<p>By way of starting a discussion, because I don't clearly see how to answer this question correctly or all the factors that may be involved.</p>

<p>If you increase $G$, then by the virial theorem, the central temperature of the ball of gas (that will contract quasti-statically, since radiation cannot escape unimpeded) will be proportional to $GM/R$. So for a fixed mass, the ignition temperature for nuclear fusion will be reached at a larger radius and stars would be physically larger and less dense.
$$T_c \simeq \frac{GMm_u}{10Rk_B},$$
assuming an ionised hydrogen gas and where $m_u$ is an atomic mass unit.</p>

<p>In order to avoid "starhood", you need to compress the gas either to within its Schwarzschild radius or to a density where it could be supported by electron degeneracy pressure without reaching a high enough temperature to ignite hydrogen fusion. </p>

<p>Considering the first of these. $R_s = 2GM/c^2$, so the central temperature reached at $R=R_s$ would be $\simeq m_u c^2/20k_B = 5\times 10^{11}$K, is <em>independent</em> of $G$ and is far higher than the $10^{7}$K required for H fusion (ignore deuterium). So direct collapse to a black hole cannot circumvent nuclear fusion.</p>

<p>The second possibility is likely to be more constraining.
There is a standard treatment given in <a href="http://adsabs.harvard.edu/abs/1983bhwd.book.....S" rel="nofollow noreferrer">Shapiro &amp; Teukolsky (1983)</a> thaat determines the radius $R$ at which electron degeneracy becomes important in supporting a star of a given mass $M$. The idea is that you work out the mass and radius for which the phase space occupied by an electron approaches $h^3$. The result is (for ionised hydrogen):
$$ R/R_{\odot} \simeq 0.07 (M/M_{\odot})^{-1/3}$$
However, digging into the maths a bit, this can be written as a proportionality
$$ R \propto G^{-1} M^{-1/3}$$
and thus the central temperature at which electron degeneracy pressure becomes important will be
$$T_c \propto GM/R  \propto G^2 M^{4/3}$$</p>

<p>Thus (for a fixed $G$) what happens is that a contracting ball of gas will reach a critical $T_c$ for fusion if the mass exceeds some critical threshold (above about $0.075M_{\odot}$ in our universe) then H fusion will begin before degeneracy sets in. If you increase $G$ then this threshold actually becomes <em>smaller</em> and smaller balls of gas will become stars.</p>

<p>But what about the Chandrasekhar mass? In our universe the threshold mass for fusion is much lower than the Chandrasekhar mass ($5.8M_{\odot}$ for pure hydrogen), so there is no problem in supporting an object just below the fusion threshold with electron degeneracy pressure. Dig a little deeper and we find that the Chandrasekhar mass is in fact $M_{Ch} \propto G^{-3/2}$. This means we can now express the critical central temperature in terms of mass <em>as a fraction</em> of the Chandrasekhar mass.
$$ T_c \propto G^2 \left(\frac{M}{M_{Ch}}\right)^{4/3} (G^{-3/2})^{4/3} \propto \left(\frac{M}{M_{Ch}}\right)^{4/3}$$</p>

<p>In other words, the threshold between a star that burns hydrogen and an object (brown dwarf) that will be supported by degeneracy pressure is a fixed fraction of the Chandrasekhar mass and does not depend on $G$.</p>

<p>My conclusion is therefore that although changing $G$ will do interesting things to stellar evolution (changing the rate at which it occurs for instance - see <a href="https://astronomy.stackexchange.com/questions/17937/solar-luminosity-proportional-to-g7?rq=1">here</a> where I show that the luminosity scales as $G^7$ and therefore lifetimes scale as $G^{-7}$ !), I do not think that changing $G$ allows direct collapse to a black hole without the central temperature becoming high enough to initiate hydrogen fusion.</p>

<p>Secondly, a failed star is possible in our universe - it just needs to be less than $0.075M_{\odot}$ and collapse is halted by electron degeneracy pressure. In a universe with an increased $G$ this threshold becomes <em>smaller</em> as $G^{-3/2}$.</p>

