What is the largest possible appearance of a celestial body in the sky?

Question

In SciFi Art we often see planets/moons/stars fill enormous amounts of the visible sky. I'm wondering what the theoretical limit for this is realistically, since there are maxima for the possible size of planets, for how close stable orbits can form, for how close you could orbit a given star, Roche Limit, etc...

--> Specifically, what is the maximum size in degrees/arc seconds at which a celestial body could appear in the sky for an observer on the surface? And what would that observed celestial body be? A very large moon? A twin-planet arrangement? Something else?

Assumptions:

• Let's assume that the observer stands on some celestial body (and they would like to survive).
• Body size/mass is such that it could be (made) habitable for humans.
• Proximity to the bodies star is such that the temperature can be manageable for humans given approx. current level of technology.
• The constellations of celestial bodies should be possible according to current science and stable at least in the short term.

E.g. standing on a sizeable moon looking up at the planet does count. Looking up at an incoming rogue planet 10 meters before impact obviously doesn't, since that arrangement is neither stable nor survivable.

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Inspired by these questions...

• What could be the largest geosynchronous moon?
• Could We See a Lagrange Giant in the Sky?

Answer 1

First thing to remember is that when you're close to a spherical object, you can only see a portion of its surface (the horizon). Therefore the anglular diamater of some body with radius $r$ viewed from a distance from the surface of $h$ will be $2\sin^{-1}\frac{r}{r + h}$.

(this also means that the maximum angular diameter of a celestial object (such as the galaxy you are in) for someone standing on some moon or planet is going to be greater than 180° unless that person has zero height, but that's not a useful answer to your question)

Clearly this tends to a maximum of 180° as $h$ tends to zero, so need to find the biggest thing you can and orbit it as low down as possible. The biggest object currently viewable by humans in something with a short-term stable orbit is the Earth, as viewed from the ISS. With an altitude of 408 km and an equatorial radius of 6378.1 km, the angular diameter of the Earth is 140°3'42.68'', and whilst the ISS is habitable you can't really stand on it as such, so it isn't quite an answer.

I'm not going to consider the Sun, as it is just a bit too hazardous.

The biggest thing that you could reasonably see in the Solar System from an ISS-like orbit would be Jupiter. You could imagine parking a suitable body just above Jupiter's exobase. Earth's exobase doesn't generally fall below 500km, so by being just above it our theoretical moon will have a more stable orbit than the ISS. Jupiter's exobase is at ~2300km. At this altitude, Jupiter will have an angular diameter of 151°18'52.44'', bigger even that Earth as seen from the ISS.

The Roche limit for a your moon is largely determined by densities and the size of the body it orbits: $l_R = R_J \sqrt[3]{\frac{2\rho_J}{\rho_m}}$ where $R_J$ is Jupiter's radius, $\rho_J$ is Jupiter's density and $\rho_m$ is the required density of the moon. This means that anything that's more than twice the density of Jupiter will have a Roche limit that's less than Jupiter's own radius! Jupiter, being a gas giant has a pretty low density of ~1326km/m³. Jupiter's own icy moons all have lower densities than this, but if your theoretical low-altitude moon were even as dense as the Moon (3346km/m³) it would not exceed the Roche limit!

It does not appear that anyone has considered any other lower limit on moon orbital radius other than the Roche limit, so my suggestion isn't currently known to be implausible (Limits on the orbits and masses of moons around currently-known transiting exoplanets).

Note that this is now right on the edge of plausibility. Long term stability of the moon's orbit is dubious, and were Jupiter close enough to the Sun for the moon to be within the habitable zone the additional heat input from solar irradiance would warm the gas giant and inflate its atmosphere, and your low-altitude moon would likely be dragged down into the atmosphere in a relatively short period of time. Because the precise relationship between atmospheric height of a gas giant and surface temperature isn't trivial to pin down, this could be handwaved as having a less-massy giant than Jupiter which has expanded to the same size in the heat... this would of course affect things like orbital period, but as you've not asked about that in your question it does not need to affect the answer in the slightest.

