<blockquote>
<p><em>[OP]</em> Nascent terrestrial planets quickly have their surfaces cool down. Formation of crust then yields formation of a figure to fossilise.</p>
<p><em>[OP, emphasis added]</em> An answer to this question is <strong>defined by how the crust (or most of it) survives the collision</strong>.</p>
</blockquote>
<p>I don't think that's a valid assumption. As <a href="https://astronomy.stackexchange.com/users/2365/james-k">James K</a> commented, planets such as Earth may seem solid to you, but their scale is such that they behave like fluids. Consider these two terrestrial planets:</p>
<ul>
<li><p><a href="https://www.aeronomie.be/en/encyclopedia/venus-internal-structure" rel="nofollow noreferrer">Venus</a> has a radius of 6051 km and a crust thickness between 10 and 30 km. The crust is thus just 0.17% – 0.5% of the radius. Most of the rest is viscous fluid.</p>
</li>
<li><p><a href="https://en.wikipedia.org/wiki/Earth#Internal_structure" rel="nofollow noreferrer">Earth</a> has a radius of 6371 km and a crust thickness between 6 km and 50 km. The crust is thus just 0.09% – 0.78% of the radius. Most of the rest is viscous fluid.</p>
</li>
</ul>
<p>The upper mantle is also part of the <a href="https://en.wikipedia.org/wiki/Lithosphere" rel="nofollow noreferrer">lithosphere</a> and thus rigid, but that doesn't change the situation much.</p>
<p>Think of a large water balloon where the skin is made of a thin silicate instead of rubber, and the inside is filled with thick syrup instead of water.  The balloon is a meter in diameter, and the skin is just a few millimeters thick. The balloon is in free fall in vacuum.</p>
<p>The silicate-syrup balloon will behave much like a regular rubber-water balloon except that the skin will fracture and break much more easily. Any added rotation will make the balloon take a oblate shape. If the skin is not flexible enough to allow that, it will break, and the viscous but still fluid material from the inside will flood the cracks and over the skin, taking an oblate shape.</p>
<p>This analogy illustrates that it is somewhat irrelevant whether the crust melts in the collision scenario described in the question. The shape of the planet is dominated by the rotation rate and not by the crust nor the lithosphere.</p>
<p>It is worth a note that tidal forces also cause shifts in the shape. Witness the twice-daily half a meter shift in Earth's crust. See <a href="https://en.wikipedia.org/wiki/Tide#Earth_tides" rel="nofollow noreferrer">Earth tides</a>.</p>

CC BY-SA 4.0
2024-11-25T12:31:17.223Z
You confuses the crust with the lithosphere. If you want to consider the rheology, then you have to take the whole lithosphere into account: the upper mantle is rigid. Your point is still valid though, it's only a few % of the radius.
Jean-Marie Prival
CC BY-SA 4.0
2024-11-25T15:17:23.333Z
@Jean-MariePrival good point. Added a link to lithosphere at Wikipedia. The question asks specifically about the crust, hence the wording in the answer.
doubleunary
CC BY-SA 4.0
2024-11-24T08:00:35.590Z
2024-11-25T15:15:40.597Z
2024-11-25T15:15:40.597Z
doubleunary
<p>Nascent terrestrial planets quickly have their surfaces cool down.<br />
Formation of crust then yields formation of a figure.</p>
<p>What happens to the already formed crust and the figure after an energetic collision sufficient to produce a small or middle-size moon (not a large one like our Moon)?</p>
<p>Will the crust and figure stay about unchanged?</p>
<p>Or will the collision shake and warm up the planet's surface layers so
that the old shape will be gone, and formation of crust and figure will start anew?</p>
<p>More specifically:</p>
<p>Let a planet be spinning at a rate <span class="math-container">$\Omega$</span> and have a dynamical oblateness <span class="math-container">$J_2$</span> corresponding hydrostatically to <span class="math-container">$\Omega$</span>.
After a hit, the planet's rotation rate becomes <span class="math-container">$\Omega^{\,\prime}$</span>. Will the planet retain the oblateness value <span class="math-container">$J_2\,$</span>, or will it acquire a new value <span class="math-container">$J_2^{\,\prime}$</span> corresponding to <span class="math-container">$\Omega^{\,\prime}\,$</span>? An answer to this question is defined by how the crust (or most of it) survives the collision.</p>

