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Stars can be crushed by gravity and create black holes or neutron stars. Why doesn't the same happen with any planet if it is in the same space time?

Please explain it in simple way. Note: I am not a physicist but have some interest in physics.

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    $\begingroup$ Related: physics.stackexchange.com/q/143166/2451 , physics.stackexchange.com/q/141865/2451and links therein. $\endgroup$ – Qmechanic Oct 20 '16 at 7:07
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    $\begingroup$ Because planets are not as massive? $\endgroup$ – Steeven Oct 20 '16 at 8:36
  • $\begingroup$ I've deleted some comments, including some (very nice) comments that would have been more appropriate as answers. $\endgroup$ – rob Oct 22 '16 at 14:05
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    $\begingroup$ The same reason a water baloon doesn't hold its shape but a basketball does under gravity. $\endgroup$ – Ali Caglayan Oct 24 '16 at 8:08
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In very simple terms which I hope you will understand.

The gravitational force of attraction depends on mass and distance.
For the atoms which make up the Earth there are two forces acting on them, the gravitational attraction due to all the other atoms and the Coulomb/electrostatic repulsive force between the electrons orbiting the atoms.
The electron shells repel one another.

As mass increases the gravitational attractive force increases and the atoms come closer together and the repulsion between the electron shells increases to balance the increased gravitational attraction.

If the mass increases even more the Coulomb repulsive force cannot balance the increased gravitational attractive force and the atom collapses with protons and electrons combining to form neutrons.
You then have an entity composed of neutrons - a neutron star.

There is still the gravitational attractive force between neutrons but now the repulsive force is provided by the strong nuclear force between the neutrons - neutrons do not like to be "squashed".

Increase the mass even more and the gravitational attractive force increases and so does the repulsive force between neutrons by the neutrons coming closer together.

Eventually if you increase the mass even more the repulsive force between the neutrons is not sufficient to balance the gravitational attractive force between the neutrons and so you get a further collapse into a black hole.

So the simple answer to your question is that the gravitational forces between the atoms which make up a planet are not large enough to initiate catastrophic collapse because the mass of a planet is not large enough.

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    $\begingroup$ There's the (electro-)degenerate matter (found in white dwarfs) stage before neutron star matter forms, which IMHO is important enough to include. Alternatively change wording so that neutron stars are not implied to form by atoms collapsing. $\endgroup$ – hyde Oct 22 '16 at 18:45
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    $\begingroup$ @hyde Thank you for your perceptive comment. I know that my answer lacks detail but I wanted to give an easy to understand answer which had some semblance of correctness. There have been lots of other comments (and where have they all gone?) with excellent suggestions which could well have been included but unfortunately brevity often means that not everything which is important is not included. So electron degeneracy pressure took the hit of omission. $\endgroup$ – Farcher Oct 22 '16 at 19:14
  • $\begingroup$ So in short, black holes don't contain electrons, protons OR neutrons, but something even denser? Until now, I was imagining it to be like a massive nucleus with smaller nuclei beside one another. $\endgroup$ – cst1992 Oct 23 '16 at 14:15
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    $\begingroup$ To add a simple information here.... from google: The Sun weighs about 333,000 times as much as Earth. That is a lot more mass. And still not enough to form a black hole. Google here says... "Wherever that cutoff between a neutron star and a black hole lies — whether it's 2.5 or 2.7 or 3.0 or 3.2 solar masses — that's where you might think the minimum mass black hole could possibly come from." Which would be roughtly a million times earth. WAAAAAYYYY not enough mass. $\endgroup$ – TomTom Oct 23 '16 at 16:52
  • $\begingroup$ This answer is totally incorrect. Firstly, gravitational pull at the massive scale (like stars) is totally different from those of the micro-scale (like electrons). Heisenberg uncertainty and wave particle duality figured much more importance at this scale. Quantum mechanical effects take precedence over relativistic effects. $\endgroup$ – Peter Teoh Oct 25 '16 at 0:55
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Planets are crushed by gravity! That's why, for example, Earth is a densely packed spherical rock rather than a loose cloud of dust.

There's just not enough crushing 'force' to do more than that.