It does however put a fairly reasonable upper bound on the apparent size of an object in the sky. It may be possible to fit in something a tiny bit bigger, but I'm having trouble finding a plausible brown dwarf/gas giant cutoff which has implications for radiative heating of your satellite. I'll update in future if I find something bigger, but for now Jupiter is a reasonable proxy for a very large non-fusing gas giant.

Here's a simple chart of angular diameter vs altitude. The orange circle marks Metis, which I believe has the biggest object in its sky of any known body in the solar system, and with its surface gravity of half a milligee you could conceivably stand on it. You'd have to attach a habitat to it in order to live there, though. At its perijove of 127974 km (equivalent to an altitude of 56482 km) Jupiter has an angular diameter of 67°55'27.29''. This marks a good proxy for the best achievable view, and one likely to remain stable even when the parent planet inflates if bought closer to the Sun.

This issue of habitability is a separate one that has been covered in many, many other questions and answers here passim ad nauseam so I shall not be going into detail there. Suffice it to say, your atmosphere scraping moon could be habitable.

• It will be tidally locked, but reflected light and radiant heat from the has giant will keep the planet-side warm and light, and with a day length of a little over 3 hours there's little risk of the non-planet side getting cold at night, eliminating the usual problems with tidally locked worlds.
• It will be well within its parent's magnetosphere, so it does not need to generate one of its own to preserve its atmosphere.
• In the absense of gravitational influence from other large moons around the gas giant you can expect its orbit to be quite circular, which means it will not suffer from excessive tidal heating like Io, which has orbital resonances with other Galilean moons which in turn drive its massive tidal heating and vulcanism.
• Without Io-like vulcanism, there's no plasma torus around the gas giant, rendering space around it much less hostile than Jupiter.

All that needs to be done is to keep the exobase of the moon's atmosphere clear of the gas giant's exobase, to prevent any drag or atmosphere-stripping problems.

Here's a slightly different diagram from the usual chart for atmospheric escape, just for variety. You can see that Mars is right at the lower limit for the Mass required to retain an N₂ atmosphere. It is also dense enough, at ~3900 kg/m³ to be safe from destruction by tidal forces. Unlike the real-life Mars, a Mars moon snuggled up to a gas giant is safe in its parent's protective magnetosphere and so should hang on to much more gas for much longer.

The scale height of a Martian atmosphere with an Earthlike average atmospheric temperature of 287K is defined as $H = \frac{k_BT}{mg}$ where $k_B$ is Boltzmann's constant, $T$ is 287K, $m$ is the mass per molecule of the atmosphere (4.65x10^-26 kg for N₂) and $g$ is the surface acceleration due to gravity, which for Mars is ~3.7m/s². $H$ is therefore ~22.9 km, not far off Jupiter's own. Earth's scale height is about 8.5 km, and its exobase is ~500 km. Given the same proportions, the Mars-moon's exobase would start at ~58.8 times its scale height, or ~1350 km. In practise you could get away with a lower surface pressure than Earth for a reduced exobase altitude, but this is an adequate start.

Your minimum altitude therefore becomes 2300 km (for the parent's exobase) + 3400 km (for the moon's radius) + 1350 km (for the moon's exobase) or ~7045 km for the barycenter, or merely 2300 km + 1350 km for the closest sub-Jovian point, giving still quite dramatic maximum Jovian angular diameter of 131°.

Answer 2

Let's break it down by type of system, as planet-moon or planet-planet systems will have different constraints from star-planet systems.

