I'm doing some writing set on a fictional earthlike planet that for reasons needs to be substantially larger than Earth but can't have substantially higher gravity. I've been able to compromise with a planet that's 6x the circumference, having a surface gravity of 1.5G, yielding a density of about 4700kg/m^3.

This is quite a bit (~15%) less than Earth and Venus, but (20%) more than Mars. It would proportionally have less heavy metal, meaning less iron for a protective magnetic field. I would like this planet to have some basis in reality - could such a planet be earth-like, and is there a definitive range of planet densities where life as we know it could exist?

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    $\begingroup$ Prominent planetary scientist David Stevenson explains that too high content of metal also destroys the magnetic field, because high conductivity cools the core faster. It is temperature differences that matter to allow for convection, which moves matter that create the magnetic field of a planet. $\endgroup$
    – LocalFluff
    Jul 18, 2015 at 5:02
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    $\begingroup$ Right now, anything people say about life elsewhere is extrapolating from a sample size of one. Extrapolating from a sample size of one is a spectacualy bad idea. $\endgroup$ Jul 18, 2015 at 6:47
  • $\begingroup$ @DavidHammen That's a view based on great simplification. Our ancestors murdered all other great ape races, so our intelligence might not have arisen only once. Stone tools were used 3.3 million years ago, before even Homo Erectus appeared. And all life on Earth is family now, maybe not because life on Earth arose once, but because we ate all the other lifeforms. Biology is immensely rich and it is possible to extrapolate from it. Earth' biosphere is not a single point of data. $\endgroup$
    – LocalFluff
    Jul 18, 2015 at 7:18
  • $\begingroup$ What is "life as we know it?". There are obviously gravity limits for a giraffe! I also can't reproduce your calculation. I get 5g. $\endgroup$
    – ProfRob
    Jul 18, 2015 at 10:07
  • $\begingroup$ My bad, I meant to say circumference. With a radius of 71000km and a g of 15m/s, ρ = 4754 kg/m^3 $\endgroup$
    – BB ON
    Jul 20, 2015 at 16:04

1 Answer 1


You could probably get away with replacing molten iron, 7.87 g/cc, in the core with aluminum, 2.70 g/cc, and still generate a substantial magnetic field: Molten Metal Magnet

It’s easy to create a magnetic field by using a battery to force an electric current through a loop of wire. But Earth’s core, a rotating mix of iron and nickel with internal flows driven by the passage of heat, has no battery and no wires. Instead, it creates magnetism by means of self-sustaining feedback. Liquid metal moving through a magnetic field generates a current, similar to that induced in the moving coil of an electric generator. That current in turn generates the magnetic field. This “self-generation” mechanism can dramatically amplify the small, random fields that always exist in magnetic materials.

Sodium was used in the linked article, however any molten metal or conductor, should do, and sodium, at 0.97 g/cc is just not dense enough to sink to a planet's core. You'd have to do some hand waving over "an iron poor/aluminum rich Bok globule", but I think you could get your planet, with a sufficient magnetic field to fend off solar winds.

  • $\begingroup$ There's handwaving and handwaving. You would need to arrange for aluminium to be more than 20 times overabundant compared to the averages seen in stars and the solar system. $\endgroup$
    – ProfRob
    Jul 18, 2015 at 9:57
  • $\begingroup$ @RobJeffries Don't need pure aluminum, an alloy would do. Maybe only a tenfold overabundance to get a low density core. Abundance: en.wikipedia.org/wiki/Abundance_of_the_chemical_elements The big trick here is to get the iron abundance down. $\endgroup$ Jul 18, 2015 at 11:13
  • $\begingroup$ Yes, my comment got mangled. I meant to say 20 times overabundant compared to the abundances seen in the solar system and exceeding the abundances of iron, magnesium and silicon - which are all far more abundant refractory elements. Never seen this anywhere to my knowledge. $\endgroup$
    – ProfRob
    Jul 18, 2015 at 11:30

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