<blockquote>
  <p>Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat</p>
</blockquote>

<p>Planets in general can have internal heat sources (like radioactive breakdown or a complex hot core).  During planetary formation there are thought to be multiple collisions with large proto-planets before things settle down to a more-or-less stable set of orbits and planets.  These collisions add energy to the planets as they develop and increase temperature (a lot), as does ongoing bombardment from boring-old asteroids.</p>

<p>Now all of this energy does basically come from gravitational sources, but as you can see it's a complex (messy) business and there are many significant events that can alter temperature.</p>

<p>It might be instructive for you to read about existing theories on the <a href="https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System" rel="nofollow noreferrer">formation of Solar system</a> as well as material related to the <a href="https://en.wikipedia.org/wiki/Structure_of_the_Earth#Core" rel="nofollow noreferrer">structure of the Earth</a>.</p>

CC BY-SA 3.0
2017-04-13T10:56:30.937Z
2017-04-13T10:56:30.937Z
StephenG - Help Ukraine
<p>It's important not to confuse the amount of heat the Earth has radiated in its lifetime (a quantity you have no access to here) with its current heat content.  This question is asking you to neglect all the heat radiated (which could be affected by things like radioactivity), and just look at the current heat content, and ask if that is much more, or much less, than the gravitational energy GM^2/R.  So find some way to get the interior T of the Earth (not the surface T, that has nothing to do with this question), and you should find that the total heat content is way <em>less</em> than the gravitational energy released.</p>

<p>The virial theorem was mentioned above, so if you know about that, you might have thought the heat content would be half the gravitational energy.  That's only true of normal stars, which are held up by the gas pressure of largely noninteracting particles, whereas the Earth is held up by forces within the rocks that are not included in the usual virial theorem and have little to do with heat content.   </p>

<p>Still, the gravitational energy gives the energy that was released in forming the Earth, so when you find how little the heat content is in comparison, it gives you a good idea of how much the Earth has cooled since forming.  It also gives you a lower bound on how much net heat it has radiated (over and above the heat of sunlight it has absorbed).  It's only a lower bound because there are also internal radioactive heat sources, but I believe their sum total effect is still fairly small compared to GM^2/R.</p>

<p>Hence the answer to your question is going to turn out to be "yes" when you do that comparison.  (I wouldn't give that spoiler except that some of the answers above might have led you to think the answer should be "no.")</p>

CC BY-SA 3.0
2017-04-16T08:38:47.480Z
Question doesn't mention the Earth. I was thinking gas giant otherwise the question doesn't contain the relevant variables.
ProfRob
CC BY-SA 3.0
2017-04-16T14:38:53.103Z
Sure it does, those are the relevant variables (they're just not quantified).  And even for a gas giant, there are significant forces not included in the virial theorem, so the answer will generally still be "yes".  I agree that all the question is really asking for is a comparison of the two quantities mentioned, but that's pretty simple by itself.   It's only interesting if it compels you to look up the values for actual planets like Earth and Jupiter, and it is then interesting that in either case the answer is "yes" by a significant excess margin.
Ken G
CC BY-SA 3.0
2017-04-16T14:56:35.877Z
No, I've looked again and "the Earth" is not mentioned.You're answering your own question.
ProfRob
CC BY-SA 3.0
2017-04-16T17:20:53.313Z
You are certainly free to replace "Earth" with "any planet you like" in my answer, it holds just as well.  Ergo, your objection is without foundation.
Ken G
CC BY-SA 3.0
2017-04-16T21:05:28.227Z
Rough estimate of GPE released for Jupiter $\leq 2\times 10^{36}$ J, so a trivial application of the virial theorem suggests a thermal energy of $\leq 10^{36}$ J. A more accurate estimate of the thermal energy of Jupiter is $6\times10^{35}$ J (http://evolution.calpoly.edu/jupiter). ie In the case of Jupiter, a gas giant, the two estimates are comparable (as expected). In the case of Earth they are an order of magnitude different. So yes, it matters what planet we are talking about.
ProfRob
CC BY-SA 3.0
2017-04-17T06:22:59.790Z
Obviously it matters what planet is chosen.   But thank you for verifying what I said-- the answer is "yes" for both Earth and Jupiter, it's just *really really* yes for Earth.
Ken G
CC BY-SA 3.0
2017-04-15T15:14:07.940Z
2017-04-15T15:14:07.940Z
Ken G
<p>So assume we have a planet of temperature T, and mass M, We can calculate it's internal energy/heat content by the formula $U=cMT$ where $c$ is the specific heat.</p>

<p>And if we also know the gravitational energy of the planet via $\frac{GM^2}{R}$  </p>

<p>how do I approach the following question?</p>

<p>Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat</p>

<p>Any help would be appreciated.</p>

CC BY-SA 3.0
2017-04-13T08:45:58.577Z
I think, you should edit your post to emphasize (and correct) the real question. Anyway, I suppose you are looking for the derivative of your gravitational energy. Is this any close to what you mean?
Py-ser
CC BY-SA 3.0
2017-04-13T13:03:22.240Z
Related to this question: Check out the [Kelvin-Helmholtz Mechanism](https://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_mechanism). That's basically what this question is asking, is if this mechanism is enough to produce all the heat a planet may produce (the answer of course is no).
zephyr
<planet><heat>
gravitational energy and internal energy
casualprogrammer
# gravitational energy and internal energy

(Asked by: username_0 on 2017-04-13)

So assume we have a planet of temperature T, and mass M, We can calculate it's internal energy/heat content by the formula $U=cMT$ where $c$ is the specific heat.

