# Quick question about the moon's orbit.

Yes, I’m thinking the original skyline photo was taken with a telephoto lens, which makes the moon look a bit ‘too big’ in both pictures.

Another way to look at it: 20 degrees is about the height of a ‘typical’ (40") widescreen tv, viewed from around 3 feet away.

Another rule of thumb: if you make a fist and extend your arm, your fist is about 10 degrees across.

Why is this?

OK, it could have been taken through a telescope. I was assuming it was meant to be a naked-eye image of a huge moon. In that case, it is way too large to be only 20 degrees.

However, note that there is another tall building in the picture and that building is not at the same distance as the Needle. If someone felt like it, they could find the height of that building and the relative locations of the two structures and then use their apparent heights in the image to find the magnification of the photo. But that’s likely a waste of time, since the info may already be on the net somewhere. At any rate, I’d be surprised if the magnification is not more than 40x.

Well, you read up on tidal acceleration, but that’s a bit long. Here’s a short version:

The moon raises a tidal bulge in the Earth (both ocean and land) immediately below its position. If the Earth is rotating faster than the moon is orbiting, the bulge will move forward of moon’s position by a little bit before it relaxes back to normal height. That bulge generates a gravitational pull (admittedly a very small one) on the moon, pulling the moon forward. An acceleration forward in an orbit means the orbiting body moves to a higher orbit.

If the moon orbits faster than the Earth rotates, that bulge will trail the moon’s position, giving a backwards pull. That will cause the orbiting body to move to a lower orbit.

Very clear, thanks.

I’ve often wondered about that; thanks for the explanation.

If the moon were even 30% closer, tides would be about twice as high. The moon was once at about a tenth the distance it is now. Not only would this have resulted in enormous tides, but the day was only about 5 hours, so you would have those tides every 2 1/2 to 3 hours. But imagine the solar eclipses! The moon is still moving away–at glacial speeds, but in few hundred million years there will be no total eclipses.

The tidal Acceleration/decelleration works both ways, which is why the moon is slowing the earth’s rotation; the pull on geological bulge also slows the earth. Plus any geological scale situation like this would eventually result in a tidally locked result - the earth and moon would always face each other… no more moonrise, some lucky continent would hav perpetual moonshine.

Odds are it would disrupt geosynchronous satellites too; they would be relegated to the trojan points instead of every 2 degrees along the equatorial orbit; but of course the “day” would be that two and a half hours, not 24, etc.

If we did have a big moon orbting closer than the geosynchronous satellites and a 24-hour day, I’m guessing it would still be seriously disruptive and geosynchronous (22,300 miles up) would not be stable… not to mention frequent eclipse blackouts as the moon blocks the signal.

I’m having a problem with that picture as a a non-composite picture. First, the “columns” of the Space Needle face the cardinal directions. The Science Center arches are a little west of SW from the Space Needle and are actually quite close, maybe 500 feet. The Space Needle is 605’ and 4" on my monitor. The Arches are 110’ and 3/4" on my monitor which would mean that they are both approximately the same distance from the camera i.e. the camera is some distance away.

The distance between the arches and the Needle is about 1.5" or about half the distance it would be if the camera were pointed perpendicular to the line intersecting the arches and the Needle. This means that the camera was offset 60 degrees from the perpendicular which actually works out perfectly with the “column” on the left facing south and the one on the right facing east. Given the orientation on of the arches as a further confirmation, I would say the picture was taken facing west for a position slightly south of the Space Needle. My best guess would be from somewhere in Denny Park.

However, Puget Sound is to the Southwest of the Space Needle so there is no way the waterfront would be in there. Also that very distinctive building? I saw it in another picture of the Seattle skyline but I can’t seem to find it again.

The Seattle picture is obviously a composite. The stars are an obvious giveaway - there’s no way you can photograph stars and a brightly lit city at the same exposure. Also the Moon is much too dim compared to the city. It should look more like this.

Holy cow. “Look, honey, I can see Tranquillity Base!”

Yes, plus he posted the original image, and you can see the reflection of the normal-sized moon in the water of the mockup, as I pointed out above. It’s a photoshop job, and clearly just for fun.

It seemed like it was more than 20 degrees to me, but I’m not quite sure how to figure that out. However, given the moon is 1/2 degree, is the radius of that circle 40 times bigger than the moon in the original? Seems to me, close enough.

I meant the “original” was also a composite, in case it wasn’t obvious.

If you’re viewing it on a sufficiently small screen, or from a sufficient distance from the screen, it would look the correct size to be a naked-eye view of Seattle and the Moon, as seen from the vantage point of the camera.

At least, on grounds of angular size. I’ll cede the arguments based on Seattle geography (about which I am not conversant) and on illumination conditions, to show that that picture is a composite.

Ah, my bad. It wasn’t obvious to some of the dimmer among us (speaking personally).

Wouldnt this optimal distance nullify the giant-impact hypothesis?

Even with the right size Moon, it doesn’t line up with its reflection.

How so?