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Planet-moon/planet-planet system

Looking at planet-moon or binary planet systems, the answer to your question will always involve standing on a decent sized moon and looking up at the host planet. A moon by definition has the centre of mass (barycentre) of its orbit inside the host planet, thus one cannot make it arbitrarily heavy. If we want a stable system, we need to be at a certain minimum distance (the Roche limit) and thus if we want the largest angular size, for which we want the closest possible distance, we need to increase the size and thus mass of the object we are looking at. For the largest angular size we will look at the most extreme case of a system where the observer stands on a moon orbiting around a giant planet.

To make calculations a bit more simple and not leave it questionable whether the place is habitable, let's assume the observer is standing on an Earth-analogue.

The (rigid body) Roche limit gives the radius at which the gravitational pull of the second body equals that of the primary, thus providing the minimum distance at which a body can orbit before being torn apart by the tidal forces.

$$d=R_2\left(2\frac{M_1}{M_2}\right)^{\frac13}$$

Setting $R_2$ and $M_2$ to Earths values shows that our minimum distance scales with the cube root of the mass of the primary. Of course the habitability of such a system becomes questionable because the gravity felt standing on the moon will be varying hugely depending on where you stand on it and could go to zero. Plus we haven't looked at the effect of tidal heating, which occurs when the tidal force of the host is strong enough to kind of knead the moon and thus warm it up by friction. You could of course use this to your advantage and let it counteract the fact that you are much farther removed from the star than Earth is.

The angular size is obtained from the radius of the primary and the distance with:

$$\delta=\arctan\left(\frac{R_1}d\right)$$

Gas giants like Jupiter don't behave quite like your average rocky planet anymore in the sense that the radius doesn't increase very much with mass. The gas will compress and settle into hydrostatic equilibrium if more mass is added. If we look at the extreme end we get to the limit of brown dwarf stars. This is defined at a mass of $13 M_{\text{Jupiter}}$, but the radius will not be very different from $1 R_{\text{Jupiter}}$. To quote Wikipedia:

Brown dwarfs are all roughly the same radius as Jupiter. At the high end of their mass range (60–90 MJ), the volume of a brown dwarf is governed primarily by electron-degeneracy pressure,[26] as it is in white dwarfs; at the low end of the range (10 MJ), their volume is governed primarily by Coulomb pressure, as it is in planets. The net result is that the radii of brown dwarfs vary by only 10–15% over the range of possible masses.

The larger radii that you might have read about of some exoplanets is due to their much hotter temperature, since these reside very close to their sun. They have their name 'hot Jupiter' for a reason. So I will not consider those and just take the size of Jupiter as our maximum planet size. We have seen that the minimum distance increases with mass, so taking a more massive planet is actually disadvantageous. Taking mass away beyond a certain point will decrease our radius again, so let's take $M_{\text{Jupiter}}$ as our realistic extreme case.

Filling all of this into the relevant formulas gives an angular size of 374000 arc seconds or 104 degrees. But you will be orbiting at a distance of 8.6 Earth radii or just 0.78 Jupiter radii!

Let's say we double that orbital distance so that we feel a bit safer, we end up with a size of 65 degrees. You might still feel a bit lighter in some places on your moon though...

If that still feels too iffy in terms of realism, we can put ourselves at a distance of $6 R_{\text{Jupiter}}$ (roughly how far Io orbits around Jupiter). Now Jupiter still only has an angular size of about 19 degrees (our moon is 0.5 degrees, so this is still quite big! :D).

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Star-planet system

With stars we have to consider the extra constraint of flux received by a planet. The Stefan-Boltzmann law tells us that the flux emitted by a black body (=star to good approximation) scales with the temperature to the fourth. If we don't want to get too toasty, we better choose a star that is cold (for stellar standards). The total power emitted by a star is the flux times the emitting area. To calculate the flux at Earth (or at any distance) we simply divide this power by the total surface area of a sphere centered at the star and with the radius equal to the distance of the planet.

Red giants are the perfect candidate here: they can be as cool as 3000 Kelvin (on the surface) and are around 200 solar radii so at first thought that might maximize the angular size.

