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edit approved Dec 21, 2021 at 15:51

Planet-moon/planet-planet system

Planet-moon/planet-planet system

$\ d=R_{2}\left(2{\frac {M_{1}}{M_{2}}}\right)^{\frac {1}{3}}}$ $$d=R_2\left(2\frac{M_1}{M_2}\right)^{\frac13}$$

Setting R_2$R_2$ and M_2$M_2$ 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.

$\ \delta =2\arctan \left({\frac {R_1}{d}}\right)$ $$\delta=\arctan\left(\frac{R_1}d\right)$$

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_Jupiter$13 M_{\text{Jupiter}}$, but the radius will not be very different from 1 R_Jupiter$1 R_{\text{Jupiter}}$. To quote Wikipedia:

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_Jupiter$M_{\text{Jupiter}}$ as our realistic extreme case.

If that still feels too iffy in terms of realism, we can put ourselves at a distance of 6 R_Jupiter$6 R_{\text{Jupiter}}$ (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).

Star-planet system

Star-planet system

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

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

Planet-moon/planet-planet system

$\ d=R_{2}\left(2{\frac {M_{1}}{M_{2}}}\right)^{\frac {1}{3}}}$

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.

$\ \delta =2\arctan \left({\frac {R_1}{d}}\right)$

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_Jupiter, but the radius will not be very different from 1 R_Jupiter. To quote Wikipedia:

If that still feels too iffy in terms of realism, we can put ourselves at a distance of 6 R_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).

Star-planet system

$\ P \propto A_R*T^4 \propto R^2*T^4$ , so: $\ F = \frac{P}{A_d} \propto \frac{R^2*T^4}{d^2}$

so if we set F_1 = F_Earth = F_2: $\ \frac{R_1^2*T_1^4}{d_1^2} = \frac{R_2^2*T_2^4}{d_2^2}$

Planet-moon/planet-planet system

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

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.

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

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:

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.

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).

Star-planet system

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

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

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created Mar 30, 2021 at 18:15

L. IJspeert

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

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.

$\ d=R_{2}\left(2{\frac {M_{1}}{M_{2}}}\right)^{\frac {1}{3}}}$

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

$\ \delta =2\arctan \left({\frac {R_1}{d}}\right)$

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_Jupiter, but the radius will not be very different from 1 R_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.

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...

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:

$\ P \propto A_R*T^4 \propto R^2*T^4$ , so: $\ F = \frac{P}{A_d} \propto \frac{R^2*T^4}{d^2}$

so if we set F_1 = F_Earth = F_2: $\ \frac{R_1^2*T_1^4}{d_1^2} = \frac{R_2^2*T_2^4}{d_2^2}$

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).

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.