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According to Einstein, energy is equal to mass. Consider a planet that is in gravitational attraction to two stars. Normally I would say that the gravitational attraction is proportional to the masses of the two stars. But if they are orbiting each other, they possess energy.

Is it correct to say that this star system therefore has a stronger gravitational pull that is greater than just the two added masses of the stars?

user985366
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Anon
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    Related: https://en.wikipedia.org/wiki/Gravitational_binding_energy – PM 2Ring Jan 28 '23 at 10:13
  • Closely related: https://physics.stackexchange.com/q/196280/226902 "Proper mass" and "gravitational binding energy" in general relativity; https://physics.stackexchange.com/q/603532/226902 Are the concepts of kinetic energy, potential energy etc not valid in general relativity? The following are also useful: https://physics.stackexchange.com/q/509036/226902 Justification for excluding gravitational energy from the stress-energy tensor; https://physics.stackexchange.com/q/41662/226902 Why does no physical energy-momentum tensor exist for the gravitational field? – Quillo Jan 28 '23 at 11:52
  • According to Einstein, energy is equal to mass If this is so, then (speaking as a complete novice), why is the famous equation e=mc^2 and not e=m? – John Gordon Jan 30 '23 at 18:37
  • I feel like the answers are to a different question, namely two stars bound or unbound. Aren't you asking about if the same mass is in a single star or separated into two distinct orbiting stars? Or did I misunderstand the question? – Michael Jan 31 '23 at 02:21

3 Answers3

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A double star has less energy than if the two stars were separated.

It is fairly easy to see why this is. If you have two stars orbiting each other you would need to add energy to separate them. That is, assuming you had some form of Star Trek-esque tractor beam you'd have to use that to grab the stars and physically pull them away from each other. Then you'd be putting work into the system and that added work means the two separated stars have a greater combined energy than the original double star system.

That means the gravitational field of a double star system is slightly smaller than you'd expect from the masses of the two stars, though in practice the difference is far too small every to be measured.

John Rennie
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    Doesn't any change in how they are orbiting each other require additional energy? I would need a tractor beam whether I wanted them to orbit closer than they currently were or whether I wanted them to orbit farther than they currently were. I guess the difference is whether I'd apply the tractor beam in the same direction as the current orbit or the opposite direction, or something? But then how does that tie into the argument about adding energy to the system because there's a tractor beam at all? – Daniel Wagner Jan 29 '23 at 17:25
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    @DanielWagner no. If you decrease the energy of the system you gain it in outside. The simplest example I can think of is the gravitational slingshot - if you fly by one star you gain energy and star system looses. As a result your spaceship will have slightly larger gravitational pull and the binary system slightly less. – Maja Piechotka Jan 29 '23 at 19:13
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    @DanielWagner To demonstrate the point cartoonishly, you could imagine placing an enormous spring in the way of the planet's orbit, which the planet would compress as they collide. As the spring reaches maximum compression, you could lock the spring and blast away with a rocket. The planet continues in an orbit of lower energy, the deficit being stored in your giant spring for you to use as you see fit. – jawheele Jan 30 '23 at 02:46
  • The first sentence doesn't make sense to me. A double star already has the two stars separated. Second, the second paragraph seems to be saying two separate stars have more energy than the same mass as a single star, so how could the gravitational field be less since the total mass+energy is greater? – Michael Jan 31 '23 at 02:20
  • @Michael "Separated" as in "not gravitationally bound to each other" not as in "not touching". – Yakk Jan 31 '23 at 17:03
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The sum of the kinetic and potential energy of a bound system is negative. This must be the case, because you would have to inject more energy into the system to separate the components to infinity.

Therefore the "gravitational mass" of the binary - what you would measure with another orbiting test mass at greater distance - would be less than you would expect from the sum of all the masses of the components of the system measured when they are far apart and stationary.

Whether this is an important effect (i.e. the relative size of the correction) can be judged from the ratio of the absolute value of binding energy to the rest mass energy: $$\alpha \simeq \frac{GM_1M_2}{2Rc^2(M_1+M_2)}\ ,$$ where $R$ is the separation and the result is exact for a circular orbit.

The effect can be important in binaries featuring compact stars (e.g. neutron stars) that have stellar masses and where $R$ can be quite small - the ratio above is a few per cent for a pair of neutron stars separated by 30 km.

As an aside, this consideration also applies to single stars, where the sum of their internal kinetic energy and gravitational potential energy is also negative. Again, this is important in white dwarf and neutron star physics.

ProfRob
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Your question is not really different from a proton and an electron joining together to form a hydrogen atom.
When such a combination takes place approximately $13.6\,\rm eV$ of energy is released and the mass of a hydrogen atom is less by the mass equivalent of $13.6\,\rm eV$ $(E=mc^2)$ than the combined mass of an isolate proton and an isolated electron.

So if you found the mass of two stars separately and then arranged for them to be in a bound state orbiting each other, the system of two orbiting stars would have a mass which was less than the combined masses of the two individual stars.

Farcher
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