And if we also know the gravitational energy of the planet via $\frac{GM^2}{R}$

how do I approach the following question?

Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat

Any help would be appreciated.

**Tags:** planet, heat

> **Comments:**
> **username_1**: I think, you should edit your post to emphasize (and correct) the real question. Anyway, I suppose you are looking for the derivative of your gravitational energy. Is this any close to what you mean?  
> **username_2**: Related to this question: Check out the [Kelvin-Helmholtz Mechanism](https://en.wikipedia.org/wiki/Kelvin%E2%80%93Helmholtz_mechanism). That's basically what this question is asking, is if this mechanism is enough to produce all the heat a planet may produce (the answer of course is no).  

---

## 2 answers

### Answer by username_3 on 2017-04-13. Score: 2

> Can the release of gravitational energy during planet formation in principle (is there enough) explain this much heat

Planets in general can have internal heat sources (like radioactive breakdown or a complex hot core). During planetary formation there are thought to be multiple collisions with large proto-planets before things settle down to a more-or-less stable set of orbits and planets. These collisions add energy to the planets as they develop and increase temperature (a lot), as does ongoing bombardment from boring-old asteroids.

Now all of this energy does basically come from gravitational sources, but as you can see it's a complex (messy) business and there are many significant events that can alter temperature.

It might be instructive for you to read about existing theories on the [formation of Solar system](https://en.wikipedia.org/wiki/Formation_and_evolution_of_the_Solar_System) as well as material related to the [structure of the Earth](https://en.wikipedia.org/wiki/Structure_of_the_Earth#Core).


---

### Answer by username_4 on 2017-04-15. Score: 0

It's important not to confuse the amount of heat the Earth has radiated in its lifetime (a quantity you have no access to here) with its current heat content. This question is asking you to neglect all the heat radiated (which could be affected by things like radioactivity), and just look at the current heat content, and ask if that is much more, or much less, than the gravitational energy GM^2/R. So find some way to get the interior T of the Earth (not the surface T, that has nothing to do with this question), and you should find that the total heat content is way *less* than the gravitational energy released.

The virial theorem was mentioned above, so if you know about that, you might have thought the heat content would be half the gravitational energy. That's only true of normal stars, which are held up by the gas pressure of largely noninteracting particles, whereas the Earth is held up by forces within the rocks that are not included in the usual virial theorem and have little to do with heat content.

Still, the gravitational energy gives the energy that was released in forming the Earth, so when you find how little the heat content is in comparison, it gives you a good idea of how much the Earth has cooled since forming. It also gives you a lower bound on how much net heat it has radiated (over and above the heat of sunlight it has absorbed). It's only a lower bound because there are also internal radioactive heat sources, but I believe their sum total effect is still fairly small compared to GM^2/R.

Hence the answer to your question is going to turn out to be "yes" when you do that comparison. (I wouldn't give that spoiler except that some of the answers above might have led you to think the answer should be "no.")

> **Comments:**
> **username_5**: Question doesn't mention the Earth. I was thinking gas giant otherwise the question doesn't contain the relevant variables.  
> **username_4**: Sure it does, those are the relevant variables (they're just not quantified).  And even for a gas giant, there are significant forces not included in the virial theorem, so the answer will generally still be "yes".  I agree that all the question is really asking for is a comparison of the two quantities mentioned, but that's pretty simple by itself.   It's only interesting if it compels you to look up the values for actual planets like Earth and Jupiter, and it is then interesting that in either case the answer is "yes" by a significant excess margin.  
> **username_5**: No, I've looked again and "the Earth" is not mentioned.You're answering your own question.  
> **username_4**: You are certainly free to replace "Earth" with "any planet you like" in my answer, it holds just as well.  Ergo, your objection is without foundation.  
> **username_5**: Rough estimate of GPE released for Jupiter $\leq 2\times 10^{36}$ J, so a trivial application of the virial theorem suggests a thermal energy of $\leq 10^{36}$ J. A more accurate estimate of the thermal energy of Jupiter is $6\times10^{35}$ J (http://evolution.calpoly.edu/jupiter). ie In the case of Jupiter, a gas giant, the two estimates are comparable (as expected). In the case of Earth they are an order of magnitude different. So yes, it matters what planet we are talking about.  
> **username_4**: Obviously it matters what planet is chosen.   But thank you for verifying what I said-- the answer is "yes" for both Earth and Jupiter, it's just *really really* yes for Earth.  

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