<h2>I am going to start with <a href="http://en.wikipedia.org/wiki/Nuclear_fusion#Nuclear_fusion_in_stars" rel="nofollow noreferrer">this paragraph from Wikipedia</a> <em>(emphasis mine)</em>:</h2>

<blockquote>
  <p>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. <strong><em>The prime energy producer in the Sun is the fusion of
  hydrogen to form helium</em></strong>, 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 <strong><em>proton-proton
  chain</em></strong> dominates. In heavier stars, the <strong><em>CNO cycle</em></strong> is more important.</p>
</blockquote>

<hr>

<h2>The <strong><em>proton-proton chain</em></strong> set of reactions look like this:</h2>

<p><a href="https://i.sstatic.net/PLExf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PLExf.png" alt="enter image description here"></a></p>

<hr>

<h2>The <strong><em>CNO</em></strong> cycle looks like this:</h2>

<p><a href="https://i.sstatic.net/VVxm1.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VVxm1.png" alt="enter image description here"></a></p>

<hr>

<h2>Net Result</h2>

<p>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 $$ \Delta M = M_{\mathsf \alpha} - 2M_{\mathsf e} - 4M_{\mathsf P}\,. $$ </p>

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

<p>Now <a href="http://physics.nist.gov/cgi-bin/cuu/Value?malsmp" rel="nofollow noreferrer">the ratio of the mass of an alpha particle to a proton is $3.9726$</a>, or $$ M_{\mathsf \alpha} = 3.9726\times M_{\mathsf P}\,. $$</p>

<p><a href="http://physics.nist.gov/cgi-bin/cuu/Value?mpsme" rel="nofollow noreferrer">The ratio of the mass of a proton to an electron is $1836.1$</a>, or $$ M_{\mathsf e} = \frac{M_{\mathsf P}}{1836.1} = 0.0005446\times M_{\mathsf P}\,. $$</p>

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

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

<hr>

<p>A more important mass loss for large stars is that from their <a href="http://en.wikipedia.org/wiki/Stellar_wind" rel="nofollow noreferrer">stellar wind</a>, which for very large main sequence stars (types O or B) removes a sizable <strong><em>fraction</em></strong> of the very large star's mass over it's lifetime.</p>

CC BY-SA 3.0
2015-11-06T13:49:48.610Z
@Aabaakawad- Well done
user5434678
CC BY-SA 3.0
2015-11-06T13:50:42.227Z
Does this also explain the flickering of a star?
user5434678
CC BY-SA 3.0
2015-11-07T21:32:21.353Z
@Mr.Cruz see http://astronomy.stackexchange.com/questions/222/why-do-stars-appear-to-twinkle
Eubie Drew
CC BY-SA 3.0
2015-11-06T03:25:25.913Z
2015-11-07T20:18:32.210Z
2015-11-07T20:18:32.210Z
Eubie Drew
<p>The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.</p>

<p>This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.</p>

<p>Radiation pressure <em>is</em> a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.</p>

CC BY-SA 3.0
2015-11-06T07:58:27.557Z
2015-11-06T07:58:27.557Z
ProfRob
<p>Some good answers, I'm going to give kind of summary, cause you touched on a few points.</p>

<blockquote>
  <p>Why does gravity increase in star formation</p>
</blockquote>

<p>Gravitation is a product of a few forces.   Mass, density and, not to be ignored, rotation speed.  </p>

<p>It's not actually the fusion process that keeps the sun from contracting, at least, not directly.   It's heat that keeps the star expanded.   That's the balancing act.  High temperature wants to expand, gravity wants to contract.  </p>

<p>The fusion process is actually pretty slow, which is why stars like our sun have a main sequence of about 10 billion years, and a lot of the heat that a star starts out with is from the heat of formation.  Potential energy gets converted to heat due to the coalescing and condensing of all that matter so stars start out hot, even before fusion begins.</p>