If we would take Earth as an example of a cozy amount of flux, we can combine these formulas to scale the flux to the new temperature and distance while dropping all constants:

$$P\propto A_R\cdot T^4\propto R^2\cdot T^4\implies F=\frac P{A_d}\propto\frac{R^2\cdot T^4}{d^2}$$

so if we set $F_1 = F_{\text{Earth}} = F_2$: $$\frac{R_1^2\cdot T_1^4}{d_1^2}=\frac{R_2^2\cdot T_2^4}{d_2^2}$$

We solve for the new distance that our planet would be at to receive the same flux from a red giant: this puts us at about 54(!) times the distance Earth-Sun. Turns out a Red giant generates a lot more flux even though it is a lot cooler (which is why Earth would be incinerated when the Sun goes red giant). We simply use the formula for angular size again with 200 times the solar radius to arrive at an angular size of: 7122 arcsec or ~2 degrees. This is still better than our Sun at ~0.5 degrees!

What would have happened if we took a small cold star? If we take a quote from Wikipedia again:

The coolest red dwarfs near the Sun have a surface temperature of ~2,000 K and the smallest have radii of ~9% that of the Sun

Using that in our equations yields a distance of merely 0.01 times the distance Earth-Sun and with angular size: 16000 arcsec or 4.4 degrees. We actually did better! Keep in mind that tidal locking might be a problem now though, due to the very close distance to the star, so again might affect habitability (not to speak of the violent flares that can occur in these types of stars).

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A closing remark:

It was a fun exercise to actually run the numbers on this, and even I hadn't realised that looking from Io, Jupiter looks quite that big! Angular size falls off really fast with distance as we saw. Distances in the universe are immense, and most celestial bodies do not actually reside very close to each other, or if they do, they are not very hospitable. This makes such beautiful big-planet-in-the-sky scenarios actually quite unlikely unfortunately.

Answer 3

Short Answer:

Many of those images might be designed to show astronomical objects though imaginary telephoto lenses. According to my reserches, the largest possible astronomical object as viewed from the surface of a habitable or potentially habitable world would have an angular diameter of 22.9183 degrees of arc, or about 82,505.88 arc seconds. Except that Part three of my aswer to: How would the reflection of sunlight change the day / night cycle of a binary planet system? Discusses another type of astronomical situation were two astronomical objects can have very large angular diameters in each other's skies. Possibly the largest angular diameters which are possible.

Long Answer in Six Parts:

I have often wondered about the subject.

Part One: Telephoto Lenses.

I note that humans have invented something called telephoto lenses for still and motion cameras.

Thus you should have often seen still photographs, or scens in movies & Tv shows where Earth's Moon appears to be vast and cover most of the sky. And Earth's Moon does cover most of the tiny angle of the sky which is visible in those scenes shot with telephoto lenses.

And we have all seen scenes in science fiction films where the moon appears to have an enormus angular diameter. LIke in E.T.: The Extraterrestrial where a bicycle flies across the face of the Moon that appears to be enormous. Those scenes in science fction movies aren't designed to show the Moon with a much greater angular diameter than it actually has, they are designed to show what the scene would look like photographed thorugh a telephoto lens that makes the moon look much larger.

And similarly many illustrations of alien worlds and movie and tv scenes show stars, or planets, or moons near the horizon that appear to have a vast angular diameter. In some cases the scenes may have been designed to show what someone would see without any magnification, in others they may have been designed to show what it would look like thorough a telphoto lens making the objects in the sky loook much larger, and in many cases nobody thought about whether the apparent angular diameter would be the actual angular diameter as seen with the naked eye.

Part Two: The Angular Size of Stars.

A small difference in the mass of main sequence stars will cause a much larger difference in their luminosities. A small increase in mass will cause a larger increase in luminosity. A small decrease in mass will cause a larger decrease in luminosity.

More massive stars also emit more light per unit of their sufraces. Less massive stars emit less light per unit of their surfaces.