CC BY-SA 4.0
2024-11-22T19:35:58.680Z
Are we starting with a planet that's roughly earth-sized? I can't see lofting anything into orbit without completely liquefying or vaporizing the mass if you're dealing with Earth-like gravity.  Maybe a tiny planetoid like Pluto or a large asteroid could yield a very small moon that's basically a whole fragment of crust?
Darth Pseudonym
CC BY-SA 4.0
2024-11-22T19:39:09.630Z
[MoonOrigin-Social-NASAWeb](https://en.wikipedia.org/wiki/File:ARC-20221004-AAV3443-MoonOrigin-Social-NASAWeb-1080p_medium.oggtheora.ogg)
doubleunary
CC BY-SA 4.0
2024-11-23T03:56:42.150Z
@doubleunary Thank you for the animation. It is mightly convincing.
Michael_1812
CC BY-SA 4.0
2024-11-23T06:55:24.653Z
Even if the impact energy isn't sufficient to fully melt all the crust, it seems likely that a moon-forming impact will temporarily make the surrounding crust *behave* like a liquid. That's what we see in impact craters.
PM 2Ring
CC BY-SA 4.0
2024-11-23T07:02:05.853Z
The Caloris Planitia impact didn't re-melt Mercury's crust, but the seismic waves affected the whole planet, and created a chaotic region at the antipodes. See https://en.wikipedia.org/wiki/Caloris_Planitia#Antipodal_chaotic_terrain_and_global_effects
PM 2Ring
CC BY-SA 4.0
2024-11-23T13:26:29.980Z
@PM2Ring Thank you for your comments. My question was not just about the surrounding crust but about the crust as a whole. Specifically, after a substantial hit the planet will change its rotation rate. Will it then adjust its oblateness ($J_2$) to the new spin, or will it retain the old value of $J_2\,$? This will depend on whether the *entire* crust becomes at some point molten or semimolten.
Michael_1812
CC BY-SA 4.0
2024-11-23T13:29:21.893Z
@DarthPseudonym Yes, the planet is earth-sized. Also, I am talking not about a Mars-sized impactor but something smaller, producing a middle-sized moon. As you rightly said, the stuff sent into orbit will be completely liquefied or vapourised. My question was if the *entire* crust would be liquified. After an impact, the planet will change its spin rate. Will it then adjust its oblateness ($J_2$) to the new rotation rate, or will it retain the old $J_2\,$? This is related to the question if the *entire* crust becomes soft after collision.
Michael_1812
CC BY-SA 4.0
2024-11-23T14:16:00.950Z
I don't think the thickness or stiffness of the crust has much effect on oblateness in Earth-like planets. It's dominated by the rotation rate. Similarly, tidal forces cause shifts in the shape. Witness the twice-daily half a meter shift in Earth's crust. See [Earth tides](https://en.wikipedia.org/wiki/Tide#Earth_tides).
doubleunary
CC BY-SA 4.0
2024-11-23T14:52:51.133Z
oh, okay, I thought you were talking about the crust *that became the new moon*.
Darth Pseudonym
CC BY-SA 4.0
2024-11-24T02:20:03.787Z
@doubleunary Yes, sure, the angular velocity. Please imagine a planet spinning at a rate $\Omega$ and having a $J_2$ value corresponding hydrostatically to that $\Omega$. After a hit, the planet's rotation rate becomes $\Omega^{\,\prime}$. My question is: will the planet retain the oblateness value $J_2$, or will it acquire a new value $J_2^{\,\prime}$ defined hydrostatically by $\Omega^{\,\prime}\,$? An answer to this question is determined by how the crust (or most of it) survives the collision -- and this is the essence of my question.
Michael_1812
CC BY-SA 4.0
2024-11-24T05:54:55.633Z
What does “a figure to fossilise” mean?
Peter Erwin
CC BY-SA 4.0
2024-11-24T11:31:14.247Z
Planets behave like fluids, not solids.  On that scale, everything flows. Whether the crust melts of not is pretty irrelevant, the planet will flow to the new oblateness value J₂ corresponding to it's new rotation rate.  The oblateness of the Earth corresponds to a 24 hour day, not to the length of the day when the moon formed.
James K
CC BY-SA 4.0
2024-11-25T10:19:41.327Z
@PeterErwin I have removed "to fossilise". Thank you.
Michael_1812
<the-moon><planet><natural-satellites><planetary-science><impact>
A planet after a moon-generating impact
Michael_1812
# A planet after a moon-generating impact

