# Quick question about the moon's orbit. **URL:** \ **Category:** Factual Questions\ **Created:** [July 1, 2013, 9:34am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328 "2013-07-01T09:34:36Z")\ **Posts on this page:** 18\ **Page:** 2
**Author:** ![blue\_infinity](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/blue_infinity/32/2817_2.png) [@blue\_infinity](https://boards.straightdope.com/u/blue_infinity)\ **Post date:** [July 2, 2013, 2:33am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/21 "2013-07-02T02:33:04Z")
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. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [July 2, 2013, 4:28am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/22 "2013-07-02T04:28:02Z")
Another rule of thumb: if you make a fist and extend your arm, your fist is about 10 degrees across. ---
**Author:** ![Quartz](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/quartz/32/267_2.png) [@Quartz](https://boards.straightdope.com/u/Quartz)\ **Post date:** [July 2, 2013, 6:48am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/23 "2013-07-02T06:48:00Z")
> [@Chronos](#): > > Above geosynchronous height, it’ll spiral out, and below geosynchronous height, it’ll spiral in. Why is this? ---
**Author:** ![dtilque](https://avatars.discourse-cdn.com/v4/letter/d/d6d6ee/32.png) [@dtilque](https://boards.straightdope.com/u/dtilque)\ **Post date:** [July 2, 2013, 9:42am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/24 "2013-07-02T09:42:21Z")
> [@Xema](#): > > I don’t think this works. You can make the current moon look small or quite big relative to the apparent size of a chosen building - the former by using a normal lens at a short distance and the latter with a telephoto lens at a considerable distance. 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. ---
**Author:** ![dtilque](https://avatars.discourse-cdn.com/v4/letter/d/d6d6ee/32.png) [@dtilque](https://boards.straightdope.com/u/dtilque)\ **Post date:** [July 2, 2013, 9:59am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/25 "2013-07-02T09:59:54Z")
> [@Chronos](#): > > Above geosynchronous height, it’ll spiral out, and below geosynchronous height, it’ll spiral in. > [@Quartz](#): > > Why is this? Well, you read up on [tidal acceleration](http://en.wikipedia.org/wiki/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. ---
**Author:** ![Quartz](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/quartz/32/267_2.png) [@Quartz](https://boards.straightdope.com/u/Quartz)\ **Post date:** [July 2, 2013, 12:08pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/26 "2013-07-02T12:08:46Z")
> [@dtilque](#): > > Well, you read up on [tidal acceleration](http://en.wikipedia.org/wiki/Tidal_acceleration), but that’s a bit long. Here’s a short version: \ Very clear, thanks. ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [July 2, 2013, 1:29pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/27 "2013-07-02T13:29:44Z")
> [@dtilque](#): > > Well, you read up on [tidal acceleration](http://en.wikipedia.org/wiki/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. 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. ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [July 2, 2013, 2:12pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/28 "2013-07-02T14:12:46Z")
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. ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [July 2, 2013, 3:38pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/29 "2013-07-02T15:38:59Z")
> [@Chronos](#): > > Above geosynchronous height, it’ll spiral out, and below geosynchronous height, it’ll spiral in. > > And it’s not meaningful to say things like “That’s far more than 20 degrees” with reference to a photograph, because there aren’t any other angle references in the photo. We don’t know how far away the camera was from the Seattle skyline. 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. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [July 2, 2013, 4:24pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/30 "2013-07-02T16:24:49Z")
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](http://www.seattlerex.com/moon-over-mikes-tavern/seattleskyline-moon1/). ---
**Author:** ![Elendil\_s\_Heir](https://avatars.discourse-cdn.com/v4/letter/e/7cd45c/32.png) [@Elendil\_s\_Heir](https://boards.straightdope.com/u/Elendil_s_Heir)\ **Post date:** [July 2, 2013, 7:10pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/31 "2013-07-02T19:10:45Z")
> [@blue\_infinity](#): > > …At the Roche limit it would subtend 20 degrees and look something like [this](http://postimg.org/image/8mzh8id0f/). Holy cow. “Look, honey, I can see Tranquillity Base!” ---
**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\ **Post date:** [July 2, 2013, 9:18pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/32 "2013-07-02T21:18:17Z")
> [@scr4](#): > > 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](http://www.seattlerex.com/moon-over-mikes-tavern/seattleskyline-moon1/). Yes, plus he posted the [original image](http://images.fineartamerica.com/images-medium-large-5/seattle-skyline-at-night-with-full-moon-valerie-garner.jpg), 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. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [July 2, 2013, 9:52pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/33 "2013-07-02T21:52:11Z")
> [@Learjeff](#): > > Yes, plus he posted the [original image](http://images.fineartamerica.com/images-medium-large-5/seattle-skyline-at-night-with-full-moon-valerie-garner.jpg), 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. I meant the “original” was also a composite, in case it wasn’t obvious. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 2, 2013, 10:34pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/34 "2013-07-02T22:34:34Z")
> [@](#): > > Quoth **dtilque** : > > 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. 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. ---
**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\ **Post date:** [July 2, 2013, 10:38pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/35 "2013-07-02T22:38:45Z")
> [@scr4](#): > > I meant the “original” was also a composite, in case it wasn’t obvious. Ah, my bad. It wasn’t obvious to some of the dimmer among us (speaking personally). ---
**Author:** ![reef\_shark](https://avatars.discourse-cdn.com/v4/letter/r/bbe5ce/32.png) [@reef\_shark](https://boards.straightdope.com/u/reef_shark)\ **Post date:** [July 2, 2013, 10:48pm UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/36 "2013-07-02T22:48:20Z")
Wouldnt this optimal distance nullify the giant-impact hypothesis? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [July 3, 2013, 1:54am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/37 "2013-07-03T01:54:35Z")
> [@scr4](#): > > I meant the “original” was also a composite, in case it wasn’t obvious. Even with the right size Moon, it doesn’t line up with its reflection. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 4, 2013, 12:09am UTC](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328/38 "2013-07-04T00:09:10Z")
> [@](#): > > Wouldnt this optimal distance nullify the giant-impact hypothesis? How so? [Previous page](https://boards.straightdope.com/t/quick-question-about-the-moons-orbit/662328.md?page=1)