...or how small can a solar system be, to sustain an Earth-like planet?

Sun is not really small, and 1AU is pretty far when you look at orbital radii of exoplanets and size of their stars.

How small a star can be, to still provide Earth's equivalent of 1.36 kW/m2 above its own Roche Limit (of Earth-sized planet)? What would the star's diameter and the orbit providing such irradiance be in such a case?

I guess a Brown Dwarf wouldn't be enough to give that much within its Roche Limit, meaning any planet receiving enough light would be torn apart by tidal forces, and likely stepping up to smallest Main Sequence stars would step up the energy output enough that a planet at the Roche Limit would be scorched like Mercury, so its orbit would need to be considerably larger... but I don't quite know how to perform the needed calculations.

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    $\begingroup$ That's an interesting mathematical exercise. There's other factors if you want the planet to actually be earth-like. When you get a star that small and a planet that close, you might find yourself in the heart of an enormous stellar-magnetosphere. Tidal forces are another problem for anything except a near circular orbit and tidal locking would seem very likely too. If the star has sun-spots, and, presumably you'd either have sunspots or a gonzo-magnetosphere, that could be a problem too. Red-dwarf sun-spots can be very intense. The practical limit would be quite a bit larger. $\endgroup$
    – userLTK
    Nov 18 '15 at 23:52
  • $\begingroup$ Update on the comment above. Reading up on it, most brown dwarf stars won't have a magnetosphere, though (perhaps) smaller or cooler ones might. X-Rays could also be a problem that close to a brown dwarf. en.wikipedia.org/wiki/… $\endgroup$
    – userLTK
    Nov 19 '15 at 1:08
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    $\begingroup$ @userLTK The X-rays are a manifestation of magnetic activity. $\endgroup$
    – ProfRob
    Nov 19 '15 at 8:04
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    $\begingroup$ @Ricky: There's a bottom limit on size of a natural celestial body producing own light of sufficient intensity. Although a sufficiently big artificial satellite made from the right radioisotopes? $\endgroup$
    – SF.
    Nov 19 '15 at 10:11
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    $\begingroup$ @ricky physics.stackexchange.com/questions/165283/… A large "rock" could not maintain its structure against its weight. It would compress and become essentially gaseous, though with an unusual composition. If it then cooled it might though then become crystalline. A white dwarf star could be considered a large "rock" if you count diamond as a rock. The definition of a planet is slippery. $\endgroup$
    – ProfRob
    Nov 19 '15 at 12:46

I consider small stars to mean main sequence stars (or brown dwarfs) with mass less than the Sun, and then also consider compact stellar remnants - white dwarfs and neutron stars.

Main sequence stars and brown dwarfs

What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also depends on the mass of the planet, so when you say earth-sized, I'll assume that means mass and radius.

So going through the calculation.

Flux at the planet is $L/4\pi r^2$, where $L$ is the luminosity and $r$ the orbital radius (assumed circular).

Let's next use an approximate relation that $$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^3$$ A more accurate numerical relationship could be obtained from evolutionary model calculations. There is a complication that the luminosity of very low mass stars and brown dwarfs does depend on age.

For a rigid body the tidal radius (Roche limit) is $$ r_t\simeq 1.4 R_E\left(\frac{M_{\odot}}{M_E}\right)^{1/3},$$ where $R_E$ and $M_E$ are the radius and mass of the Earth.

So setting $r=r_t$, the flux to $1400$ W m$^{-2}$ and replacing $L$ in terms of mass, we find $$\frac{M}{M_{\odot}} = \left[\frac{4\pi \times 1400}{L_{\odot}} (1.4 R_E)^2 \left(\frac{M_{\odot}}{M_E}\right)^{2/3}\right]^{1/3}.$$

All that remains is to put the numbers in and I get $M = 0.026 M_{\odot}$.

So, a small brown dwarf - but there are caveats. First, and most importantly, as I said, the luminosity-mass relation is a bit rough and ready, and is certainly age dependent for something as low mass as this. Second, the expression for the Roche limit depends a bit on structural properties of the planet, but I think this introduces relatively little uncertainty.

If you use that mass and the tidal radius formula, you find that the orbital radius is only $5R_E$. As the minimum size of a brown dwarf is about a Jupiter radius, it appears that tidal breakup would not be the limiting factor for an Earth-like planet. Also note that the spectrum of radiation from a brown dwarf might contain the same power per unit area as sunlight, but it has a very different spectrum; mostly infrared with almost nothing in the visible band.

Compact stellar remnants

Smaller "stars" are possible in the form of the collapsed remnants known as white dwarfs or neutron stars. Planets have been detected around neutron stars.

White dwarf stars are "born" very hot, cool rapidly and then more slowly. To answer your question requires an addendum, which is a minimum time you want this illumination for. A newborn WD has about a solar luminosity and has a similar diameter to the Earth. So you planet could be 1 au away from this and still receive a sun-like power per unit area. BUT, it would be short-lived and the spectrum of the light would be nothing like the Sun. The WD would have a temperature in excess of 50,000K and most of the light would be in the UV.

Cooling curves (and observations) suggest they cool to $10^{-4}$ solar luminosities in less than a billion years and have more sun-like temperatures, but to receive sun-like illumination at this point, your planet would need to be only 0.01 au from the white dwarf. White dwarfs have a mass a little less than the Sun. Using the same tidal radius formula as above we get $r_t \sim 0.004$au. So I conclude that yes, it might be possible to have a long-lived planet with Earth-like illumination around a white dwarf. But note (a) conditions change rapidly (compared with the Sun) and (b) you have to arrange to get the planet that close, since an orbit of that size would have been inside the red giant progenitor of the white dwarf.

Neutron stars are born in the cores of supernovae explosions and have a radius of about 10 km and mass about 1.5 times that of the Sun. If your planet can survive the supernova (or be constructed afterwards) it faces a similar set of problems to a planet around a white dwarf.

Neutron stars are born with surface temperatures of billions of degrees, emitting copious gamma ray radiation. They cool very rapidly - within a million years the surface temperature is down to an X-ray emitting million degrees and perhaps to UV-emitting temperatures of tens of thousands within 100 million years. It may be that they never cool much below this, since they are heated by their rotational spin-down, the decay of their strong magnetic fields and accretion of gas from the interstellar material. Even at 50,000K, because of its small size, the neutron star has a luminosity of only a millionth that of the Sun. That means your planet would have to be moved within 0.001 au of the neutron star to receive the same power and this would be well inside the tidal breakup radius of $\sim 0.005$ au.

I conclude that there really is no long-term possibility for sun-like illumination from a neutron star. Note also, that I have ignored magnetospheric and pulsar-like emission from the neutron star. These are also short-lived phenomena only seen in the first million years or so of a neutron stars existence.

  • $\begingroup$ "the minimum size of a brown dwarf is about a Jupiter radius" yet Jupiter isn't a brown dwarf. This is an odd thing that you explained well on Physics SE. As you add mass to a jupiter-like body, its radius hardly increases at all, it just gets denser. $\endgroup$ Nov 19 '15 at 9:51
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    $\begingroup$ @steveverrill Yes, electron degeneracy pressure means that the radii of "cold" objects are almost constant between about 0.1 Jupiter mass and 10 Jupiter masses and actually get a little bit smaller for 50 Jupiter mass objects. physics.stackexchange.com/questions/165283/… $\endgroup$
    – ProfRob
    Nov 19 '15 at 12:41

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