<p>In fact, a star in formation can be many times brighter than the star is during it's main sequence due to the high heat of formation.   Here's an <a href="http://cseligman.com/text/stars/mldiagram.htm" rel="nofollow noreferrer">article</a> that says the forming sun was 200 times brighter than it is now.  </p>

<p>Young proto-stars, as a result of conservation of angular momentum, tend to <a href="https://en.wikipedia.org/wiki/Stellar_rotation" rel="nofollow noreferrer">rotate very fast</a> and that fast rotation can create a bulge and increases <a href="https://en.wikipedia.org/wiki/Stellar_rotation#During_formation" rel="nofollow noreferrer">ejection of matter</a>.   The formation process is pretty chaotic compared to the main sequence stage.  Lots of ejected matter, much bigger solar storms, lots of lheat from formation, etc.</p>

<p>Once the main sequence stage is underway and rotation is slowed down, then there's more of a balance between heat and gravity mentioned above.  The fusion process continues to add heat to the core of star which the star, convects or conduct heat away from the core into the outer layers and then, radiates from it's surface, but during the main sequence, in general, the core of the star gradually heats up and in most cases, the energy added from fusion isn't nearly strong enough to blow apart the star, unless the star is enormously large like over 150 or 200 solar masses, then the star doesn't really work without blowing off a bunch of matter.  See: <a href="https://astronomy.stackexchange.com/questions/328/is-there-a-theoretical-maximum-size-limit-for-a-star">here</a>.</p>

<blockquote>
  <p>I get that the fusion of hydrogen atoms releases energy... fine...</p>
  
  <p>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 )?</p>
</blockquote>

<p>As others have said, mass loss by solar wind is a bigger factor especially for young and smaller stars, but there's a few factors at play.   The short answer to this question is that the mass loss, at least by fusion, is quite very compared to the total mass of the star.   Another factor, as hydrogen becomes helium, the core of the star becomes denser and greater density tends to be smaller and that increases gravity, but there are competing factors.   The inner core grows denser as it becomes more hydrogen rich and the fusion tends to expand outwards on the outside of the helium core, so a star like our sun gets a denser inner core over time, but the layers around the core can grow hotter and larger, even as they lose mass.</p>

<blockquote>
  <p>Wouldn't the radiation overpower the force of gravity and tear the
  star apart?</p>
</blockquote>

<p>As mentioned above, this happens if you have 150 or 200 solar masses.   lower mass stars, the fusion isn't nearly powerful enough to blow the star apart.   Stars and white dwarfs blow apart when they go supernova, but that's different than the main sequence fusion process.  </p>

<p>Our sun will blow off some of it's matter when it has it's helium flash, so there are examples of what you're describing happening, but not during the main sequence for stars like our sun when material is expelled primarily by magnetic storms causing coronal mass ejections.  Fusion is, generally speaking, more like a slow burn, than a big explosion when it's up against the enormous gravitational binding energy of a star.</p>

CC BY-SA 3.0
2015-11-06T08:01:20.543Z
@RobJeffries  It has something to do with it.   Certainly type1.  The Oxygen and Carbon converts to heavier elements and that creates a lot of energy.   An Iron white dwarf collapse without any energy from fusion might look quite different.   But, I'll re-word that section.
userLTK
CC BY-SA 3.0
2015-11-06T09:25:19.217Z
What mass is the Sun expected to lose as a result of the He flash?
ProfRob
CC BY-SA 3.0
2015-11-06T10:53:59.970Z
Are you quizzing me?   I don't know the specific numbers and it would vary with the size of the star.    The helium flash is associated with the formation of the planetary nebula.     http://www.universetoday.com/25669/the-sun-as-a-white-dwarf-star/    our sun is expected to lose about half it's mass, though some of that probably happens before the Helium flash.   I should probably change "A lot" to "some of", that's probably more accurate.
userLTK
CC BY-SA 3.0
2015-11-06T13:00:43.170Z
I'm quizzing you because there is *not* expected to be any major mass loss episode associated with the He flash. In fact quite the opposite.  As the star ascends the giant branch (H-shell burning), it loses some mass (not a lot compared with the AGB phase). The He flash, terminates the red giant ascent, and  is accompanied by a reduction in the size of the star, higher surface gravity and less wind. The He flash has nothing to do with planetary nebula formation.
ProfRob
CC BY-SA 3.0
2015-11-06T06:53:35.187Z
2015-11-06T10:55:36.887Z
2017-04-13T12:59:07.017Z
userLTK
<p>Here's the bare bones reason for stars like our sun.  The full story is much more...full.</p>

