In my world, there are 3 astrological bodies of significance, the sun, the moon, and a 3rd, geostationary object, believed to be the home of the gods. I want all three of them to have roughly the same size perspective (angular diameter), such as we have on earth. This is to give an awesome looking eclipse which happens once every very long time, for plot reasons. The reason the sun and the moon appear to be the same size is that the sun is 400 times more massive than the moon, but is also 400 times further away. With a bit of calculations, and internet searches, I have the following data;

Sun distance - 149.6 Million Km Sun size (Radius) - 695510 Km Suns angular diameter - 31'27'' to 32'32'' arc.

Moon distance - 363,104 to 405,696 km Moon size - 1737.1 km Moon Angular Diameter - 29'20'' to 34'6'' arc.

Geostationary orbit distance - 35786 km,

using average moon dist/Geostationary dist - 384391 km / 35786 km gives me 10,74, and dividing moon size by this gives me a total radius of Geostationary object radius - 161,72 Km. then, Using the formula x= 2arctan(d/2D) where x is radians, d is diameter of object and D is the distance between observer and object. x=2arctan(323,44/71572) gives me a total Angular Diameter of - 31'07'' of arc.

SO What I want to find out, 1. Would this setup be stable in the long term, and if it is then 2. Would this object be large enough to achieve hydro-static equilibrium.

I have found that Saturns moon Mimas, at 397 km, appears to be, but is not in hydro-static equilibrium, while the largest non equilibrium object is the rocky asteroid Vesta at 525 km.

If needed, I would make the object a watery, icy object, but then I think I will run into problems with it melting, but thats for another time.

The earth, moon and sun are as is in reality, Earth weight is 5.972 × 10^24 kg, Moon weight is 7.35 x 1022 kg,

  • $\begingroup$ What are mass and radius of the planet? $\endgroup$
    – L.Dutch
    Dec 9, 2019 at 16:24
  • $\begingroup$ @L.Dutch masses added in, all as in real earth situation. Earth radius is 6371 km $\endgroup$
    – Umbra
    Dec 9, 2019 at 16:50

2 Answers 2


To have the same angular size as the Moon at a bit less than a tenth the distance would require your stationary moonlet to be a bit less than a tenth the diameter -- that would be close to 300 km. Such a satellite, in geostationary orbit, wouldn't have an appreciable effect on the Moon's orbit because, if of similar composition, it would have less than 1% of the Moon's mass; you could easily handwave the Moon keeping its same orbit in the presence of such a small body at such a distance.

However, one of the issues with geostationary communication satellites is that, for most arbitrarily chosen points in the orbit, they won't actually remain stationary. Perturbation by the Moon, Sun, and Jupiter are part of the cause, but the Earth's non-homogeneous crust is the bigger one. The result is that there are a limited number of equatorial locations where a "stationary" moonlet would actually remain stationary.

The good news is that the tendency is for satellites to drift into those more stable locations, and thereafter just wobble a bit (barring active stabilization and station keeping, of course). If in fact the gods do live on this moonlet, it's quite reasonable for them to have chosen such a location and stabilized their home in it, making it really stationary relative to the Earth's surface.

A rocky body only 300 km diameter generally won't be spheroidal due only to gravity -- but there are ways to help it along. If the body was melted through at some time in the past, even for only a few years, it would have assumed a spheroid -- or if it's composed of small enough particles, poorly enough bound, and has a low spin rate and impact rate it would tend toward the spherical. Note that this latter condition would have to be artificial, else vacuum welding would prevent the easy "flow" of particles.

As you note, there isn't a sensible, natural-seeming way to keep a body made mainly of water ice frozen in the Earth's "climate" -- we get too much sunlight. Comets start to sublimate as far out as Jupiter's orbit. However, past near-complete melting is very possible for a body formed by accretion, or by ejection from a larger melted body (similar to our Moon's formation, according to current hypothesis). A "leftover" from a Theia impact event would be a very plausible way to create a rocky, spherical moonlet of the size you want.

  • $\begingroup$ if it where to be a body of approximately 75-80% ice, and leaving a layer about 30-40 Km of very cold, slightly slushy water on the surface, it should form a rough sphere, but would that leave it vulnerable to being ripped apart by the earths gravity? $\endgroup$
    – Umbra
    Dec 9, 2019 at 18:39
  • $\begingroup$ Honestly, I don't have the ability to calculate that (at least while at work). You should be able to calculate the difference in Earth's gravity from one side to the other without much difficulty, but knowing the object's surface gravity would also be necessary to determine its Roche limit. You'd still have an actively sublimating body at Earth's orbit, though -- black body temp is above triple point of water, as I recall. $\endgroup$
    – Zeiss Ikon
    Dec 9, 2019 at 19:10

This is just another way of asking the n body problem (n=4)

The n body problem is generally considered unstable for n=3 and larger. There are special cases of stability, your situation would certainly require a computer simulation in order to analyse it.

  • $\begingroup$ Balderdash. As long as all central bodies are far heavier than their satellites, such systems are generally stable. The solar system is full of examples. $\endgroup$
    – Karl
    Dec 9, 2019 at 23:11

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