CC BY-SA 3.0
2017-08-03T15:06:04.193Z
I haven't had the time to dig into your well written answer as much as it deserves, but I have no doubt it's as well reasoned as well. Thanks for taking the time to post it, hope you'll be still open to a question or two in a few days. Thank you!
uhoh
CC BY-SA 3.0
2017-08-01T16:48:51.313Z
2017-08-02T10:53:08.487Z
2017-08-02T10:53:08.487Z
ProfRob
<p>There is no such limit, at least not a meaningful one.</p>

<p>If the gravity would be more strong, smaller hydrogene clouds could evolve to stars, even Jupiter or Earth sized, or much smaller. As you increase the strength of the gravity, the required mass for star formation decreases, but stars will still exist.</p>

CC BY-SA 3.0
2017-06-28T20:07:18.773Z
More analysis required. Fusion could/would be prevented by degeneracy pressure in smaller objects.
ProfRob
CC BY-SA 3.0
2017-08-01T16:51:03.057Z
Turns out you're right because the radius at which degeneracy sets in is $\propto G^{-1}$,
ProfRob
CC BY-SA 3.0
2017-08-01T19:27:54.757Z
@RobJeffries Wow, thank you! It was a much more complex thing as I thought. :-)
peterh
CC BY-SA 3.0
2017-06-28T18:38:38.333Z
2017-06-28T18:38:38.333Z
peterh
<p><a href="https://astronomy.stackexchange.com/questions/21422/why-does-gas-form-a-star-but-not-a-black-hole">This</a> question got me thinking.</p>

<p>In another universe, with a different value for the gravitational constant G, how much bigger would G need to be for star formation to be prevented?</p>

<p>In other words in a hypothetical universe, the star would attempt to ignite, but the force of gravity would overwhelm the star's outward pressure and you would just get a black hole instead.</p>

<p>I'm not looking for a precise answer here... just ROM... 10 times bigger? 10,000 times bigger?</p>

CC BY-SA 3.0
2017-08-01T15:04:14.523Z
@RobJeffries I'll be happy to find out the answer no matter what it is. I can't edit the comment in the bounty box, but it does end in a question mark meaning any right answer is welcome!
uhoh
<black-hole><gravity>
How strong would gravity need to be to prevent star formation
Reginald Blue
# How strong would gravity need to be to prevent star formation

(Asked by: username_0 on 2017-06-28)

[This](https://astronomy.stackexchange.com/questions/21422/why-does-gas-form-a-star-but-not-a-black-hole) question got me thinking.

In another universe, with a different value for the gravitational constant G, how much bigger would G need to be for star formation to be prevented?

In other words in a hypothetical universe, the star would attempt to ignite, but the force of gravity would overwhelm the star's outward pressure and you would just get a black hole instead.

I'm not looking for a precise answer here... just ROM... 10 times bigger? 10,000 times bigger?

**Tags:** black-hole, gravity
> **Comments:**
> **username_1**: @RobJeffries I'll be happy to find out the answer no matter what it is. I can't edit the comment in the bounty box, but it does end in a question mark meaning any right answer is welcome!  

---

## 2 answers

### Answer by username_2 on 2017-08-01. Score: 5

By way of starting a discussion, because I don't clearly see how to answer this question correctly or all the factors that may be involved.

If you increase $G$, then by the virial theorem, the central temperature of the ball of gas (that will contract quasti-statically, since radiation cannot escape unimpeded) will be proportional to $GM/R$. So for a fixed mass, the ignition temperature for nuclear fusion will be reached at a larger radius and stars would be physically larger and less dense.
$$T\_c \simeq \frac{GMm\_u}{10Rk\_B},$$
assuming an ionised hydrogen gas and where $m\_u$ is an atomic mass unit.

In order to avoid "starhood", you need to compress the gas either to within its Schwarzschild radius or to a density where it could be supported by electron degeneracy pressure without reaching a high enough temperature to ignite hydrogen fusion.

