A significant amount of heat would be lost from the lack of tidal forces. When you knead the earth by a few meters every day, it gets pretty hot.

A much less amount from reflected sunlight.

A probably insignificant amount from radioactive elements within the moon.

I'll leave it to some other bright spark to do the math for (A) the steady state end solution, and (B) integrating from the moon's current position out to infinity as the aliens tow it away. I hope they get a job with Al Gore for their troubles!

A significant amount of heat would be lost from the lack of tidal
forces. When you knead the earth by a few meters every day, it gets
pretty hot.

But how does the heat generated by the tidal force compare to the heat generated by radioactivity? Recall that the age of the Earth was vastly underestimated because of the high heat of the Earth (since it is expected gradually to cool down from the time of its formation), but only because radioactivity had not been discovered. If the moon accounted for the lion's share of that heat then I would imagine that the ignorance of radioactivity would not have made such a big difference.

... I meant the fraction of heat from lunar tidal forces compared to the total heat from the moon was large, not the fraction of heat from lunar tidal forces compared to the total heat of the earth.

I'll hold off judgment until someone does the calculations, but until then, some googling turned up this statement from the USGS which does indeed suggest the amount of heat from lunar tidal forces compared to the total heat of the earth is small enough for them to ignore:

Heat within the Earth comes from two main sources: radioactive decay and residual heat.

As I ponder the calculations, I think the amount of tidal heat in the earth and moon combined would be equal to the amount of energy they lose in angular momentum due to their slowing rotations (earth days are gradually getting longer and the moon is gradually drifting away). So if you can get the length of a terrestrial day and distance of moon at an earlier time, you could probably work out the heat generated during that time. The changes in earth-moon distance may even be derived from the slowing day, perhaps by throwing away some less significant terms.

I start to get more fuzzy when I think about how to separate lunar heat from terrestrial heat.

Don't you want put me out of my misery and do the calculations, Constant? Then I can get back to work...

Nope. In fact, the aliens are smart enough to move a moon around, therefore I take it they are smart enough to take advantage of the gravitational energies between the Earth,Moon, and Sun. They need to add energy to the moon to get it to Solar escape, they'll take as much as possible from the Earth. As Earth's energy drops so does perehelion.

Now, how long do they take accelerating the moon to Solar escape velocity? If they do it fairly quickly, the drop in the Earth's orbit will be small and won't make any difference to global mean temperature. If they take a long time and grab up lots of Earth's orbital energy, then Alien global warming will make anthropogenic global warming look like kids play.

## Roughly...

I would guess:

I'll leave it to some other bright spark to do the math for (A) the steady state end solution, and (B) integrating from the moon's current position out to infinity as the aliens tow it away. I hope they get a job with Al Gore for their troubles!

## Heat

A significant amount of heat would be lost from the lack of tidalforces. When you knead the earth by a few meters every day, it gets

pretty hot.

But how does the heat generated by the tidal force compare to the heat generated by radioactivity? Recall that the age of the Earth was vastly underestimated because of the high heat of the Earth (since it is expected gradually to cool down from the time of its formation), but only because radioactivity had not been discovered. If the moon accounted for the lion's share of that heat then I would imagine that the ignorance of radioactivity would not have made such a big difference.

## By "significant" ...

... I meant the fraction of heat from lunar tidal forces compared to the total heat from the moon was large, not the fraction of heat from lunar tidal forces compared to the total heat of the earth.

I'll hold off judgment until someone does the calculations, but until then, some googling turned up this statement from the USGS which does indeed suggest the amount of heat from lunar tidal forces compared to the total heat of the earth is small enough for them to ignore:

As I ponder the calculations, I think the amount of tidal heat in the earth and moon combined would be equal to the amount of energy they lose in angular momentum due to their slowing rotations (earth days are gradually getting longer and the moon is gradually drifting away). So if you can get the length of a terrestrial day and distance of moon at an earlier time, you could probably work out the heat generated during that time. The changes in earth-moon distance may even be derived from the slowing day, perhaps by throwing away some less significant terms.

I start to get more fuzzy when I think about how to separate lunar heat from terrestrial heat.

Don't you want put me out of my misery and do the calculations, Constant? Then I can get back to work...

## Presumably pulling the moon

Nope. In fact, the aliens are smart enough to move a moon around, therefore I take it they are smart enough to take advantage of the gravitational energies between the Earth,Moon, and Sun. They need to add energy to the moon to get it to Solar escape, they'll take as much as possible from the Earth. As Earth's energy drops so does perehelion.

Now, how long do they take accelerating the moon to Solar escape velocity? If they do it fairly quickly, the drop in the Earth's orbit will be small and won't make any difference to global mean temperature. If they take a long time and grab up lots of Earth's orbital energy, then Alien global warming will make anthropogenic global warming look like kids play.