<p>Expansion means cooling.  Cooling means less fusion.  Less fusion means less energy driving expansion, meaning the outward pressure is going down.  Eventually gravity is pulling inward more strongly than radiation is pushing outward.  So the material collapses again.  Collapsing means heating.  Heating means more fusion.  More fusion means more radiation pushing outward on the star.  Produce enough energy, and you'll overcome gravity and the star expands.  </p>

<p>Rinse and repeat.  </p>

<p>The star naturally sits at an equilibrium where gravity and radiation balance each other. Deviations from this are self-correcting.</p>

CC BY-SA 3.0
2015-11-06T07:42:10.243Z
2015-11-06T07:42:10.243Z
zibadawa timmy
<p>As a star runs out of hydrogen fuel, the fusion slows, causing the gravity to overpower the outward force of pressure, thus contracting. Contraction of the star causes high temperature and pressure, to the extent that it is enough to fuse helium into carbon, then the energy released is stronger than the gravity, increasing the size of the star into a red giant. The following paragraph from an article explains this:</p>

<blockquote>
  <p>Over its life, the outward pressure of fusion has balanced against the
  inward pressure of gravity. Once the fusion stops, gravity takes the
  lead and compresses the star smaller and tighter. Temperatures
  increase with the contraction, eventually reaching levels where helium
  is able to fuse into carbon. Depending on the mass of the star, the
  helium burning might be gradual or might begin with an explosive
  flash. The energy produced by the helium fusion causes the star to
  expand outward to many times its original size.</p>
</blockquote>

<p>The amount of mass lost is more due to stellar wind, rather than to fusion. To answer your second question, the pressure will never overcome completely the force of gravity. When a star reaches the iron nickel stage of fusing, it'll stop, unable to go further. This causes a rapid contraction for either a star to go supernova (which is really tearing most of the star apart except for its core), or cool down to a black dwarf.</p>

CC BY-SA 3.0
2015-11-06T07:02:04.200Z
The fusion process tends to speed up as the star burns, this is cause, as it adds heat, the speed of the nuclei increases and that increases the number of interactions.   It's kind of counter-intuitive, but as our sun burns hydrogen it increases it's hydrogen fusion rate, until the hydrogen is nearly run out that is.
userLTK
CC BY-SA 3.0
2015-11-06T03:39:24.443Z
2015-11-06T03:39:24.443Z
CipherBot
<p>When a star ignites ( ie. fusion starts ), the star maintains its form by balancing gravity's inward pressure, and radiation's outward pressure.</p>

<p>I get that the fusion of hydrogen atoms releases energy... fine...</p>

<p>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 )?</p>

<p>Wouldn't the radiation overpower the force of gravity and tear the star apart?</p>

<gravity><star-formation><radiation>
Why does gravity increase in star formation?
user2738698
# Why does gravity increase in star formation?

(Asked by: username_0 on 2015-11-06)

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?

**Tags:** gravity, star-formation, radiation

---

## 5 answers

### Answer by username_1 on 2015-11-06. Score: 3 | Accepted answer

## I am going to start with [this paragraph from Wikipedia](http://en.wikipedia.org/wiki/Nuclear_fusion#Nuclear_fusion_in_stars) *(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
> chain*** dominates. In heavier stars, the ***CNO cycle*** is more important.

