Why doesn't the moon leave earth's orbit?

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Jim

How do you know the moon is receding at 4cm per year?
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The Apollo astronauts left retro-mirrors on the moon which the boys at NASA fire lasers at to check out recession data. NASA's fact sheet -

[url] nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html [/url]

has the recession at 3.8cm a year.

This is the NASA page that details the laser approach -

[url] sunearth.gsfc.nasa.gov/eclipse/SEhelp/ApolloLaser.html [/url]

These are very old links and in my opinion not very reliable. See the disclaimer that the measurements are within 3cm of actual distance. And then consider the Earth crust is rotating and bouncing up and down more than 3cm. The 3.8cm recession guess work is just not good science. NASA should not post such stuff.

This stuff's been measured over a period of 30 years. The Moon fact sheet is dated April 2002; the sheet describing the setup is "old"(?) - 1999, why they'd need to update that, I don't know. I don't think a 3cm error rate is significant when the recession has been measured at over a metre. I've never seen any data regarding the bouncing crust (what's the period of this bounce?), but fair comment. I'll stick with the observations for now until someone sticks some hard counter-evidence in my face (and I'm not talking speculation/theory/personal belief a la zero/infinity).

It is not measured at a meter over 30 years and if the data shows this I'll accept that this is happening. Anyway, I cannot prove it is not so. Not just yet anyway. The idea that the moon is receeding because of tidal friction makes little sense either way. What about the other points of lunar theory? Why is the period not from new moon to new moon? If the inclination of the orbit was 90 degrees rather than 7 degrees the period would be 29.4 days. So why is it 27.5 days just because of the inclination?

There is another point about tidal theory that makes little sense even though it works. The magnetude of tides is proportional to the 3rd power of the distance of the sun and moon from Earth. If the 3rd power of the distance is a real factor why is there very little in the tide when the moon is near and far from Earth? The moon's orbit is quite eccentric and if a 3rd power of distance law existed the tides should be much higher when the moon is nearest to Earth.

Jim

It's difficult to argue the whys and wherefores of tidal effects. At the last count, about 400 variables related to observer's position, weather, and motion of the earth and other objects in the solar system have been identified as contributing to the exact height of the tide, although admittedly, above all this, the Moon has the most significant effect by far. Also it's next to impossible to produce an equation that will calculate the tides - each location has a set of variables, or "harmonic constituents", as the tidal buffs like to call them. Having said that, here's an excellent link that explains it all, at least from the point of view of your point regarding the seeming third power paradox -

[url] mb-soft.com/public/tides.html [/url]

Regarding the 3.8cm/1m thing, it's too late on a Saturday night to be worrying about it, but if you insist, I'll write to NASA in the morning for confirmation of the source data used.
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