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If the moon were three times the distance from Earth would it still be visible?

I mean: how far away would it need to be to fade out into the darkness of space?

My guess is that it would no longer be able to be seen during the day if it were three times the distance from Earth it is today. Does this sound correct?

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  • $\begingroup$ Welcome to WorldBuilding Jamie! If you have a moment please take the tour and visit the help center to learn more about the site. Have fun! $\endgroup$ – Sec SE - clear Monica's name Dec 31 '17 at 10:26
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    $\begingroup$ What's your actual question? Three times, or how far? What's your definition of "visible"? Because Mercury, Venus, Mars, Jupiter and Saturn can be seen with naked eye. $\endgroup$ – Mołot Dec 31 '17 at 10:30
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    $\begingroup$ The moon will never be far enough away to be completely invisible unless it was behind something else. If you want it far enough away to where its little more than a speck in the sky, lets go ahead and say .75 AU. At that range it'll probably look like a tiny star on a good night and it'll probably be considered its own planet and not our moon. The moon appears less visible during the day for various reasons and distance isn't one of them. $\endgroup$ – DMQ Dec 31 '17 at 10:46
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    $\begingroup$ Mercury is not that much larger than the Moon, you know... $\endgroup$ – Mołot Dec 31 '17 at 11:29
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    $\begingroup$ If the moon was 3x as far away, it'd be about 1/3 the size, but still with the same surface brightness - that's determined by how far from the sun it is, and the shift is a tiny fraction of the sun-moon distance. And we do see the entire lit section of the moon that's facing towards us in the daytime. $\endgroup$ – JerryTheC Dec 31 '17 at 12:12
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Three times isn't far enough

tldr;

- Three time further away is magnitude -10.5 (brighter than Comet Ikeya–Seki which was visible by day)

- If the moon was further than 60.3 times further away it wouldn't be visible by day (and also wouldn't be orbiting Earth)

The light reaching us would be reduced with the inverse square law, assuming the moon is still orbiting Earth the amount of sunlight hitting it should be roughly the same and the intensity then reduced to a ninth that of the current day. $$r=\frac{r_{0}}{3}$$ $$I=\frac{I_{0}}{3^{2}}=\frac{I_{0}}{9}$$

So your moon would still be well in the visible range. Rearranging our equation and taking $\frac{I_{1}}{I_{ref}}$ as a ratio of distances squared from earth we get $\frac{r_{new}^{2}}{r_{old}^{2}}$ or $\frac{1}{N^{2}}$ where $N$ is the number of times further away our new moon is. Now take -4 as its apparent magnitude: $$10^{\frac{-4 +12.9}{2.5}} = N^{2} = 3631$$ $$N = \sqrt{3631} = 60.257...$$

So the moon has to be 60 times further away, thats $384,400 km \times 60 = 23,064,000 km$. Thats nearly half way to mars (56 million km) at closest approach, it would be unlikely that the moon would still be our moon for long. In fact, as bendl pointed out, this is well outside the Earth's Hill sphere and so a stable orbit cannot exist.

This would still be visible at night, however. There are several limits to choose from in this scale, naked eye standard dark skies, naked eye darkest skies, binocular....etc. Just enter them into the equation:

$$N = \sqrt{10^{\frac{M+12.9}{2.5}} }$$

Where the $M$ is the magnitude of your choice.

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  • $\begingroup$ The Hill Sphere for Earth is ~1.5 million km, so it wouldn't be our moon at all. $\endgroup$ – bendl Jan 4 '18 at 14:35
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If the moon were three times further away it would still be visible in daylight under some conditions such as morning and evening times. It would still be visible as a disc although a very small one.

For the moon to be invisible to the naked eye it would have to be tens of millions of miles away. The planet Mercury is visible to the naked eye under the right conditions and is not that much larger than the moon.

A moon of earth cannot have an orbit of tens of millions of miles as this would be gravitationally unstable. Such an orbit would be very rapidly distorted by the influence of the sun and inner solar system planets and the moon would probably end up in orbit around the sun rather than the earth.

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  • $\begingroup$ Heck, the moon is already trying to leave Earth orbit (it gets farther away every year by some minuscule fraction I can't be bothered to look up) $\endgroup$ – Draco18s Jan 2 '18 at 19:20
  • $\begingroup$ For sure it is, but that wasn't the question. $\endgroup$ – Slarty Jan 3 '18 at 1:55
  • $\begingroup$ Oh I know it wasn't. :) $\endgroup$ – Draco18s Jan 3 '18 at 3:44
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By the time the Moon is far enough away to be invisible to the naked eye, it wouldn't be orbiting the Earth anymore. That could also be exciting, though :)

Brightness reduces with the inverse square of distance, and brightness magnitude (wiki link: see the 'Examples' table) is measured on a special log scale. Put together, we get:

r/r0 = 2.5^(BrightnessMagnitudeChange/2)

where r/r0 is the relative increase in distance. As an example, if it's 2, that would mean the Moon would be twice as far away as it is now. Calculating

Full Moon invisible at day r/r0 = 61 times farther away = 2.5^((-4--13)/2)

Full Moon invisible at night r/r0 = 6031 times farther away = 2.5^((6--13)/2)

The Moon would start to get dragged away from Earth by Solar gravity at an r/r0 of about (roughly) 2.5, which doesn't allow for a huge amount of change in brightness. Search 'Earth SOI' for more. The r/r0 of 61 (invisible at day) puts the Moon about a third of the way to Mars. At this point the distance to the sun would start to dim it as well. Meanwhile that 6031 value for invisible at night is right out there!

But of course you could always change its albedo (Paint it Black!)

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  • $\begingroup$ Love the paint job idea. $\endgroup$ – ohwilleke Jan 4 '18 at 9:29

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