So a planet that has the sem temperature as Earth orbiting a more massive and brighter star will have to orbit it farther than the distance at which the star it would have the same angular diameter as the Sun. A planet with the same temperature as Earth orbitng a less massive and less luminous star would have to orbit it closer than the distance at which the star would have the same angular diameter as the Sun.

Astronomers have calculated the sizes of the circumstellar habitable zones of main sequence stars of various spectral types. All one has to do is find the luminosity of the star comparared to that of the Sun and then multiply the inner and outer edges of the Sun's circumstellar habitable zone by that ratio.

Unfortunately, different calculations of the inner and outer edges of the Sun's circumstellar habitable zone vary greatly, as this list shows:

https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates[1]

A writer who wants to have several habitable planets orbiting the same star will have to see if he can fit them in within a habitable zone that seems reasonable and plausible to him. A writer who wants one and only one habitable planet in his star system can play it safe and calculate the Earth Equivalent Orbit for that star, an orbit where a planet would receive exactly as much radiation from its star as Earth receives from the Sun. Thus he would know that a planet in that orbit could be habitable.

The answer by user177107 to the queston:

https://astronomy.stackexchange.com/questions/40746/how-would-the-characteristics-of-a-habitable-planet-change-with-stars-of-differe%5B2%5D[2]

Has a table listing main sequence stars of various spectral types.

For each spectral type of star the mass, radius (half of the diameter), luminosity, etc. is listed, as well as the distance a planet would have to orbit to received exactly as much radiation as the Earth receives from the Sun.

The smallest type of star listed is spectral type M8V, with a mass of about 0.082 the mass of the sun, and a diameter of about 0.111 of the Sun, and a planet in an Earth Equivalent Orbit would orbit it at a distance of 0.0207 of an Astronomical UNit (AU), the distance of Earth from the Sun.

Because the orbit of the Earth is elliptical, the angular diameter of the Sun as seen from earth varies between 31.6 and 32.7 arc minutes, or 1,896 to 1,962 arc seconds. Since there are 60 arc minutes in a degree of arc, we can simplify the calculation by assuming that the Sun has an average angular diameter of 0.5 degrees of arc, or about 1,800 arc seconds.

Since a planet in an Earth Equivalent Orbit around a M8V star would beonly 0.0207 as far as Earth is From the Sun, the angular diameter of the Sun would have to be multiplied by 1 divided by 0.0207, or by 48.309, to get an angular diameter of 24.1545 degrees, and then divided by 1 divided by 0.111, or 9.009, the ratio of the diameters of the two stars, to get an angular diameter of 2.681 degrees, or about 9,651.6 arc seconds,.

So a M8V star would look about 5.3 times as wide seen from a planet in an Earth equivalent orbit as the Sun looks from Earth. Which would be quite noticable for a human on the planet, but probably not too spectacular.

Of course, if the planet orbits a bit closer to its star than the Earth Equivalent Orbit, the star would appear to have a larger angular diameter.

According to the estimate by Kopparapu et al in 2013, the inner edge of the Sun's circumstellar habitable zone is 0.99 AU, or only 0.01 AU closer than the Sun's orbit. However, Zsom et al in 2013 estimated that the inner edge of the Sun's habitable zone could be ast 0.38 AU. Their estimate is probably for a planet with atmospheric condidtions suitable for some forms of Earth-like life and not for humans, who might, however, be able to live on such a planet with protective gear.

At such a distance around a M8V star, the star would appear to have an Angular diameter of about 7.055 degress or about 14.11 times teh angular diameter of the Sun as seen from Earth.

The closer a planet orbits to its star, the stronger the gravity and tides of the star will be upon the planet. Because the luminosity of low mass stars decreases faster than their mass, planets in the circumstellar habitable zones of low mass stars will experience intense gravity and tides, which will tend to slow down the rotation of those planets.

