# B Apparent magnitude

1. Jul 24, 2016

### RingNebula57

Hello everyone!

What would've been the apparent magnitude of the moon if it were 1000 times closer? Can we even talk about an apparent magnitude in the case of a large solid angle?

2. Jul 24, 2016

### phinds

If it were 1000 times closer it would crash into the Earth and we'd all be dead and would not see anything.

3. Jul 25, 2016

### RingNebula57

When I say a 1000 times closer I am referring to the distance between the surface of the earth and the surface of the moon.

4. Jul 25, 2016

### Staff: Mentor

Sure. The light simply comes from a surface instead of a point.

5. Jul 25, 2016

### phinds

As am I. Do you think somehow that the surface of the moon could get 1000 times closer and the center of mass of the moon would stay where it is now?

6. Jul 25, 2016

### RingNebula57

It is a thought experiment... let's say that I am dreaming

7. Jul 25, 2016

### phinds

Yes, I understand that. Post #2 was my attempt at humor. You have taken it too seriously and thus my followup to your serious reply.

8. Jul 25, 2016

### Janus

Staff Emeritus
It's not an easy question to answer. Putting the surface of the Moon 1000 times closer to the Earth puts the nearest point ~375 km away from the surface of the Earth. At this distance, you would never have a full Moon, because the Moon would be eclipsed by the Earth's shadow. To avoid that, the Moon would have to be lit from the side so the most you will ever see is a half-moon. Also, with the Moon so close, you will not be able to see all of the side facing the Earth as much of it will be hidden behind the Moon's own horizon. In addition, the relative distance of different points of the surface you can see will vary quite a bit. From 375 km straight above you to ~1200 km to the furthest limb of the Moon that you can see.
All this would have to be taken into account to determine just how much light you would receive from the Moon if it were that close.

9. Jul 25, 2016

### Staff: Mentor

I too thought you were serious, my furry little friend.

10. Jul 25, 2016

### phinds

Well, yeah, but we all know you have no sense of humor.

11. Jul 26, 2016

### Chronos

I figured like 350km at aphelion, but, that is not particularly relevant. More to the point is tidal effects would be hugely annoying. We are talking more than 1000m on average in coastal regions, or, similar to a tsunami twice per day.

12. Jul 26, 2016

### OmCheeto

I calculated an orbital period of 2.15 hours to keep the moon in orbit @ 384 km.
hmmm... Phil Plait said it was only 1.5 hours at a distance of 420 km.
Is it safe to assume he just used the orbital period of the ISS, and forgot to account for the added distance to the center of the moon?
No matter, I guess, as he says we'd all be dead, and wouldn't have to worry about arguing the maths.

Anyways, it's a fun article, and directs to an interesting video:

If the Moon were at the same distance as the ISS​

13. Jul 26, 2016

### sophiecentaur

The calculation of apparent magnitude assumes the inverse square law applies. But the ISL only applies to point sources (a condition that's met by which all other stars). The visibility / magnitude of larger objects like the major planets, Sun, Moon can't be calculated merely from the distance but must involve the angle subtended. I would suggest that those magnitude figures are arrived at by some kind of general agreement between 'those people who deal with these things'.

14. Jul 27, 2016

### RingNebula57

The thing is that this problem was actually given at an olympiad and their answer was that the brightness ration for the moon before and after the displacement is given by the ratio of the solid angles subtended by the moon. But I don't agree. I think that if we wuold like to generalise the Pogson law ratio for 2 non-point sources we shold write it like this:
(E/E')*(Ω'/Ω), where E represents the brightness of the object(moon) and Ω the solid angle subtended by the object (moon). So I am basically saying that the intensity ratio gives the diffrence between magnitudes.
My formula is diffrent from the solution formula because I say that when the solid angle grows the magnitude grows. I came up with this formula thinking about the apparent magnitude of the milky way.
What do you think?

Last edited: Jul 27, 2016
15. Jul 27, 2016

### sophiecentaur

@Ring Nebula
Your have clearly spotted that the ISL formula cannot apply for large solid angles.
That definitely makes sense when there is a given flux leaving the object and what comes through the lens is spread over an increasing number of sensors as the angle increases. But the detailed result of this depends upon the resolution of the sensor array (or the telescope optics). Looking at the Moon through obscure glass would not, I suggest, show a difference as its distance alters.
You mention the Milky Way as an interesting case and so is the sky at dusk. More stars become visible as dusk progresses so you could attribute a 'magnitude' to the sky at any time, relating to which stars (known magnitudes) are visible. Of course, we are not dark adapted at the time but the comparison would still be valid. I have had similar thoughts whilst trying to see the Andromeda Galaxy. The 'magnitude' increases with distance from the centre and, with the Moon being very present at the moment, and the general gunk in the atmosphere, you could award it an apparent magnitude more like the 'lower magnitude' stars around it.
What is needed (and it probably exists) is a more meaningful quantity for objects with large subtended angles but it would be harder to come up with a simple formula for it if you wanted to apply it to observing the night sky.

16. Jul 27, 2016

### Janus

Staff Emeritus
With the surface 350 km away, the center of the Moon is ~1/45 the distance from the center of the Earth than it is now. Tidal forces vary by the cube of the distance, so you are looking at tides over 91,000 times higher, or a mid ocean tide ~91 Km high.

Of course, if you are going to factor in tides, this becomes a moot point, because this puts the Moon well within the Roche limit and it would have been torn apart long before it could ever get this close.

17. Jul 27, 2016

### Chronos

Under the giant impactor theory, the initial distance to the moon was probably in the range of 20,000-30,000 kilometer [3-5 earth radii]. Much closer and tidal forces would have kept the infant moon fragmented. Depending on the [uncertain] details of the impact, simulations suggest collisional debris was probably not flung any further out than about 5 earth radii. Oceanic tides [assuming oceans had yet formed] would still have been awesome at such distances.

18. Jul 27, 2016

### sophiecentaur

The OP's question was about apparent magnitude and not the mechanics of how to achieve the condition. Let's just assume the Moon has been supported on three long pillars and get on with the optics of the situation - eh?
We could have avoided all this is the OP had limited the change in distance of the Moon to within the range of values that currently apply. The answer would still have been the same - just a bit less extreme.