Here it has been discussed why the moon is moving away from earth at a rate of 1.5 inches per year. Is there any way we can estimate the value (e.g, 1.5 inches per year) mathematically?
$\begingroup$ The Moon's orbital energy is lost in friction when it generates tides on Earth, and so the orbit gets more and more loose and wide. If you could somehow estimate the energy lost in tides, you could then apply that to the Moon. BTW, the implication is - when we're building generators powered by tides, we're actually stealing from the Moon's kinetic energy. $\endgroup$– Florin AndreiAug 29, 2017 at 23:55
$\begingroup$ Thanks for the nice argument. I was actually looking for a quantitative argument for it. $\endgroup$– rainmanAug 30, 2017 at 0:29
1$\begingroup$ I don't want to post "Wiki" as an answer, so I'll drop it here, but here are a few mathematical explanations and references. rationalwiki.org/wiki/Recession_of_the_Moon $\endgroup$– userLTKAug 30, 2017 at 5:45
1$\begingroup$ Dunno how to break it to you, but there is no way other than "mathematically" to estimate the moon's orbit. Even if the math involved consists of fitting a curve to collected observations and attempting to extrapolate. $\endgroup$– Carl WitthoftAug 30, 2017 at 13:13
This is such a complex problem that I think your chances of getting an accurate answer are almost nil.
First, the equations are likely to be highly complex if you want to be accurate down to the tenth of an inch. For example, you'd need to model the exact shape of the Earth due to the Moon's tides, most likely accurate down to the meter or so. Scientists have done this on average, but finding how these types of models minutely change on an hourly basis is near impossible. The data just doesn't exist.
Second, filling in the variables you'll have in your equations is likely going to be impossible because many won't be known. For example, do you know the flexure/elasticity/tensile strength/etc. of the entire Earth's crust at every point on Earth over time? My guess is you don't. You'd need to know that though to do really accurate calculations.
The only way we know the recession rate of the Moon is through measurements. Crude models might get you crude order of magnitude numbers, but to match the precision of observations would require models that are far more informed than we currently are.
$\begingroup$ It's also a three-body problem - because of the Sun. So no non-numerical solution exists. $\endgroup$ Aug 30, 2017 at 15:39
1$\begingroup$ @adrianmcmenamin Another good point. I also wonder, if you want to get down to tenths of inches in accuracy, would you have to include perturbations on the level of Jupiter, making the problem even more unwieldy? $\endgroup$– zephyrAug 30, 2017 at 15:47
Easy. Bounce a laser off the moon at a specific time of day and do this everyday for years and years. You could in fact get the distance it moves away from earth down to the day just by the time it takes for the laser light to come back to your device.
Multiply the time it takes for the light to come back with the speed of light and divide by 2. Note down the time each day and check the differences over time. You'll be able to get the change of rate down to the day.
Heck, we could even fly back to the moon, house a computer with some batteries and a few solar panels to charge them and ping it. On the trip back just drop off a few satellites 22,000 miles apart back to earth to boost signal so it could connect to the internet lol. Fun aside.
$\begingroup$ Yes, it's math, but I don't think this is what the OP meant with mathematically. See the other answer. $\endgroup$– user1569Jun 18, 2018 at 11:53
$\begingroup$ People do regularly bounce laser beams off retro-reflectors on the Moon. That's why we know its orbit so well. $\endgroup$– PM 2RingJun 18, 2018 at 16:01