So as the stellar habitable zones around stars get smaller and smaller for less and less massive stars, eventually the inner edges of their cicumstellar habitable zones reach the distance at which a planet would swiftly become tidally locked to its star. With stars of even less mass, the Earth equivalent orbit will eachethe distance at which the planet will become tidally locked to the star. With stars of even less mass than that, the outer edges of their habitable zones will reach the distance at which a plaent become tidally locked and it will be impossible for any planet witin their habitable zones. to avoid being tidally locked.

Stephen H. Dole in Habitable Planets for Man, believed that a tidally locked planet would be unhabitable. He also calculatled that at 0.88 the mass of the Sun, the inner edge of the circumstellar habitable zone, or "ecosphere" as he called it, would reach the distance where a planet would be tidally locked, and at a stellar mass of 0.72 that of the Sun, the outer edge of the habitable zone would reach the distance at which a planet would be tidally locked.

https://www.rand.org/content/dam/rand/pubs/commercial_books/2007/RAND_CB179-1.pdf[3]

Dole also noticed a potential escape clause, where a planet which became tidally locked to a large moon, or a companion planet, could avoid becoming tidally locked to it's star. Dole estimated that could enable habitable palnets to exist n the habitable zones of stars down to a stellar mass of 0.35 of the Sun, before the sun would raise too high tides on the planet.

O.35 the mass of the Sun would be somewhere between a M5V and a M2V star; 0.72 the mass of the Sun is approximately a K1V star; and 0.88 the mass of eh Sun is approximately a G9V star.

But there have been calculations indicating that a tidally locked planet could be habitable if it had enough atmosphere and water to distribute heat from the day side to the night side. This is rather controversial.

Part Three: The Angular sizes of Brown Dwarfs.

Brown dwarfs are objects with masses greater than about 13 times that of Jupiter and up to about 75 to 80 times that of Jupiter. Brown dwarfs are intermediate in mass and conditins between planets and stars, and could be orbited by natural satellites, perhaps large enough for life.

Considering how dim brown dwarfs are, any theoeretical habitable moons or planets or whatever obiting them would have to orbit very close. Thus it is possible that A brown dwarf would appear several times as wide in the sky of a habitable world orbiting it as any star appers in the sky of its habitable planets. And possibly humans in space suits or lesser environmental protection could work on the surfaces of nonhabitable worlds with large brown dwarfs in the sky.

I am unable to clculate the maximum possible angular diameter of a brown dwarf as seen from a world which humans might want to land on.

Part four: Planets With Neighboring Orbits.

The star TRAPPIST-1 is a dim M8V class star, noted for having several potentially habitable planets orbiting in its circumstellar habitable zone.

The orbits of the TRAPPIST-1 planetary system are very flat and compact. All seven of TRAPPIST-1's planets orbit much closer than Mercury orbits the Sun. Except for b, they orbit farther than the Galilean satellites do around Jupiter,[43] but closer than most of the other moons of Jupiter. The distance between the orbits of b and c is only 1.6 times the distance between the Earth and the Moon. The planets should appear prominently in each other's skies, in some cases appearing several times larger than the Moon appears from Earth.[42] A year on the closest planet passes in only 1.5 Earth days, while the seventh planet's year passes in only 18.8 days.[40][37]

https://en.wikipedia.org/wiki/TRAPPIST-1#Planetary_system[4]

The potentially habitable palnets of TRAPPIST-1 are d, e, f. and g. They could be habitable, and maybe if not they could be terraformed to be habitable. Anyway, habitable or not, people should be able to walk around on them in spacesuits and possibly less evironmental protection.

The semi-major axis of the orbit of the Moon is about 384,399 kilometers, or about 0.002569 AU. At that distance Earth has an angular diameter of approximately 2 degrees of arc or 7,200 arc seconds.