(Asked by: username_0 on 2024-11-22)

Nascent terrestrial planets quickly have their surfaces cool down.  
Formation of crust then yields formation of a figure.

What happens to the already formed crust and the figure after an energetic collision sufficient to produce a small or middle-size moon (not a large one like our Moon)?

Will the crust and figure stay about unchanged?

Or will the collision shake and warm up the planet's surface layers so
that the old shape will be gone, and formation of crust and figure will start anew?

More specifically:

Let a planet be spinning at a rate $\Omega$ and have a dynamical oblateness $J\_2$ corresponding hydrostatically to $\Omega$.
After a hit, the planet's rotation rate becomes $\Omega^{\,\prime}$. Will the planet retain the oblateness value $J\_2\,$, or will it acquire a new value $J\_2^{\,\prime}$ corresponding to $\Omega^{\,\prime}\,$? An answer to this question is defined by how the crust (or most of it) survives the collision.

**Tags:** the-moon, planet, natural-satellites, planetary-science, impact

> **Comments:**
> **username_1**: Are we starting with a planet that's roughly earth-sized? I can't see lofting anything into orbit without completely liquefying or vaporizing the mass if you're dealing with Earth-like gravity.  Maybe a tiny planetoid like Pluto or a large asteroid could yield a very small moon that's basically a whole fragment of crust?  
> **username_2**: [MoonOrigin-Social-NASAWeb](https://en.wikipedia.org/wiki/File:ARC-20221004-AAV3443-MoonOrigin-Social-NASAWeb-1080p_medium.oggtheora.ogg)  
> **username_0**: @username_2 Thank you for the animation. It is mightly convincing.  
> **username_3**: Even if the impact energy isn't sufficient to fully melt all the crust, it seems likely that a moon-forming impact will temporarily make the surrounding crust *behave* like a liquid. That's what we see in impact craters.  
> **username_3**: The Caloris Planitia impact didn't re-melt Mercury's crust, but the seismic waves affected the whole planet, and created a chaotic region at the antipodes. See https://en.wikipedia.org/wiki/Caloris_Planitia#Antipodal_chaotic_terrain_and_global_effects  
> **username_0**: @PM2Ring Thank you for your comments. My question was not just about the surrounding crust but about the crust as a whole. Specifically, after a substantial hit the planet will change its rotation rate. Will it then adjust its oblateness ($J_2$) to the new spin, or will it retain the old value of $J_2\,$? This will depend on whether the *entire* crust becomes at some point molten or semimolten.  
> **username_0**: @DarthPseudonym Yes, the planet is earth-sized. Also, I am talking not about a Mars-sized impactor but something smaller, producing a middle-sized moon. As you rightly said, the stuff sent into orbit will be completely liquefied or vapourised. My question was if the *entire* crust would be liquified. After an impact, the planet will change its spin rate. Will it then adjust its oblateness ($J_2$) to the new rotation rate, or will it retain the old $J_2\,$? This is related to the question if the *entire* crust becomes soft after collision.  
> **username_2**: I don't think the thickness or stiffness of the crust has much effect on oblateness in Earth-like planets. It's dominated by the rotation rate. Similarly, tidal forces cause shifts in the shape. Witness the twice-daily half a meter shift in Earth's crust. See [Earth tides](https://en.wikipedia.org/wiki/Tide#Earth_tides).  
> **username_1**: oh, okay, I thought you were talking about the crust *that became the new moon*.  
> **username_0**: @username_2 Yes, sure, the angular velocity. Please imagine a planet spinning at a rate $\Omega$ and having a $J_2$ value corresponding hydrostatically to that $\Omega$. After a hit, the planet's rotation rate becomes $\Omega^{\,\prime}$. My question is: will the planet retain the oblateness value $J_2$, or will it acquire a new value $J_2^{\,\prime}$ defined hydrostatically by $\Omega^{\,\prime}\,$? An answer to this question is determined by how the crust (or most of it) survives the collision -- and this is the essence of my question.  
> **username_4**: What does “a figure to fossilise” mean?  
> **username_5**: Planets behave like fluids, not solids.  On that scale, everything flows. Whether the crust melts of not is pretty irrelevant, the planet will flow to the new oblateness value J₂ corresponding to it's new rotation rate.  The oblateness of the Earth corresponds to a 24 hour day, not to the length of the day when the moon formed.  
> **username_0**: @PeterErwin I have removed "to fossilise". Thank you.  

