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How much of an effect does fusion have on a star's gravity?

  1. Feb 29, 2012 #1
    I am not a student, unfortunately, just a curious layperson who gets random, weird questions sometimes, so my apologies if this is a basic question generally covered in astrophysics for dummies, but ... here goes ...

    Nuclear fusion is the process of atoms combining to form heavier elements, and in the process some of the components of the original atoms are converted to energy based on the much beloved e = mc^2, correct? So the new atom, created by fusion, actually has slightly less mass than the two atoms that combined to make it?

    Gravity is based on mass, so wouldn't this mean that as fusion in a star progresses over the lifetime of the star, the star loses mass in this process and, thus, gravity? Acknowledging the fusion process casts off particles anyway, and things like coronal mass ejections probably cast off more mass than the loss of mass from fusion anyway, how much of an effect would this have on a star the size of, say, our local sun over its lifetime? And does this gradually lowering gravity affect the orbits of planets orbitting the star?

    Thanks for answering my silly question.
  2. jcsd
  3. Feb 29, 2012 #2


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    The sun loses only 10^11kg/s due to fusion reactions, which, over the entire solar lifetime, would work out to be 10^27kg, about 0.1% of the Sun's total mass.

    In general, for a solar-type star, the effects of mass loss are less important than the evolutionary changes accompanied by the sun's journey on the main sequence (increasingly becoming hotter) and its departure from the main sequence (puffing up into a red giant).
  4. Mar 2, 2012 #3
    Just to add a little to Nabeshin's great response:
    The energy lost from mass in fusion is deposited into the core of stars in the form of radiation (photons) and kinetic energy----i.e. the total energy has to be conserved. For this reason, the total gravitational mass of the sun stays the same after nuclear reactions.

    Gravitational-mass is lost in the form of luminosity at the sun's surface (and additional the solar-wind). The luminosity on the surface is very close to the same amount of energy as is produced in the core via nuclear reactions (this is required for equilibrium); and thus the overall gravitational mass lost from radiation should be about the same as the mass lost in the nuclear reactions.

    A slightly more complete picture... hope that doesn't confuse the situation.
  5. Mar 2, 2012 #4
    Okay, I've read a number of arguments along this line, and it confuses me from what little I remember of my high school physics courses some twenty years ago. I thought it was a law of mass/energy conservation, which superceded previously believed individual laws of mass conservation and energy conservation, as Einstein uncovered a relationship between matter and energy culminating in the equation even non-physicists have heard and rememberd: energy equals mass times the speed of light squared, suggesting matter can become energy or vice-versa. That's what happens in the process of nuclear fusion, right? Part of the mass of the component atoms becomes energy.

    From your statement, it seems to indicate energy converted from mass still has mass itself to contribute to generating a gravitational field if the energy is trapped within the mass. Any chance I could get you to explain this better to me? I thought energy was not mass the way matter (and antimatter, exotic matter, or other particles of mass), but merely converted from mass and does not generate gravity itself, except perhaps indirectly by interacting with mass, causing it to accelerate and increasing the mass of matter through relativity (mass increasing with velocity as the speed of light is approached by an object possessing mass)?
  6. Mar 2, 2012 #5

    This is also true. In Newtonian gravity, there is the concept of mass 'm' which a unique property of matter---that is conserved. In general relativity, there are numerous ways of thinking about 'mass'---generally the term 'mass' equates to the 'rest mass' of particles, a property which is invariant for a given particle. At the same time, the source of gravity is given by something called the 'stress-energy tensor' which includes not only rest mass 'm', but also any other type of energy.

    This is a good example. Speaking technically, physicists generally try to say that a particles 'mass' (per se) is invariant---it is constant. I.e. they try to use the term 'mass' only to refer to 'rest mass'. According to this, you wouldn't just say "e equals m, c squared", you would say, [itex] E = \gamma m c^2[/itex] or better yet, [itex]E^2 = p^2 c^2 + m^2 c^4 [/itex] where '[itex]p = \gamma m v[/itex]' is the momentum. In these equations the mass 'm' is always constant; but the total energy (or momentum) increase.

    That being said, energy and mass are somewhat different. Mass is an invariant property of particles; where-as energy can be transfered, and changed, etc etc.
    This is different than the concept of 'gravitational mass' - which refers to the mass required to create a given gravitational field.... I hope that's not too confusing.
    For example, when people talk about the mass of a neutron star, they're usually talking about the 'gravitational mass' which is noticeably different than the combined rest-mass of all of the constituent particles (for the same reason as in your initial question about the sun). So the terminology is confusing, and not used entirely consistently.
  7. Mar 2, 2012 #6


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    What he's saying is that in the core of the sun four hydrogen nuclei are fused into a Helium nucleus which weighs slightly less than the original 4 hydrogen. This mass deficit is released as energy, primarily in the form of photons and neutrinos. But Einstein's General theory of Relativity tells us that even things like energy and pressure contribute to gravitation, so while those photons and neutrinos are still within the sun, it's gravitational mass is effectively the same.

    What happens next is that the neutrinos immediately zip out of the sun, and thus the sun has lost whatever gravitation those may have provided. The photons, on the other hand, take tens of thousands of years to get from the core to the surface, but once they do, they too leave the system and diminish the effective gravitational mass of the sun. The net effect is that the energy lost through these photons and neutrinos is ~ the energy released when you fuse four hydrogen atoms into a helium.

    Edit: He beat me to it :P
  8. Mar 2, 2012 #7
    Haha, sorry Nabeshin---just too fast for you. Good addition on the neutrinos!
  9. Mar 3, 2012 #8


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    May I just make sure I understand what you are saying here; so the magnitude of the gravitational field from a massive body will also depend (albeit very infinitesimally) on the temperature of that body?
  10. Mar 3, 2012 #9
    Yes. For a relativistic fluid (e.g. neutron-star) it would be far from infinitesimal.
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