How far away would the moon need to be to not be visible anymore?

Question

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?

Answer 1

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. $$r=\frac{r_{0}}{3}$$ $$I=\frac{I_{0}}{3^{2}}=\frac{I_{0}}{9}$$

• In its current position the Moon has an apparent magnitude of -12.9
• An apparent magnitude of -4 or less is needed to be visible in the daytime sky
• Apparent magnitude can be calculated by: $$m_{1}-m_{ref} = -2.5 \log_{10}\frac{I_{1}}{I_{ref}}$$ So lets take $m_{1}$ as our new magnitude (for the moon three times further away) and $I_{1}$ is then the intensity of light from the moon in its new position but $I_{ref}$ is the current position of the moon and $m_{ref}$ is the current magnitude. $$m_{ref} = -12.9$$ $$\frac{I_{1}}{I_{ref}} = \frac{1}{9}$$ So: $$m_{1} = -12.9 - 2.5 \log_{10}\frac{1}{9} = -10.5...$$

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.

Answer 2

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.

Answer 3

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!)