<p>The short answer: yes it's possible. If you pull a diamond hard enough it will break. If the black hole is to small it will aleady be evaporated before the diamond is pulled in copletely (but a small part will be torn off). If the hole is huge, the diamond will pass the horizon without much happening to it. It will get torn apart later on while falling to the singularity. For a black hole with a size somewhere in between, the diamond is torn apart already before entering
the hole.<br />
It should be noted however that the tearing apart is due to the diamond being spread out in time.</p>

CC BY-SA 4.0
2021-06-22T16:32:43.180Z
2021-06-22T16:32:43.180Z
Deschele Schilder
<p>It depends on the size of the black hole and the size of the bit of material you are trying to pull apart, as well as what it is made of. What is important is the <em>difference</em> in gravitational pull between one end of your sample and the other. This difference force grows as the inverse of distance <em>cubed</em> as one moves in towards the black hole.</p>

<p>However, nothing special marks crossing event horizon and if we were talking about a <em>supermassive</em> black hole (a million solar masses or more), you might not even notice it had happened, since the force difference, sometimes called the <a href="https://en.m.wikipedia.org/wiki/Tidal_force" rel="noreferrer">tidal force</a>, would be quite small. At the event horizon of a <em>smaller</em> black hole, the tidal force would be much larger. To see this, we note that the difference in force between two ends of a bar of length <span class="math-container">$\Delta r$</span> is approximately
<span class="math-container">$$ F = \frac{8GM}{r^3}\Delta r,$$</span>
where <span class="math-container">$M$</span> is the black hole mass. At the event horizon <span class="math-container">$r = 2GM/c^2$</span>, so
<span class="math-container">$$ F = \frac{c^6}{G^2M^2}\Delta r.$$</span></p>

<p>The spaghettification force continues to increase as the object heads towards <span class="math-container">$r=0$</span>, which will occur in a finite proper time for the falling object. So it will always increase sufficiently to rip something apart, no matter what it is made of; the question is merely at what radius does this happen.</p>

<p>It is of interest however to calculate whether it would happen to an object <em>outside</em> the event horizon, simply because then an external observer (at a safe distance) might witness it. This in turn depends on the material making up the object, it's size and the mass of the black hole.</p>

<p>Consider a material of tensile strength <span class="math-container">$T$</span>, size <span class="math-container">$\Delta r$</span> and cross-sectional area <span class="math-container">$\sim (\Delta r)^2$</span>. For this to break up <em>before</em> reaching the event horizon then the tidal force at the event horizon must exceed <span class="math-container">$T (\Delta r)^2$</span> and thus
<span class="math-container">$$ \frac{c^6}{G^2M^2}\Delta r &gt; T (\Delta r)^2 $$</span>
<span class="math-container">$$ M &lt; 2\times10^5 \left({T \Delta r}\right)^{-1/2}\ M_{\odot}$$</span></p>

<p>Diamond has <span class="math-container">$T\sim 10^{11}$</span> N/m<span class="math-container">$^2$</span>. If it was of size 1 cm, then a black hole would need to be <em>larger</em> than 6 solar masses for this chunk of diamond to survive being ripped apart <em>outside</em> the event horizon. Other materials or larger bits of material are easier to break and would not survive outside the horizon of even more massive black holes. e.g. The tensile strength of bone is about <span class="math-container">$10^{8}$</span> N/m<span class="math-container">$^2$</span>, so a 0.1m piece of bone would be pulled apart prior to falling through the event horizon if <span class="math-container">$M&lt; 60M_{\odot}$</span>.</p>

CC BY-SA 4.0
2016-12-14T20:56:04.740Z
2019-01-17T12:40:14.243Z
2019-01-17T12:40:14.243Z
ProfRob
<p>Is it possible for a Black Hole to easily stretch or turn any hard object (Like: Diamond, Gold bar, Platinum bar, Brass) into spaghettification state in just a fraction of seconds or minute once it gets near or has reached the Black Hole's event horizon?</p>

CC BY-SA 4.0
2021-06-22T18:33:34.067Z
TIL https://en.wikipedia.org/wiki/Spaghettification
Tim Holt
<black-hole>
Spaghettification of hard-things/Elements
Jeffson
# Spaghettification of hard-things/Elements

(Asked by: username_0 on 2016-12-14)

Is it possible for a Black Hole to easily stretch or turn any hard object (Like: Diamond, Gold bar, Platinum bar, Brass) into spaghettification state in just a fraction of seconds or minute once it gets near or has reached the Black Hole's event horizon?

**Tags:** black-hole
> **Comments:**
> **username_1**: TIL https://en.wikipedia.org/wiki/Spaghettification  

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## 2 answers

### Answer by username_3 on 2016-12-14. Score: 12 | Accepted answer

It depends on the size of the black hole and the size of the bit of material you are trying to pull apart, as well as what it is made of. What is important is the *difference* in gravitational pull between one end of your sample and the other. This difference force grows as the inverse of distance *cubed* as one moves in towards the black hole.

However, nothing special marks crossing event horizon and if we were talking about a *supermassive* black hole (a million solar masses or more), you might not even notice it had happened, since the force difference, sometimes called the [tidal force](https://en.m.wikipedia.org/wiki/Tidal_force), would be quite small. At the event horizon of a *smaller* black hole, the tidal force would be much larger. To see this, we note that the difference in force between two ends of a bar of length $\Delta r$ is approximately
$$ F = \frac{8GM}{r^3}\Delta r,$$
where $M$ is the black hole mass. At the event horizon $r = 2GM/c^2$, so
$$ F = \frac{c^6}{G^2M^2}\Delta r.$$

The spaghettification force continues to increase as the object heads towards $r=0$, which will occur in a finite proper time for the falling object. So it will always increase sufficiently to rip something apart, no matter what it is made of; the question is merely at what radius does this happen.

It is of interest however to calculate whether it would happen to an object *outside* the event horizon, simply because then an external observer (at a safe distance) might witness it. This in turn depends on the material making up the object, it's size and the mass of the black hole.

Consider a material of tensile strength $T$, size $\Delta r$ and cross-sectional area $\sim (\Delta r)^2$. For this to break up *before* reaching the event horizon then the tidal force at the event horizon must exceed $T (\Delta r)^2$ and thus
$$ \frac{c^6}{G^2M^2}\Delta r > T (\Delta r)^2 $$
$$ M < 2\times10^5 \left({T \Delta r}\right)^{-1/2}\ M\_{\odot}$$

Diamond has $T\sim 10^{11}$ N/m$^2$. If it was of size 1 cm, then a black hole would need to be *larger* than 6 solar masses for this chunk of diamond to survive being ripped apart *outside* the event horizon. Other materials or larger bits of material are easier to break and would not survive outside the horizon of even more massive black holes. e.g. The tensile strength of bone is about $10^{8}$ N/m$^2$, so a 0.1m piece of bone would be pulled apart prior to falling through the event horizon if $M< 60M\_{\odot}$.


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### Answer by username_2 on 2021-06-22. Score: 0

The short answer: yes it's possible. If you pull a diamond hard enough it will break. If the black hole is to small it will aleady be evaporated before the diamond is pulled in copletely (but a small part will be torn off). If the hole is huge, the diamond will pass the horizon without much happening to it. It will get torn apart later on while falling to the singularity. For a black hole with a size somewhere in between, the diamond is torn apart already before entering
the hole.  
It should be noted however that the tearing apart is due to the diamond being spread out in time.


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