At their nearest, TRAPPIST-1 f and TRAPPIST-1 g are about 0.00834 AU apart, which is about 3.246 times the distance between the Earth and the Moon. So if those planets had the same diameter as the Earth, they would appear to have an angular diameter of 0.616 degrees or 2,218.114 arc seconds. They are both larger than the Earth. The larger, TRAPPIST-1 g, has a diameter 1.129 that of the Earth, and so would have an angular diameter of 0.695 degrees or 2,504.251 arc seconds when seen from TRAPPIST-1 f at its closest.

The innermost potentially habitable planet, TRAPPIST-1 d, orbits only 0.00647 AU beyond the orbit of TRAPPIST-1 c, which is 2.5184 times the Earth-Moon distance, which would make TRAPPIST-1 c look up to 0.7941 degrees wide from TRAPPIST-1 d, except that TRAPPIST-1 c has 1.308 times the diameter of Earth. So TRAPPIST-1 c would have an angular diameter of about 1.0387 degrees of arc of 3,739.517 arc seconds when the two planets were closest.

The absolute maximum possible angular diameter of a planet seen from a habitable or potentially terraformable planet in a neighboring orbit should be at least that much.

The best is yet to come when I complete this answer.

Part four: The Ultimate retorgrade solar System.

The planetPlant blog has a section called the Ultimate solar system, with attempts to design imaginary solar systems with as many habitable planets as possible.

The post called "The Ultimate Retrograde Solar System" cites a paper demonstrating that more planetary orbits can fit inside the circumstellar habitable zone of a star if half of the planets orbit in a backwards or retrograde direction compared to the other planets.

It makes a big difference. You can fit about twice as many planets into a given stretch of orbital real estate. The requirement is simply that every other planet must orbit in the opposite direction. So, planets 1, 3, 5, and 7 orbit the star in a clockwise direction, and planets 2, 4, 6, and 8 orbit counter-clockwise.

Take Earth-mass planets orbiting a star like the Sun. On prograde orbits, 4 Earths fit within the habitable zone. For alternating prograde and retrograde orbits, 8 Earths fit.

https://planetplanet.net/2017/05/01/the-ultimate-retrograde-solar-system/[5]

So depending on how close the TRAPPIST-1 system comes to being as closely packed as posssible, a retrograde TRAPPIST-1 system could have 7, 8, or 9 planets orbiting within the circumstelllar H habiable zone e if each palnet orbited in the opposite diretion to the planets in the next inner and outer orbits to it.

So possibly a retrograde solar system could have planets which can have an angular diameter of about 2 degrees of arch, or 7,200 arc seconds, when seen from potentially habitable planets in the circumstellar habitable zone.

But:

With the Retrograde Ultimate Solar System we are now swimming in impossible waters. Two planets can end up orbiting the same star in opposite directions, but only if their orbits are widely separated. I don’t know of any way that nature could produce a system of tightly-packed planets with each set of planets orbiting in the exact opposite direction of its immediate neighbors.

This means that the Ultimate Retrograde Solar System would have to be engineered. Created on purpose by some very intelligent and powerful beings.

Writers who use such a solar system will have to show that it must be artificial, created by a very advanced society.

Part Five: Giant Planets Seen from their Habitable Moons.

It should be obvious that a moon could orbit a large enough planet closely enough that the planet would almost fill an entire hemisphere of the sky of the moon. But unfortunately the question asks for the viewing to be from worlds shich are habitable or which could be terraformed to become habitable. Habitable moons are not very common.

It is possible that a planetary mass exomoon could orbit a giant exoplanet and be habitable. But such an exommon would have to meet the requirements for an exoplanet to be habitable, such has having a mass in the proper range. And the exomooon would also have to meet special requirements due to being a moon of a giant planet.

For example, there is the concept of a habitable edge, a minimum distance from a planet that a potentially habitable exomoon would have to be in order to avoid too much tidal heating from the planet. Too much tidal heating would cause the exomoon to suffer a runaway greenhouse effect and become lifeless.

