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I thought that a large, close moon might help generate enough heat and stress to raise the average temperature on my world and stave off a Snowball Earth scenario. According to this article and this article, I was incorrect. (Thanks very much for the help on that problem, though--I couldn't have done it without you!)

This leaves me with a missing link. How should I generate the seismic and volcanic activity necessary to keep my planet warm, despite its distance of 2 AU or so from my star of 1.7 LSol? If it comes to a total remodel I will gladly do so, because I would prefer to keep my world an ocean world rather than a dirty snowball.

Thanks for all the help these last couple days--I'm new to Stack Exchange and this has been a very welcoming community so far. You're the best. <3 R

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  • $\begingroup$ Does the planet need to be habitable or have native life? Does the temperature have to be directly due to volcanism or is a planet with heavy volcanism that is habitable due to heavy green house effect acceptable? I think it would be easiest to simply have the planet be too young to have cooled down, and life and green house gasses in the atmosphere be supplied by terraforming. $\endgroup$
    – Ville Niemi
    Commented Feb 16, 2018 at 21:36
  • $\begingroup$ Yes! I would like the planet to be habitable and to develop its own complex life much like Earth; however, it is an ocean world with a very thick atmosphere high in CO<sub>2</sub> and O<sub>2</sub>. $\endgroup$
    – Rúnatál Davino
    Commented Feb 16, 2018 at 21:40
  • $\begingroup$ So it would be ok to chalk up habitable temperature to green house gasses and lower albedo of an ocean planet without he volcanism having a direct effect? (It would help recycle carbon and keep CO2 high.) $\endgroup$
    – Ville Niemi
    Commented Feb 16, 2018 at 21:49
  • $\begingroup$ @VilleNiemi - he has his atmosphere with around 16% CO2 in the other questions. $\endgroup$
    – Twelfth
    Commented Feb 16, 2018 at 22:18
  • $\begingroup$ The hard-science tag comes with some specific restrictions. It means you have to use equations and cite scientific papers to answer. You should be careful when using it on a question, since it makes answers much harder to come by. For something as speculative as this, I don't think hard-science is the best tag; stick with science-based. $\endgroup$
    – kingledion
    Commented Feb 16, 2018 at 22:50

2 Answers 2

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Frictional warming!

perseid meteor shower http://www.sun-gazing.com/perseid-meteor-shower-peaks-august-12th-13th-predicted-best-ever/

First let us figure out how much energy you need to make up for what your dim sun does not provide. How much does Earth get from the sun?

Earths radius = 6,371 km * 1.5 for radius of sphere of atmopshere = 9556 which we will round to 10,000km.

Area of a sphere = 4πr2 = 1256000 km2 = 1.256e+12 m2

Divide in half because only half in sunlight = 628000000000 m2

https://en.wikipedia.org/wiki/Solar_irradiance#Total_irradiance

Annual irradiance of of earth = 21.6 Mj (megajoules) / m2 * 628000000000 = 1.35648E+13 Mj

Star in fictional world is 1.7 sol but 2 AU away. By inverse square law 2 AU = 25% as bright so 1.7 * .25 = 0.425 sol on this planet. To get to full solar energy that means one must provide 0.575 sol some other way.

0.575 sol * 1.35648E+13 Mj = 7.79976E+12 Mj or 7.79976e+18 j to be provided.

By incoming meterorites! Solve for needed mass.

Physics stack! How to calculate the energy of a meteor. https://physics.stackexchange.com/questions/136970/how-to-calculate-the-impact-of-a-meteor

kinetic energy in joules = mv2/2
mass in kg, velocity in m/s 7.79976e+18 j = mv2/2
https://www.amsmeteors.org/fireballs/faqf/ Assume velocity of 50 km/second or 50,000 m / s. 50,000 ^ 2 = 2500000000

7.79976e+18 = ( m * 2500000000)/2 1.55995E+19 = m * 2500000000

6239808000 kg = m

Let us round to a clean 6 billion kg or 6 e+9 kg to provide enough energy to warm the atmosphere up by the needed 0.575 sol.

Does that pass the sniff test?

https://arxiv.org/abs/1403.6391 Assessments of the energy, mass and size of the Chicxulub Impactor

The mass is in the range of 1.0e15 kg to 4.6e17 kg.

More than we need!

But what mass of asteroids actually hit Earths atmosphere? https://science.nasa.gov/science-news/science-at-nasa/2011/01mar_meteornetwork

“Every day about 100 tons of meteroids… enter earths atmosphere”.

100 tons * 365 days = 36500 tons / year = 33112243 kg. Just 33 million kg.

So - if your planet were warmed by the kinetic energy of 6 billion kg of meterorites each year (200 times the amount that hits earth on a good year, 0.00006% of the amount that hits earth on one very bad year) you could warm the atmosphere by the desired amount. Bear in mind that these meteoroid do not have to whack the ground - they can almost all dissolve away into the atmosphere, dissipating their energy as heat.

Math, you temptress. I am sure those orders of magnitude threw me somewhere along the way. I will not have feelings hurt if someone finds a math error and points it out. I am certain meteroids can warm an atmosphere but less certain I calculated the required amount.

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Can I offer you some radiation?

http://physicsworld.com/cws/article/news/2011/jul/19/radioactive-decay-accounts-for-half-of-earths-heat

About 50% of the heat given off by the Earth is generated by the radioactive decay of elements such as uranium and thorium, and their decay products. That is the conclusion of an international team of physicists that has used the KamLAND detector in Japan to measure the flux of antineutrinos emanating from deep within the Earth. The result, which agrees with previous calculations of the radioactive heating, should help physicists to improve models of how heat is generated in the Earth.

...

As a result of this model, scientists believe that about 20 TW is generated by radioactive decay – 8 TW from the uranium-238 decay chain; 8 TW from the thorium-232 decay chain and the final 4 TW from potassium-40. Fortunately, these decay chains also produce anti-electron-neutrinos, which travel easily through the Earth and can be detected, thereby giving physicists a way to measure the decay rates and ultimately the heat produced deep underground.

It's a topic we don't know too much on, but one could venture the guess that more radioactive material inside the earth would increase this heat (doubt it's a linear increase though).

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  • $\begingroup$ I like it! Something I forgot about entirely, which is a little silly. Thanks very much :) $\endgroup$
    – Rúnatál Davino
    Commented Feb 16, 2018 at 22:31
  • $\begingroup$ Heh there is a counter-intuitive piece here...outer circle planets won't contain as many of these radioactive elements and the planet being further away from the sun than earth would suggest it should contain less of these materials. That being said, orbital bombardment by asteroids or even having the planet drift away from where it was formed are both likely solutions to the radioactive material content question. $\endgroup$
    – Twelfth
    Commented Feb 16, 2018 at 22:34
  • $\begingroup$ Yes, I was thinking it would probably be bombarded by asteroids. My system's main asteroid belt is rather nearby, and the moon is so large that it could only have been formed by collision. $\endgroup$
    – Rúnatál Davino
    Commented Feb 16, 2018 at 22:36
  • $\begingroup$ No, thank you. I already have more than enough radiation :D $\endgroup$
    – Xavon_Wrentaile
    Commented Feb 17, 2018 at 0:01

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