Far, far away in the universe (specifically, 542 million kiloparsecs from here), a galaxy stands as an enduring legacy of a long-extinct hyperadvanced alien civilization, high among the Kardashev scale. At the center is a supermassive black hole one trillion times as massive as our sun orbited by multitudes of giant stars, their lifespans artificially prolonged to last as long as red dwarves.

  1. First ring is a co-orbital of nine blue hypergiant binaries, each star is 56 times as wide, 250 times as massive and over six-and-a-half million times as bright as our sun.
  2. Second ring is a co-orbital of nine mixed binaries, each one a blue hypergiant (same as the first ring) orbited by a yellow hypergiant (1575 times as wide, 40 times as massive and 630,000 times as bright as our sun)
  3. Third ring is a co-orbital of nine yellow hypergiant binaries, all of which of the same parameters as on the second ring
  4. Fourth ring is a co-orbital of nine mixed binaries, each one a yellow hypergiant (same as the second and third rings) orbited by a red supergiant (2,069 times as wide, 20 times as massive and 589,000 times as bright as our sun)
  5. Fifth and final ring is a co-orbital of nine red supergiant binaries of the same parameters as the fourth ring

This arrangement creates a galactic habitable zone big enough to engulf our entire Milky Way. There are nine orbital rings within this habitable zone, each one consisting of a co-orbital of nine white dwarf stars, each one as wide as our moon, yet one-tenth of a percent as bright and 140% as massive as our sun, as per the Chandrasekhar Limit. Orbiting each of the white dwarves are nine orbital rings, and on each ring is a co-orbital of nine Earth-like planets, complete with its own large moon.

This is actually a benefit, as that since the white dwarves' purpose is strictly gravitational, sunlight comes only from all those oodles of massive stars orbiting the black hole. But then there is one issue that comes with orbiting a smaller but more massive star--the Roche Limit.

Basically speaking, the Roche Limit defines the minimum distance that one body can orbit a larger body without getting torn apart by gravity. Our sun's Roche Limit is 556,397 kilometers. How far would the Roche Limit of a star 140% as massive as our sun be?

  • 4
    $\begingroup$ How to compute Roche Limit $\endgroup$ Oct 31, 2021 at 3:43
  • 1
    $\begingroup$ Reality check: Doesn't work. The short lifespan of big stars is because they're burning fuel so fast to put out all that energy. Turn down the flame and you turn down the brightness. A red dwarf burns all it's fuel, any star that lasts as long can be no brighter per unit of mass. $\endgroup$ Oct 31, 2021 at 3:47
  • 3
    $\begingroup$ @JohnWDailey I don't think it can be easier unless I give you it already computed. But I cannot because missing data. $\endgroup$ Oct 31, 2021 at 3:53
  • 2
    $\begingroup$ There's a lot of unneeded info in that question. The Roche limit of a White Dwarf Compact Star is exactly the same as the Roche limit of an undercooked lemon meringue of the same mass. ALL that matters is the mass of the gravity source, and the physical structural nature of the thing orbiting it. Follow the link Erdel von Mises gave, and the calculation is (mostly) trivial. $\endgroup$
    – PcMan
    Oct 31, 2021 at 6:15
  • 2
    $\begingroup$ @StarfishPrime it does not matter what the composition of the central body is, as long as its gravity field (and thus shape) is roughly spherical. The orbiting body's composition matters a great deal, as its physical resistance to deformation affects the Roche limit by a good amount. But the center body can be a black hole, or a neutron star, or a gas giant, or a rocky planet, as long as it packs the same mass into the effective point-source that is the mathematical center of its gravity. (obviously it cannot be allowed to extend out past the considered orbit height) $\endgroup$
    – PcMan
    Nov 1, 2021 at 15:36

2 Answers 2


We need more information to answer this question. To calculate the answer yourself, you can use this formula. $$ D = R_M(\frac{p_M}{p_m})^(1/3)$$ In this case the radius of the star, $R_M$ = 8919. But now you'll also need to know the densities of the dwarf star $p_M$ and its hypothetically orbiting planet $p_m$. The Roche limit changes depending on the density of the planet involved so you would need to calculate it again for every planet in your solar system.

Alternatively if you wanted to seem really accurate, you could decide the resonant frequency of your Sun based on its increased size, and then place the orbiting planets at points of resonant harmony along the orbital radius. For example, in our system the ratio of Earth's orbital period to Mars' orbital period is 3:2, and Venus:Earth is 8:5 which are nice ratios and go together like notes in a chord. All you need to do is figure out which key your sun is singing in and you can draw on music theory to support it at every step. Your readers won't be able to tell the difference because it will seem totally analogous to how our own skies move.

  • $\begingroup$ Note that pretty much rocky planets are rocky planets, you won't see big differences. $\endgroup$ Mar 6 at 6:29

The Roche Radius is not important.

enter image description here

The Ferrero Roche radius for an Earth-like planet orbiting your white dwarf is about 640,000 km. About a hundred times the radius of the Earth.

On the other hand, the habitable zone of a White dwarf is claimed to be about 0.01 Astronomical units (AU). One AU is about 150,000,000 km and is the distance from the Earth to the sun.

Your planet tears itself apart when within 640,000 km of the star. But all life on the planet is burnt to a crisp long before that, once you get within 1,500,000 km.

The big number is nearly three times larger than the small number. So don't worry about the Roche radius.

Instead you should worry about whether your nested stars-orbiting-stars-orbiting stars constellation has any chance to be stable in the long term. To my knowledge three body configurations are only stable if they are a Sun-Earth-Moon configuration. There are no stable orbits in a binary star system unless maybe the orbit is so large you can treat the two stars as one mass. And you have a lot more than three bodies.

If you are not too busy, something else to worry about is how long the larger stars last before burning themselves out. Big stars don't last as long you see.

If there is time left over in the day, you might worry about whether there is an atmospheric Roche radius which is larger than the normal Roche radius, and within which tidal forces tear off the gas around the planet without pulling the planet apart. This might lead to everything on the planet suffocating before burning to a crisp.

Estimate the Roche Radius $r$ using the formula

$$r = R_m\left(2\frac{M_M}{M_m} \right)^{1/3}$$

for $R_m$ the planet radius; $M_M$ the star mass; and $M_m$ the planet mass. The above formula is equivalent to the one Devio52 gave, only the numbers are easier to find online.

Since the planet is earth-like and you give the star mass we have

$$R_m \simeq 6400 \text{km} \simeq 6.4 \times 10^6 \text{m}$$

$$M_M \simeq 1.4 \times 2 \times 10^{30} \text{kg} = 2.8 \times 10^{30} \text{kg}$$

$$M_m \simeq 6 \times 10^{24} \text{kg}$$

So the Roche radius is

$$ R_m\left(2\frac{M_M}{M_m} \right)^{1/3} = 6.4 \times 10^6 \left(2\frac{2.8 \times 10^{30} }{6 \times 10^{24} } \right)^{1/3}$$

$$ \simeq 6.4 \times 10^6 \left(\frac{10^{30} }{ 10^{24} } \right)^{1/3} = 6.4 \times 10^6 \times 10^{6/3} $$

$$ = 6.4 \times 10^6 \times 10^2 = 6.4 \times 10^8 \text{metres}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .