# Quick question about the moon's orbit.

Given the mass of the two objects, how much closer could the moon get to Earth and still maintain a steady orbit? At its closest possible distance, how much larger would it appear in the sky?

You’re talking about the Roche Limit.

So the Moon is 41 times farther out than it’s Roche Limit. Were it just beyond its Roche Limit, it would appear just under 41x larger.

Whoa. I’d really like to see a mockup of what that would look like. Does anyone know of such a picture?

Like this?

I was thinking about something similar the other day. I wondered if we could have far more dramatic tides with a closer or heavier moon. I’d be very interested to know what the results of the Moon being 41x closer to us would be.

Normally the moon subtends half a degree and looks something like that.

At the Roche limit it would subtend 20 degrees and look something like this.

Much higher tides, and greater volcanism. Both due to the greater gravitational force of the moon.

Well, you can create a similar effect with the moon as it is now, using a telephoto lens. See this excellent video.

If it were that close, would atmospheric drag become a factor?

Much higher tides. Tides vary inversely with the cube of the distance. So if the Moon were only 1/41 times as far away as it is, tides would be 68921 times as much. A one foot* tide would translate into 13 mile tide. That’s more than twice the height of Everest, so I suspect the volume of water in the Earth’s oceans is going to start to have an effect on the actual tide height. But it seems quite possible that no part of the Earth would be dry all day long.

*The theoretical tide is 21 inches. Actual tides of course vary. One foot is a conservative number which is a bit more than the theoretical neap tide of 11 inches.

Probably not. It’s still >3000 km above the earth’s surface, almost 10x higher than low-earth orbit satellites. Also, the larger the object, the less it’s affected by atmospheric drag, because drag is proportional to area which is proportional to square of object diameter, while mass is proportional to cube of diameter. (Similarly, if you dropped an elephant and an ant from an airplane, atmospheric drag would slow down the ant much more than the elephant.)

… except for the small reflection in the water

Still, that’s a stunning image, thanks!

That would be over 9,000 km - there’s no atmosphere at that distance.

How much of that gravitational energy would go into the mantle and distorting the rock surface of the world?

Note that the Moon cannot have attained its current mass at less than ~1/20th its current distance as it would initially have been a fluid (of molten rock), so tides would ‘only’ have been 8000x as high. ‘Only’ 1.5 miles high.

That’s far more than 20 degrees.

If I did my calculations right, the orbital period of the moon at the Roche limit would be just over 2.5 hours. Since that’s less than the length of a day, the moon would spiral inward and soon[sup]1[/sup] cross the Roche limit. At that point it would break up and form a ring. The pieces would eventually spiral in and hit the Earth.
[sup]1[/sup] That is “soon” in astronomical terms. Depending on how close it is to the Roche limit, it might take thousands of actual years.

At the Roche limit, the Moon gets destroyed, but assuming it’s just beyond, why would it spiral in, and not spiral out or orbit stably?

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.

And geosynch height is well above the Roche limit, so if the moon is below the geosynch height (about 42,000 km) it’s already in an orbit that’s not stable in the long-term - the spiraling in will eventually lead to the moon meeting its demise through being shredded, or due to its collision with the Earth.

There are structures that are a known height, especially the Space Needle. That can be used to get an estimate of the distance.

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.