---

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

[![enter image description here](https://i.sstatic.net/PLExf.png)](https://i.sstatic.net/PLExf.png)

---

## The ***CNO*** cycle looks like this:

[![enter image description here](https://i.sstatic.net/VVxm1.png)](https://i.sstatic.net/VVxm1.png)

---

## 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 $3.9726$](http://physics.nist.gov/cgi-bin/cuu/Value?malsmp), or $$ M\_{\mathsf \alpha} = 3.9726\times M\_{\mathsf P}\,. $$

[The ratio of the mass of a proton to an electron is $1836.1$](http://physics.nist.gov/cgi-bin/cuu/Value?mpsme), 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](http://en.wikipedia.org/wiki/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.

> **Comments:**
> **username_2**: @Aabaakawad- Well done  
> **username_2**: Does this also explain the flickering of a star?  
> **username_1**: @Mr.Cruz see http://astronomy.stackexchange.com/questions/222/why-do-stars-appear-to-twinkle  

---

### Answer by username_3 on 2015-11-06. Score: 6

The Sun's luminosity is $3.8\times 10^{26}$ W. Application of mass energy equivalence tells you it loses mass at a rate of 4.25 million tonnes per second as hydrogen turns into helium.

This is practically nothing as far as the structure of the star goes. Over its lifetime, the Sun has lost about 0.03% of its mass in this way.

Radiation pressure *is* a feature in stellar evolution calculations. It is almost negligible in the solar interior (at the 1% level compared with thermal pressure). However, it does become more important in more massive stars with hotter interiors and higher luminosities.


---

### Answer by username_4 on 2015-11-06. Score: 3

Some good answers, I'm going to give kind of summary, cause you touched on a few points.

> Why does gravity increase in star formation

Gravitation is a product of a few forces. Mass, density and, not to be ignored, rotation speed.

It's not actually the fusion process that keeps the sun from contracting, at least, not directly. It's heat that keeps the star expanded. That's the balancing act. High temperature wants to expand, gravity wants to contract.

The fusion process is actually pretty slow, which is why stars like our sun have a main sequence of about 10 billion years, and a lot of the heat that a star starts out with is from the heat of formation. Potential energy gets converted to heat due to the coalescing and condensing of all that matter so stars start out hot, even before fusion begins.

In fact, a star in formation can be many times brighter than the star is during it's main sequence due to the high heat of formation. Here's an [article](http://cseligman.com/text/stars/mldiagram.htm) that says the forming sun was 200 times brighter than it is now.

Young proto-stars, as a result of conservation of angular momentum, tend to [rotate very fast](https://en.wikipedia.org/wiki/Stellar_rotation) and that fast rotation can create a bulge and increases [ejection of matter](https://en.wikipedia.org/wiki/Stellar_rotation#During_formation). The formation process is pretty chaotic compared to the main sequence stage. Lots of ejected matter, much bigger solar storms, lots of lheat from formation, etc.

Once the main sequence stage is underway and rotation is slowed down, then there's more of a balance between heat and gravity mentioned above. The fusion process continues to add heat to the core of star which the star, convects or conduct heat away from the core into the outer layers and then, radiates from it's surface, but during the main sequence, in general, the core of the star gradually heats up and in most cases, the energy added from fusion isn't nearly strong enough to blow apart the star, unless the star is enormously large like over 150 or 200 solar masses, then the star doesn't really work without blowing off a bunch of matter. See: [here](https://astronomy.stackexchange.com/questions/328/is-there-a-theoretical-maximum-size-limit-for-a-star).

> 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 )?

As others have said, mass loss by solar wind is a bigger factor especially for young and smaller stars, but there's a few factors at play. The short answer to this question is that the mass loss, at least by fusion, is quite very compared to the total mass of the star. Another factor, as hydrogen becomes helium, the core of the star becomes denser and greater density tends to be smaller and that increases gravity, but there are competing factors. The inner core grows denser as it becomes more hydrogen rich and the fusion tends to expand outwards on the outside of the helium core, so a star like our sun gets a denser inner core over time, but the layers around the core can grow hotter and larger, even as they lose mass.

