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In many series (Elder scrolls for example) they have a huge moon. It is rare though when they explain how it can be larger. There are only two choices both of which couldn't work;

One is to bring the moon closer, but that can get risky, how close can it get for example never mind the tides.

Two is to keep the distance but make it bigger but then how big can I make it?

Either way it seems impossible to make the moon very large (for your answers lets assume I want it to appear twice the size of the sun) But are there any ways I can make the moon appear that large without the world ending side effects

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  • $\begingroup$ Making the moon bigger will also mean bigger waves. Surfers will enjoy that. A lot. :D $\endgroup$ – Zizouz212 Oct 1 '15 at 1:10
  • $\begingroup$ Do you mind how bright it is (e.g. can we make it look larger, but not reflect as much sunlight)? $\endgroup$ – Mikey Oct 1 '15 at 1:14
  • $\begingroup$ worldbuilding.stackexchange.com/questions/25675/… $\endgroup$ – JohnWDailey Oct 1 '15 at 1:28
  • $\begingroup$ @Mikey no It can be any brightness $\endgroup$ – TrEs-2b Oct 1 '15 at 1:45
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    $\begingroup$ We do have a huge moon. $\endgroup$ – JDługosz Oct 1 '15 at 3:43
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One is to bring the moon closer, but that can get risky, how close can it get for example never mind the tides.

The answer to this is the Roche Limit. Closer than this and gravitational forces will rip it apart. The Roche Limit is about 2 1/2 times the radius of the central body. Just to make things easy, figure 3 times the radius of the earth to the center of the moon, or about 12,000 miles. Since the current average distance to the moon is about 240,000 miles, this is 1/20 the current value.

When the moon is directly overhead, the distance will be 2 radii, or 8,000 miles. Since the moon's radius is 0.237 earth, or 938 miles, the apparent size of the moon, in degrees, is $$\theta = 2 tan^{-1}(\frac{948}{2\times 3960}) = 13.7 degrees$$ Comparing this to the current size, about 0.23 degrees, the close moon will be about 60 times bigger (apparent diameter, not area) than is true now.

Something else to consider is that, as the distance to the moon changes, the orbital period changes, too, and the change goes as (Kepler's Third Law)$$T^2 = K\times R^3$$ Reducing the radius by a factor of 20 will reduce the length of a month by a factor of 90. So the moon will orbit the earth about every 7 1/2 hours.

Two is to keep the distance but make it bigger but then how big can I make it?

That's pretty straightforward: you can make it as big as you like. Of course, if it gets much bigger than the earth, the earth will be the moon. Since the moon doesn't have an iron core it is less dense than earth, so if the makeup and densities are kept the same, the moon will have to get somewhat larger than earth before it becomes more massive, which is what counts in determining which is primary and which is satellite. Since the current diameter ratio of moon to earth is 0.237, the moon can get about 4 times bigger than it is now, and still be the moon. Of course, if it does get that big it will probably have an atmosphere and won't look at all like it does, but rather more like earth.

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The gravitational force between any two bodies, such as the Earth and the Moon, is proportional to their mass, but inversely proportional to the square of their distance. Apparent angular diameter, however, is roughly radius over distance. That is to say, $F\propto m/r^2$, $\theta\approx R/r$. What does this mean? Suppose we multiply the radius of the Moon by a factor of $a$. The mass is multiplied by a factor of $a^3$. If we multiply the distance from Earth by a factor of $a^{\frac{3}{2}}$, the gravitational force stays constant. The apparent size is multiplied by $a^{-\frac{1}{2}}$. What this implies is that if we shrink the Moon, decrease the mass, and move it closer, the apparent size could increase. For example, if we halve the radius of the Moon, and multiply the distance by 0.35 or so, then the gravitational force stays the same, but the apparent size is increased by about 40%. So to double the size of the Moon, make the radius 1/4 its current value, and cut the distance to an eighth of its current value.

Let me emphasize something: this does not change the gravitational force. That is, except for a somewhat more heterogeneous field, because the point mass approximation is somewhat less valid, the gravitational pull of this new smaller, closer Moon is the same.

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  • $\begingroup$ Jonah, I added $\LaTeX$ to make the exponents more readable, and I made the capitalization consistent. Did you not like that, or did your subsequent edit simply accidentally roll over mine? $\endgroup$ – HDE 226868 Oct 1 '15 at 2:24
  • $\begingroup$ Maybe I rolled over yours. $\endgroup$ – Obie 2.0 Oct 1 '15 at 2:24
  • $\begingroup$ Does this take ordinary $\LaTeX $? $\endgroup$ – Obie 2.0 Oct 1 '15 at 2:24
  • $\begingroup$ It should, yes, using dollar signs ($). \LaTeX $\to$ $\LaTeX$. $\endgroup$ – HDE 226868 Oct 1 '15 at 2:25
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The moon could be much closer. The closer it is the faster it needs to orbit to stay up. If its orbital period was exactly 1 planetary day then it would be tidally locked. (ie: high tide would stay on the same side of the planet so that nobody experiences tides.) but it would only be visible form 1 side of the planet.

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