radiation eventually erode the entire spacecraft into dust? If it remained in cold insterstellar space, never to encounter anything but cosmic radiation, could it remain intact for the projected life of the universe?

The point of my question is this: Is there (or are there) any process(es) which is/are entirely unavoidable by which the recorded content is doomed to eventually become lost, perhaps even while some semblance of the original spacecraft still remains? Is the entire vehicle utimately doomed to certain destruction within the projected life of the universe? Conversely, if not vaporized, smashed, or otherwise damaged in some sort of encounter(s), would the recordings be more or less guaranteed to remain intact and playable if the craft is found by some intelligence, no matter how far into the future that might be?

• I'm assuming the disks are protected inside the probe. Only an unlikely event like an asteroid collision would damage them. – this Jun 9 '14 at 21:37
• They're actually outside the probe, but facing towards it. – simontemplar Jun 12 '14 at 4:08

There are, in principle, three ways the record might be destroyed: impact with another body; heating/melting; or atomic deterioration.

I have no idea as to what the ultimate trajectory of Voyager might be, namely whether it might eventually either A. run into an icy body in an Oort Cloud equivalent of another solar system; or B. make a close enough pass to another star to cause the thermal load to become high enough to melt the record. Someone may have run a rough calculation on Voyager's far-distant trajectory, but I'm not aware of it if so. But the galaxy is very much mostly empty space so this is unlikely. But since either of these seem unknowable for now let's just address the possible impacts of cosmic rays or other particles.

The record is constructed of gold-plated copper. Stable isotopes of both metals do not decay atomically. Since the record is analog, not digital, the information on the disc is macro-scale (bumps on a groove) and thus not subject to the kinds of losses from cosmic ray bombardment that data encoded in digital bits would be.

So given that - and again, assuming that there are no direct impacts or close heat encounters with another body, I think the answer as to how long the disc will remain playable (or at least recoverable by a civilization sophisticated to recover it in the first place) is: indefinitely.

• I like most of your answer. But the part about digital data loss from cosmic rays is nonsense. If digital data can be corrupted this way, then surely analog data will gain noise this way too. While I don't have a suggestion for how to correct your text, I urge you to please think about it. – Nayuki Mar 12 '16 at 4:20
• @Nayuki: That's too many cosmic rays. I haven't done the math but I suspect proton decay kicks in first. – Joshua Dec 11 '18 at 18:36
• It won't be possible to do a calculation for the far-distant trajectory due to limitations imposed by Lyapunov time. – forest Jul 22 at 10:06
• @Joshua And if proton decay doesn't kick in, quantum tunneling will eventually reduce it to a perfect sphere (and then a black hole). Of course, by that time, no one will be around to play the disc anyway... – forest Jul 22 at 10:06

The disks on Voyager 1 and Voyager 2 (they are the same) are copper disks, plated in gold. The disks are encased in an aluminum housing to protect them from the "elements" of space. Barring any impacts with space rocks or other events like that, the records will last and be playable pretty much forever. The two space craft will not make a near approach to any stars in the foreseeable future--V1 will get about 1.4ly from another star in about 40,000 years. If anyone's out there to play the records, surely they will have found them and/or made friends with us way before then--unless they think we are delicious.

• Or they think the Voyage probe is supposed to be sentient, fix it, and send it back to us as V-Ger. – RoboKaren Jul 9 '15 at 11:53

Assuming the spacecraft isn't destroyed by a collision or radiant energy from closely passing a star, it isn't immediately clear to me if the maximum theoretical lifetime would be limited by the theoretical lifetime of protons (>$10^{25}$ years) or the process of sublimation. I was curious about the answer so I started to do some digging (I would post more reference links, but my lack of reputation limits the number of links I can include)...

In a vacuum over along enough period of time, anything above 0°K tends to eventually become a gas via sublimation. The practical effect of this is that the entire spacecraft will eventually deteriorate into a rubble pile. The period of time that takes is extremely long (which is also an extreme understatement).

But how long? Well... We know the record was gold-plated copper so we can first start with how long it would take for the gold plating to sublimate. First we would need to calculate the vapor pressure of gold. We use the temperature of the interstellar medium (2.73K) as our temperature. By my calculations¹, the vapor pressure is around $1.1725×10^{-40939}$ atmospheres or $1.1572×10^{-40944}$ pascals. That pressure is exerted over the surface area of the record, which is $2827cm^{2}$.

At this point I need to stop because I need to re-familiarize myself with the appropriate physics equations to calculate the actual material loss rate. I will update this post when I do that.

¹ Using data from here. Will post links to my calculations once I have more reputation points.

• You can add the links already, just make sure they're text, not links (e.g. write dot instead of an actual period). Somebody else will edit it. – Jan Doggen Aug 24 '17 at 7:38
• 2.73K? That's the current temperature, but the universe is cooling (at the timescales involved). The biggest problem with that insane low vapor pressure is that you're down in quantum mechanical scales. Vapor pressure is the result of molecules bouncing off a surface. At such low temperatures and pressures, that simply doesn't apply anymore. An atom that takes off, is lost to space without ever colliding again. And furthermore, you get quantum tunneling effects. – MSalters Aug 24 '17 at 13:47
• It's 0K, not 0°K. Kelvin is an absolute unit, not a degree. – forest Jul 22 at 10:09

Unless as you state something large impacts with the disc, which is certainly unlikely until they enter into another solar system. The disc should remain readable, there are some processes which may distort the disc but they wouldn't damage the possibility of recovering the data stored on the disc if this distortion is taken into account when trying to recover the data.