Rene Heller and Jorge Zuluaga, in Magnetic Shelding of Exomoons Beyond the Planetary Habitable Edge" discuss a maximum possible distance of a habitable exomoon from its planet.

https://arxiv.org/pdf/1309.0811.pdf[6]

They discuss exomoons whch are large enough for life but significantly smaller than Earth, and which thus wouldn't have their own magnetic fields to shield them from high energy particles from space. Those moons would have to be within the magnetic fields of their planets and be shielded by those planetary magnetic shields.

They believe that there is no safe zone around Neptune sized planets. Their exomoons would orbit either within the habitable edge and suffer runaway greenhouse effects, or else orbit outside the protection of planetary magnetic fields.

But moons around Jupiter sized planets could be habitable in orbits at distances between 5 and 20 planetary radii, or 2.5 to 10 planetary diameters.

According to my rough calculations, an exomoon at the outer edge, or about 20 planetary radii, would see the planet as having an angular diameter of about 5.7295 degrees of arc, or 20,626.2 arc seconds.

And according to my rough calculations, an exomoon at the inner edge, or about 5 planetary radii, would see the planet as having an angular diameter of about 22.9183 degrees of arc, or about 82,505.88 arc seconds.

Part Six: A large Moon Seen from its Planet.

The planet Earth is habitable for humans at the present, and and has a lorge Moon, which at the present time orbits the Earth in an elliptical orbit with a semi-major axis of 384,399 kilometers.

At that distance the Moon has an angular diameter of 29.3 to 34.1 arc minutes, or 1,758 to2,046 arc seconds, depending on its current distance from Earth.

But this was not always the case. Earth only acquired enough oxygen in the atmosphere to be brethable for humans a few hundred million years ago, after existing for billions of years.

And the Moon was not always at its present distance from Earth.

According to present theory, the Moon formed after Earth collided with another planet, called Theia, over four and a half billion years ago. Most of the mass of the two planets merged, much of it was ejected into space and lost, and part of it formed a ring around Earth, which eventually formed the Moon. tidal interactions between the Earth and the Moon caused the Moon to gradually move to a higher and higher orbit over billions of years.

When the Moon formed out of the ring it was abourt 4 Earth radii or 15,000 to 20,000 miles from Earth. It would have appear 15 times as wide as it now appears, and it would have been glowing dull red from the hot lava it was made of. So the Moon would have had an angular diameter of about 7.5 degrees or about 27,000 arc seconds.

About four billion years ago, after 500 million years, the Moon would have moved to an orbit about 80,000 miles from Earth and would have had about three tiems its present angular diameter, and so about 1.5 degreees of arc or about 5,400 arc seconds.

https://www.forbes.com/sites/quora/2018/07/11/what-did-the-moon-look-like-from-earth-4-billion-years-ago/?sh=1a465ddf1151[7]

And I presume that hypothetical aliens who found a similar system of a young planet with a young moon receeding from it could have decided to terraform theplanet to make it habitable at a point where the moon was a lot closer than 80,000 miles and had a much greater angular diameter than 1.5 degrees or 5,400arc seconds.

And I suppose that sometime the future humans could find such a young planet with a moon that appears very large in its sky, and decide to terraform that planet to make it habiable for humans.

Conclusion:

As I said in part one, many of those images of astronomical objects looking very large in the sky could be depicting how large those objects would look seen through a telephoto lens, instead of how large they would appear with the naked eye.

The greatest angular diameter of an astronomical object as seen from the surface of a habitable or terraformable astronomical object I could find would be a giant planet viewed from the surface of a large habitable or potentially habitable moon, with an angular diameter of up to 22.9183 degrees of arc, or about 82,505.88 arc seconds. Except that Part three of my aswer to: How would the reflection of sunlight change the day / night cycle of a binary planet system? discusses another type of astronomical situation were two astronomical objects can have very large angular diameters in each other's skies. Possibly the largest angular diameters which are possible.