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    $\begingroup$ As an example, of planets being crushed by gravity, the density of the Earth's iron/nickel inner core is about 13.1 g/cc. Compare that with an iron-nickel meteorite of presumably similar composition (but not crushed by gravity), in which the density is about 7.8 g/cc. $\endgroup$ – David Hammen Oct 21 '16 at 16:12
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    $\begingroup$ This phenomenon is also present to a ridiculous degree in the cores of gas planets, where it supposedly causes the formation of metallic hydrogen. $\endgroup$ – MauganRa Oct 22 '16 at 14:39
  • $\begingroup$ also they round due to gravity $\endgroup$ – jk. Oct 24 '16 at 9:35
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The particles which make up atoms are electrically charged, and they repel each other when they get too close to each other. Gravitational forces only attract one particle to another, and never repel, but they're extremely weak compared to the electrical force. To create a black hole, the gravitational force needs to overcome these repulsive forces between particles. For objects like the earth and the sun, the repulsive forces are much greater than the gravitational force.

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There have been several answers already but as a synthesis attempt :

Gravity is attractive, and in absence of repulsive counter force it causes the collapse of a massive object. The order of magnitude of the pressure needed to resist against gravitational collapse is roughly of the order of $GM^2/R^4$ where $M$ is the mass of the object and $R$ its radius.

In the case of a planet such as the Earth, the repulsive forces are of electrostatic nature (their electrons tend to repel). For the earth, $GM^2/R^4 \sim $ 1000 GPa.

If the mass is much larger, gravity is too strong and electrostatic forces are too weak to counter it. When the density is high enough, nuclear reactions can occur, emitting a high amount of radiation. In this case the object is a star and it is held by thermal pressure. For the sun, $GM_{\odot}^2/R_{\odot}^4 \sim 10^{6} $ GPa, but this pressure can vary a lot from one star to another.

After some time, nuclear reactions no longer release enough energy, for example when iron starts being produced (iron is the most stable nucleus, and reactions that transform it would be endothermic). In this case, the object can collapse to a higher density form of matter, this time stabilized thanks to Pauli's exclusion principle.

This principle states that two fermions cannot occupy the same quantum state, resulting in a very strong repulsive force between them. In white-dwarfs, these fermions are electrons. In neutron stars, they are mostly neutrons. The strong force also contributes to resisting gravity in neutron stars. In these cases, the pressure can be extreme. A neutron star mass is usually $\gtrsim 1.2 M_{\odot}$, and its radius of the order of 10 km. This yields $P \sim 10^{25}$ GPa.

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    $\begingroup$ Neutron star radii are $\sim 10$ km. $\endgroup$ – Rob Jeffries Oct 20 '16 at 22:28
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You must understand that there are two factors involved here, first one is gravity that is trying to bring the planet closer and crush it and the second factors tries to resist this crushing e.g. pauli exclusion principle leads to repulsion sometimes, nuclear reaction also resist crushing in stars . So this play of two different factors leads to crushing in some but not all cases.

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  • $\begingroup$ @KevinWells Jupiter is almost entirely supported by electron degeneracy pressure. The "electromagnetic force, the nuclear strong force and thermal energy [density]" are small, irrelevant and minor contributors respectively in the core of a giant planet. $\endgroup$ – Rob Jeffries Oct 20 '16 at 22:35
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There are plenty of good answers here already, but one way of thinking about these things has perhaps not been emphasized enough. That is that the mass inside a sphere (assuming uniform density) increases as the cube of the radius. So if you double the radius you have eight times as much mass inside it. The gravitational force therefore increases as objects get bigger. Hence small objects may not even be spherical, but above a certain size planets will be spherical, and as they continue to increase in size, the gravitational effects become more and more significant.

The other concept that is related and useful is of the schwartzchild radius. Simplistically, you can think of this as follows. Suppose you have a certain mass ($m$) and you put it inside a sphere of radius $r$. As the radius gets smaller and smaller, so the escape velocity gets bigger and bigger. At a certain point, the escape velocity is equal to the speed of light and so nothing can escape. This radius at which this occurs is the schwartzchild radius, and it is proportional to the mass $m$. Per the discussion in the first paragraph, it is gets easier and easier for a mass to fit inside its schwartzchild radius the larger it gets. Something like earth would need to fit inside a 1cm sphere, which is basically impossible given the repulsion of the atoms. But for much larger objects, as the radius increases, the mass increases as the cube, and so the schwartzchild radius increases as the cube, and so eventually for larger enough objects it becomes easier to be a black hole.

Similar physics applies in other situations.

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protected by Qmechanic Oct 20 '16 at 11:39

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