Hypothetically speaking (once again), if the Sun stays the same size forever (i.e. it somehow maintains a stable hydrogen supply and the helium product just disappears), how long would it take before the decay of the Earth's orbit sends it plummeting into the Sun and disintegrating?

  • $\begingroup$ Peter, it’s preferable that you wait a day or two before awarding the “tick” for your preferred answer. Mark’s answer is a good one, but a few hours isn’t enough time for the Astronomy SE community to comment on any errors or provide alternative or more detailed answers. :-) $\endgroup$ – Chappo Hasn't Forgotten Monica May 30 '18 at 6:09
  • $\begingroup$ @Chappo Aight, ty for the heads-up. $\endgroup$ – Peter Johnmeyer May 30 '18 at 6:22
  • $\begingroup$ You could also write why you think the Earth should spiral into the sun. If you have a certain mechanism in mind, that would help answer it. $\endgroup$ – AtmosphericPrisonEscape May 30 '18 at 16:28

As has already been said, the major sources of change to Earth's orbit are interactions with other planets and passing stars.

We're ruling out mass loss of the Sun, so the next consideration is probably tidal interactions between the Earth and Sun. This paper suggests that the Earth is receding from the Sun by about 15cm/yr for this reason. This is 150 km per million years, so over a trillion years or so, would move Earth quite a bit further from the Sun. The effect would reduce as the Earth got further away from the Sun, but I think it would always be slowly receding. I haven't done the calculation, but I'm not sure it would ever reach tidal lock -- the angular momentum of the Sun is too high.

Independently of this, though, orbital energy and angular momentum would slowly be lost by gravitational radiation. This wikipedia page gives a formula

$$t={\frac {5}{256}}\,{\frac {c^{5}}{G^{3}}}\,{\frac {r^{4}}{(m_{1}m_{2})(m_{1}+m_{2})}}.$$

for the time to collision ignoring tidal effects.

We can plug in known values for the parameters $r = 1.5 \times 10^{11} m$, $c = 3\times 10^8 m/s$, $G = 6.7 \times 10^{-11} m^3 kg^{-1} s^{-2}$, $m_1 = 6\times 10^{24} kg$, $m_2 = 2 \times 10^{30} kg$ and get about $3.3\times 10^{30} s$ as the time to collision. I suspect that's the time until the centres of the Earth and Sun coincide, so the actual collision would be a bit earlier, but probably not enough to notice at this precision. So that's about 10 million million times the lifetime of the universe so far.

I can't find enough information to work out how these two effects would interact. Clearly in the first $10^{12}$ years tidal evolution would be more important, but that conserves angular momentum and slows as the Suns rotation slows, whereas gravitational radiation actually carries angular momentum away, so might dominate in the long term. On the other hand the radiation drops as a high power of the separation, so we might have the Earth moving ever more slowly away from the Sun, while radiating gravitational waves of ever weaker strength, and never reaching tidal lock or starting to move back in.

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    $\begingroup$ Surely tidal effects are bigger than this? $\endgroup$ – Rob Jeffries May 30 '18 at 18:31
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    $\begingroup$ Yes. Almost anything is bigger than this I admit. Since the Sun rotates considerably more than once per year, the tidal effects would transfer angular momentum from the Sun to the Earth, which would slowly be driven further from the Sun, so if we allow for tidal effects (while still ignoring other perturbations) it will never collide. $\endgroup$ – Steve Linton May 31 '18 at 9:57
  • $\begingroup$ and that is the answer. $\endgroup$ – Rob Jeffries May 31 '18 at 10:11
  • $\begingroup$ I've added a discussion to the answer. I can't work out which will dominate in the very long term $\endgroup$ – Steve Linton May 31 '18 at 10:14
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    $\begingroup$ @nicoleSharp the moon is receding due to tidal effects. Gravitational raisin head the opposite effect (but much, much, more slowly) $\endgroup$ – Steve Linton May 31 '18 at 17:33

This is unpredictable. The solar system is a chaotic system and it's long-term behavior can't be predicted. But it's also a very stable chaotic system and the planetary orbits have been stable for 4 billion years and will remain stable for some millions of years into the future. See the Wikipedia article for an overview.

When we numerically integrate the orbits of the planets into the future in an ensemble (doing many integrations, each with small differences in starting values that are still within experimental error), a small fraction of the ensemble members diverge chaotically. (E.g., Mercury is eventually flung into a wild orbit which may impact Venus.)

The key point here is that it is intrinsically unpredictable more than some millions of years into the future -- we're definitely safe for much more than a thousand human lifetimes -- other than statistically. But base don those statistics, it seems likely that the solar system will be disrupted due to chaotic dynamics before the Earth's orbit decays due to tidal forces or anything like that.

(Another presently unpredictable future is an interloping star coming close enough to disrupt planetary orbits. Stars pass through the Oort cloud on a regular basis -- one each million years or so -- but getting inside the orbit of Pluto is much less likely and hasn't happened yet as far as we can tell. Nonetheless, this could happen any time.)

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  • $\begingroup$ +1, but you’re last sentence isn’t accurate. The local stellar neighbourhood is well mapped, and there’s no star coming closer than 3-4 light years of us in the foreseeable future. $\endgroup$ – Chappo Hasn't Forgotten Monica May 30 '18 at 6:17
  • $\begingroup$ @Chappo That was what I thought too, unless a star gets catapulted towards us from one of our group of local galaxies. $\endgroup$ – Peter Johnmeyer May 30 '18 at 6:23
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    $\begingroup$ @Chappo - What you wrote in your first comment quite incorrect. Gliese 710 is predicted to come within less than a quarter of a light year of the Sun in 1.3 million years. Scholz's Star is thought to have come within 0.8 light years of the Sun about 70000 years ago. $\endgroup$ – David Hammen May 30 '18 at 12:26
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    $\begingroup$ The local neighborhood is not well-mapped towards the dim red end of the main sequence where stars are most numerous. We haven't even completely excluded the possibility that the Sun has a distance brown dwarf companion! (Though that's pretty unlikely.) And in a million years stars with the low peculiar velocity of 10 kps will travel thirty light years and a high velocity star can travel a light year in just a 500-1000 years. $\endgroup$ – Mark Olson May 30 '18 at 12:28
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    $\begingroup$ That the Sun has a distant brown dwarf companion has been excluded. $\endgroup$ – Rob Jeffries May 31 '18 at 10:17

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