---

## 1 answers

### Answer by username_2 on 2024-11-24. Score: 2 | Accepted answer

> *[OP]* Nascent terrestrial planets quickly have their surfaces cool down. Formation of crust then yields formation of a figure to fossilise.
>
> *[OP, emphasis added]* An answer to this question is **defined by how the crust (or most of it) survives the collision**.

I don't think that's a valid assumption. As [username_5](https://astronomy.stackexchange.com/users/2365/james-k) commented, planets such as Earth may seem solid to you, but their scale is such that they behave like fluids. Consider these two terrestrial planets:

* [Venus](https://www.aeronomie.be/en/encyclopedia/venus-internal-structure) has a radius of 6051 km and a crust thickness between 10 and 30 km. The crust is thus just 0.17% – 0.5% of the radius. Most of the rest is viscous fluid.
* [Earth](https://en.wikipedia.org/wiki/Earth#Internal_structure) has a radius of 6371 km and a crust thickness between 6 km and 50 km. The crust is thus just 0.09% – 0.78% of the radius. Most of the rest is viscous fluid.

The upper mantle is also part of the [lithosphere](https://en.wikipedia.org/wiki/Lithosphere) and thus rigid, but that doesn't change the situation much.

Think of a large water balloon where the skin is made of a thin silicate instead of rubber, and the inside is filled with thick syrup instead of water. The balloon is a meter in diameter, and the skin is just a few millimeters thick. The balloon is in free fall in vacuum.

The silicate-syrup balloon will behave much like a regular rubber-water balloon except that the skin will fracture and break much more easily. Any added rotation will make the balloon take a oblate shape. If the skin is not flexible enough to allow that, it will break, and the viscous but still fluid material from the inside will flood the cracks and over the skin, taking an oblate shape.

This analogy illustrates that it is somewhat irrelevant whether the crust melts in the collision scenario described in the question. The shape of the planet is dominated by the rotation rate and not by the crust nor the lithosphere.

It is worth a note that tidal forces also cause shifts in the shape. Witness the twice-daily half a meter shift in Earth's crust. See [Earth tides](https://en.wikipedia.org/wiki/Tide#Earth_tides).

> **Comments:**
> **username_6**: You confuses the crust with the lithosphere. If you want to consider the rheology, then you have to take the whole lithosphere into account: the upper mantle is rigid. Your point is still valid though, it's only a few % of the radius.  
> **username_2**: @Jean-MariePrival good point. Added a link to lithosphere at Wikipedia. The question asks specifically about the crust, hence the wording in the answer.  

---