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

As mentioned above, this happens if you have 150 or 200 solar masses. lower mass stars, the fusion isn't nearly powerful enough to blow the star apart. Stars and white dwarfs blow apart when they go supernova, but that's different than the main sequence fusion process.

Our sun will blow off some of it's matter when it has it's helium flash, so there are examples of what you're describing happening, but not during the main sequence for stars like our sun when material is expelled primarily by magnetic storms causing coronal mass ejections. Fusion is, generally speaking, more like a slow burn, than a big explosion when it's up against the enormous gravitational binding energy of a star.

> **Comments:**
> **username_4**: @RobJeffries  It has something to do with it.   Certainly type1.  The Oxygen and Carbon converts to heavier elements and that creates a lot of energy.   An Iron white dwarf collapse without any energy from fusion might look quite different.   But, I'll re-word that section.  
> **username_3**: What mass is the Sun expected to lose as a result of the He flash?  
> **username_4**: Are you quizzing me?   I don't know the specific numbers and it would vary with the size of the star.    The helium flash is associated with the formation of the planetary nebula.     http://www.universetoday.com/25669/the-sun-as-a-white-dwarf-star/    our sun is expected to lose about half it's mass, though some of that probably happens before the Helium flash.   I should probably change "A lot" to "some of", that's probably more accurate.  
> **username_3**: I'm quizzing you because there is *not* expected to be any major mass loss episode associated with the He flash. In fact quite the opposite.  As the star ascends the giant branch (H-shell burning), it loses some mass (not a lot compared with the AGB phase). The He flash, terminates the red giant ascent, and  is accompanied by a reduction in the size of the star, higher surface gravity and less wind. The He flash has nothing to do with planetary nebula formation.  

---

### Answer by username_5 on 2015-11-06. Score: 1

Here's the bare bones reason for stars like our sun. The full story is much more...full.

Expansion means cooling. Cooling means less fusion. Less fusion means less energy driving expansion, meaning the outward pressure is going down. Eventually gravity is pulling inward more strongly than radiation is pushing outward. So the material collapses again. Collapsing means heating. Heating means more fusion. More fusion means more radiation pushing outward on the star. Produce enough energy, and you'll overcome gravity and the star expands.

Rinse and repeat.

The star naturally sits at an equilibrium where gravity and radiation balance each other. Deviations from this are self-correcting.


---

### Answer by username_6 on 2015-11-06. Score: -1

As a star runs out of hydrogen fuel, the fusion slows, causing the gravity to overpower the outward force of pressure, thus contracting. Contraction of the star causes high temperature and pressure, to the extent that it is enough to fuse helium into carbon, then the energy released is stronger than the gravity, increasing the size of the star into a red giant. The following paragraph from an article explains this:

> Over its life, the outward pressure of fusion has balanced against the
> inward pressure of gravity. Once the fusion stops, gravity takes the
> lead and compresses the star smaller and tighter. Temperatures
> increase with the contraction, eventually reaching levels where helium
> is able to fuse into carbon. Depending on the mass of the star, the
> helium burning might be gradual or might begin with an explosive
> flash. The energy produced by the helium fusion causes the star to
> expand outward to many times its original size.

The amount of mass lost is more due to stellar wind, rather than to fusion. To answer your second question, the pressure will never overcome completely the force of gravity. When a star reaches the iron nickel stage of fusing, it'll stop, unable to go further. This causes a rapid contraction for either a star to go supernova (which is really tearing most of the star apart except for its core), or cool down to a black dwarf.

> **Comments:**
> **username_4**: The fusion process tends to speed up as the star burns, this is cause, as it adds heat, the speed of the nuclei increases and that increases the number of interactions.   It's kind of counter-intuitive, but as our sun burns hydrogen it increases it's hydrogen fusion rate, until the hydrogen is nearly run out that is.  

---

