In a lot of media one can see moons or other planets that cover a significant portion of the sky (like the background on this page for example). This could be because the moon is actually large or because it's close. Presumably, though, if a moon is too close or too large it will cause catastrophic gravity related destruction.

I'm wondering if it would be possible to have a planet that's more or less exactly like Earth, whose moon would appear to be much larger in the sky? Can it be, say 90 angular minutes without coming within Roche limit or causing deadly tidal effects?

  • $\begingroup$ Are you specifically talking about Earth and its moon? What do you count as "gravity makes its toll" and yes, its going to have something to do with the composition of the moon, density dictates your apparent size vs gravitational effect. $\endgroup$ Feb 9, 2018 at 15:56
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    $\begingroup$ I think that 'gravity takes its toll' is referring to either the Roche limit (too close) or to binary orbits (like pluto and Charon.) $\endgroup$ Feb 9, 2018 at 16:09
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    $\begingroup$ This question might be better asked at Astronomy.SE or Physics.SE. However, you need to be a lot more specific. An object the size of a softball will be invisible until it's so deep inside the planet's gravity well that it will definitely fall to the ground. On the other hand, living on a gas giant moon will result in the gas giant periodically filling the entire sky. So, you must either (a) specify the size of the object or (b) specify the amount of the sky you want to fill. Without specifying one of these, the question cannot be answered. $\endgroup$
    – JBH
    Feb 9, 2018 at 16:10
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    $\begingroup$ You have not added the specifications. If you tell us the size of your moon, we can tell you how close it can orbit and, therefore, how much of the sky it will fill. If you tell us how much of the sky you want filled, we can tell you the size of the moon and orbit necessary to achieve that. Please pick one. $\endgroup$
    – JBH
    Feb 9, 2018 at 16:20
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    $\begingroup$ @JBH Turns out you don't need specifics. I tried to solve it as a set of linear equations, but it turns out the units cancel beautifully. Who knew? Science is wonderful. $\endgroup$
    – kingledion
    Feb 9, 2018 at 18:39

2 Answers 2


69 degrees of arc

Apparent size in the sky

The formula for the visual angle of an object is

$$\alpha = 2\arctan{\frac{2r}{2d}},$$ where $\alpha$ is the visual angle of the object, $r$ is the radius of the object (for a sphere, which is what we will be talking about), and $d$ is the distance from the viewer to the object.

Limit of $d$: how close can a moon be?

The Roche limit for a rigid spherical moon, while taking its synchronous rotation into account is

$$d_{roche} = \left(\frac{9M_M}{4\pi\rho_m}\right)^{1/3},$$

where $d_{roche}$ is the Roche limit, $M_M$ is the mass of the planet (Earth in this case, $5.97\times10^{24}$ kg), and $\rho_m$ is the density of the moon.

Limit of $r$: how big can the moon be?

The limit for the size of the moon is the point where the moon becomes more massive than the planet. Thus, the mass of the moon must be, at the most, equal to the mass of the Earth.

There are many ways we can express this mass, but wish to solve for $r$ in terms of something we are already using as a variable; namely, $\rho_m$. Thus

$$\begin{align}M_M &= M_m \\ &= \frac{4}{3}\pi r^3\rho_m \\ r &= \left(\frac{3M_M}{4\pi\rho_m}\right)^{1/3}. \end{align}$$

That looks surprisingly familiar...

Putting it together

We now plug our minimum $d$ and maximum $r$ into the visual angle equation

$$\begin{align} \alpha &= 2\arctan{\frac{2r}{2d}} \\ &=2\arctan{\frac{\left(\frac{3M_M}{4\pi\rho_m}\right)^{1/3}}{\left(\frac{9M_M}{4\pi\rho_m}\right)^{1/3}}} \\ &=2\arctan{(1/3^{1/3})} \\ &= 1.213 \text{ radians}. \end{align}$$ The density terms conveniently cancel out, so we are left with a maximum value of 1.213 radians.


If there are two Earth-mass planets orbiting each other as binary planets, at the closest possible distance based on their Roche limits; they would appear to be 69 degrees of arc wide in the sky.

There are actually infinite solutions to this problem. A less dense Earth-mass 'moon' would be larger in radius, but would have to be farther away to not be pulled apart by gravity. So you can have some variation here, within the reasonable bounds of planetary density. For example, an Earth-mass moon at the distance of Luna would have to be 1320 kg/m$^3$ to be the maximum visual size. That is barely more than the density of water, so it is unlikely unless it is mostly of gas like a gas giant (gas midget?).

So this speaks to your gravitational effects problem. Since gravity falls off as the square of distance, a lower density object farther away would have minimal gravitational effects.

Also note that this minimum orbit is almost certainly not stable in the long run. Any reasonable moon would have to be smaller. But if artificial means are acceptable, then you can have a short term stable satellite take up nearly half of the sky.

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    $\begingroup$ I wish I could upvote twice, once for the very comprehensive answer and another for "gas midget" :) Also, nice to know that you can have some huge-looking moons $\endgroup$
    – Maxim
    Feb 9, 2018 at 20:19
  • $\begingroup$ I also appreciate your huge looking moons, king. $\endgroup$
    – Willk
    Feb 9, 2018 at 21:22

Well... here are some back of the napkin calculations for you.


  • M = larger body's mass
  • m = smaller body's mass
  • R = larger body's radius
  • r = smaller body's radius
  • d = distance between the center of the two bodies
  • Theta = apparent angular size of the moon/planet in the sky

From scratching at a piece of paper, the apparent size of a moon in the sky is:
Theta = 2*tan^-1(r/(d-R))

From Wikipedia the Roche limit for a rigid body (this is the extreme, but the math is a little easier):
d = 1.26r(M/m)^1/3

I pulled some numbers from this table and plugged them in to solve for Theta.

  • Earth/Moon = 58.4 deg
  • Earth/Earth = 150.9 deg
  • Jupiter/Earth = 110.6 deg
  • Sun/Earth = 103.4 deg

For the last two, I swapped the r and R in the Theta equation since it's the larger body's Roche limit we care about but the observer is still standing on the Earth.

Well those are fun answers. For reference the Moon is about half a degree or 30' (60 arc minutes in a deg). The take away would be that the biggest thing you can put in the sky is going to be a binary planet of equal mass. If you decide to have the Earth as the smaller body you get diminishing returns. These are closest pass numbers, most orbits aren't perfect circles so the moon doesn't have to look this huge all the time. Finally the real Roche limit is going to be somewhere between the rigid and fluid satellite numbers, I just did the rigid above and that lets the moon come a lot closer. So some in between distance isn't going to look nearly as impressive in terms of moon's apparent size. I'll look at the fluid stuff later and add it as an edit.


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