Why does gravity increase in star formation?

  • When a star ignites ( ie. fusion starts ), the star maintains its form by balancing gravity's inward pressure, and radiation's outward pressure.

    I get that the fusion of hydrogen atoms releases energy... fine...

    How does gravity keep it together if the mass is lessening as a result of fusion( mass being converted into energy from fusion) while gravity is weakening( as mass lessens )?

    Wouldn't the radiation overpower the force of gravity and tear the star apart?

  • Eubie Drew

    Eubie Drew Correct answer

    7 years ago

    I am going to start with this paragraph from Wikipedia (emphasis mine):

    The most important fusion process in nature is the one that powers
    stars. In the 20th century, it was realized that the energy released
    from nuclear fusion reactions accounted for the longevity of the Sun
    and other stars as a source of heat and light. The fusion of nuclei in
    a star, starting from its initial hydrogen and helium abundance,
    provides that energy and synthesizes new nuclei as a byproduct of that
    fusion process. The prime energy producer in the Sun is the fusion of
    hydrogen to form helium
    , which occurs at a solar-core temperature of
    14 million kelvin. The net result is the fusion of four protons into
    one alpha particle, with the release of two positrons, two neutrinos
    (which changes two of the protons into neutrons), and energy.
    Different reaction chains are involved, depending on the mass of the
    star. For stars the size of the sun or smaller, the proton-proton
    dominates. In heavier stars, the CNO cycle is more important.

    The proton-proton chain set of reactions look like this:

    enter image description here

    The CNO cycle looks like this:

    enter image description here

    Net Result

    Either way, the net result is 4 protons ($^1\!$H nuclei) are turned into 1 alpha particle ($^4\!$He nucleus) plus 2 positrons (e$^+$). The 2 positrons go on to annihilate 2 electrons, so altogether we have a mass change of $$ \Delta M = M_{\mathsf \alpha} - 2M_{\mathsf e} - 4M_{\mathsf P}\,. $$

    Let's find out the fractional change in mass: $$ f_\Delta = \frac{\Delta M}{4M_{\mathsf P}} = \frac{M_{\mathsf \alpha} - 2M_{\mathsf e} - 4M_{\mathsf P}}{4M_{\mathsf P}}\,. $$

    Now the ratio of the mass of an alpha particle to a proton is $3.9726$, or $$ M_{\mathsf \alpha} = 3.9726\times M_{\mathsf P}\,. $$

    The ratio of the mass of a proton to an electron is $1836.1$, or $$ M_{\mathsf e} = \frac{M_{\mathsf P}}{1836.1} = 0.0005446\times M_{\mathsf P}\,. $$

    Substituting into the $f_\Delta$ equation, $$ f_\Delta = \frac{3.9726\times M_{\mathsf P} - 0.0011\times M_{\mathsf P} - 4\times M_{\mathsf P}}{4\times M_{\mathsf P}} = \frac{-0.0285}4 = -0.007125 = -0.7125\%\,\,.$$

    So obviously, Even if all of the hydrogen were converted (only a fraction actually is) the loss of mass to the star would be too negligible to matter.

    A more important mass loss for large stars is that from their stellar wind, which for very large main sequence stars (types O or B) removes a sizable fraction of the very large star's mass over it's lifetime.

    @Aabaakawad- Well done

    Does this also explain the flickering of a star?

License under CC-BY-SA with attribution

Content dated before 7/24/2021 11:53 AM