0
$\begingroup$

updated 8/27/2020

While the recession of our Moon from the Earth may slow and even stop,

(see When will the Moon reach escape velocity?)

binary star systems will

(1) never stop experiencing mutual tidal forces, (2) continuously experience loss of mass and gravitational attraction, (3) lose energy by gravitational wave radiation.

(see On lifetime of binary stars due to orbital decay by gravitational wave radiation)

Assuming all these three mechanisms are extant, can binary stars break away from each other?

The one comment so far notes that LIGO measures effects of the convergence of gravitational bodies, that larger orbits have more energy, and dissipation of energy will reduce the orbital distances. I would appreciate an appropriate reference.

The moon is retreating from the Earth which is said to be mostly due to tidal-related dissipation. This phenomenon seems opposite the commentor's view.

A mechanism relating to binary stars, especially those that whose orbits bring them close together, is a possible exchange of mass. An expert might tell us how this exchange might affect orbits. If indeed the stars are close, then we might expect a decay in orbit due to simple drag.

I assume "expansion of the universe" is not a factor in these scales of distance and time. Thanks to all for your time on this question.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ Larger orbits have more energy, not less. LIGO detects black holes that have merged because they spiral toward each other as they radiate energy away. $\endgroup$
    – G. Smith
    Commented Aug 14, 2020 at 18:34
  • $\begingroup$ I misinterpreted "orbital decay" for gravitational wave radiation GWR). Thanks for the note. Still, measurements show recession of the Moon which is attributed to dissipation due to tidal effects, a clear loss of system energy. If one calculates this recession back in time, it suggests a much closer Moon, which I believe is the common thinking. On further thinking, perhaps the GWR is a result rather than a cause of merging. $\endgroup$
    – Incredible II
    Commented Aug 14, 2020 at 19:30

1 Answer 1

Reset to default
2
$\begingroup$

No, binary stars will not escape each other. The Moon is receding from the Earth due to tidal braking. In turn this is happening because the Earth spins faster than the Moon orbits. The Moon will not recede forever - once the Moon is tidally locked to the Earth that will be the end of recession.

Stars can also become tidally locked.

More pertinently: two systems are only bound (and systems such as the Earth/Moon, or two binary stars are definitely bound) if their total energy is negative. If they become unbound, then their total energy becomes positive. By conservation of energy, this cannot happen unless a 3rd body is involved that injects energy into the system.

Improve this answer
$\endgroup$
2
  • $\begingroup$ (a) I understand and agree with the third body argument. (b) My original question noted that stars lose mass over their lifetimes: "dissolve" is a word I've seen to describe this phenomenon. Thus, the binary system will lose mutual gravitational attraction, and the potential energy will become less negative (U = -GmM/R). The kinetic energy will decrease (K = GmM/2R). Is there a point where the system energy is no longer negative? (c) The Prophet Isaiah states "All the stars of heaven will be dissolved." [Isaiah 34:4, NIV]. $\endgroup$
    – Incredible II
    Commented Sep 4, 2020 at 20:10
  • $\begingroup$ A star is a fluid ball with a magnetic field. Binary stars will change shape as they circulate around the system center of gravity. They can be tidally locked only if all the fluid mass becomes immobile and the magnetic fields are static. There always will be tidal friction and dissipation. It is difficult to imagine a star that has its mass tightly ordered. 3-body star systems include the one closest to us, implying commonness. The 2 largest are tens of AU apart, close enough that grav and EM interactions are significant. My question seems still unanswered. $\endgroup$
    – Incredible II
    Commented Sep 11, 2020 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.