Considering the first of these. $R\_s = 2GM/c^2$, so the central temperature reached at $R=R\_s$ would be $\simeq m\_u c^2/20k\_B = 5\times 10^{11}$K, is *independent* of $G$ and is far higher than the $10^{7}$K required for H fusion (ignore deuterium). So direct collapse to a black hole cannot circumvent nuclear fusion.

The second possibility is likely to be more constraining.
There is a standard treatment given in [Shapiro & Teukolsky (1983)](http://adsabs.harvard.edu/abs/1983bhwd.book.....S) thaat determines the radius $R$ at which electron degeneracy becomes important in supporting a star of a given mass $M$. The idea is that you work out the mass and radius for which the phase space occupied by an electron approaches $h^3$. The result is (for ionised hydrogen):
$$ R/R\_{\odot} \simeq 0.07 (M/M\_{\odot})^{-1/3}$$
However, digging into the maths a bit, this can be written as a proportionality
$$ R \propto G^{-1} M^{-1/3}$$
and thus the central temperature at which electron degeneracy pressure becomes important will be
$$T\_c \propto GM/R \propto G^2 M^{4/3}$$

Thus (for a fixed $G$) what happens is that a contracting ball of gas will reach a critical $T\_c$ for fusion if the mass exceeds some critical threshold (above about $0.075M\_{\odot}$ in our universe) then H fusion will begin before degeneracy sets in. If you increase $G$ then this threshold actually becomes *smaller* and smaller balls of gas will become stars.

But what about the Chandrasekhar mass? In our universe the threshold mass for fusion is much lower than the Chandrasekhar mass ($5.8M\_{\odot}$ for pure hydrogen), so there is no problem in supporting an object just below the fusion threshold with electron degeneracy pressure. Dig a little deeper and we find that the Chandrasekhar mass is in fact $M\_{Ch} \propto G^{-3/2}$. This means we can now express the critical central temperature in terms of mass *as a fraction* of the Chandrasekhar mass.
$$ T\_c \propto G^2 \left(\frac{M}{M\_{Ch}}\right)^{4/3} (G^{-3/2})^{4/3} \propto \left(\frac{M}{M\_{Ch}}\right)^{4/3}$$

In other words, the threshold between a star that burns hydrogen and an object (brown dwarf) that will be supported by degeneracy pressure is a fixed fraction of the Chandrasekhar mass and does not depend on $G$.

My conclusion is therefore that although changing $G$ will do interesting things to stellar evolution (changing the rate at which it occurs for instance - see [here](https://astronomy.stackexchange.com/questions/17937/solar-luminosity-proportional-to-g7?rq=1) where I show that the luminosity scales as $G^7$ and therefore lifetimes scale as $G^{-7}$ !), I do not think that changing $G$ allows direct collapse to a black hole without the central temperature becoming high enough to initiate hydrogen fusion.

Secondly, a failed star is possible in our universe - it just needs to be less than $0.075M\_{\odot}$ and collapse is halted by electron degeneracy pressure. In a universe with an increased $G$ this threshold becomes *smaller* as $G^{-3/2}$.

> **Comments:**
> **username_1**: I haven't had the time to dig into your well written answer as much as it deserves, but I have no doubt it's as well reasoned as well. Thanks for taking the time to post it, hope you'll be still open to a question or two in a few days. Thank you!  

---

### Answer by username_3 on 2017-06-28. Score: 1

There is no such limit, at least not a meaningful one.

If the gravity would be more strong, smaller hydrogene clouds could evolve to stars, even Jupiter or Earth sized, or much smaller. As you increase the strength of the gravity, the required mass for star formation decreases, but stars will still exist.

> **Comments:**
> **username_2**: More analysis required. Fusion could/would be prevented by degeneracy pressure in smaller objects.  
> **username_2**: Turns out you're right because the radius at which degeneracy sets in is $\propto G^{-1}$,  
> **username_3**: @RobJeffries Wow, thank you! It was a much more complex thing as I thought. :-)  

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