antigravity spoon

So, as my daughter dropped her spoon at a meal yesterday, my son suggested we get her an anti-gravity spoon. I am a blessed man. 🙂

I started thinking about the fact that people rarely think through the implications of suddenly defeating gravity’s pull on a single object. Namely, that it’s gravity that’s providing the centripetal force to keep it moving with the planet.

If my daughter has a spoon, and my son turns on it’s antigrav device at the lower-left of my graph, then my daughter will continue following the round green path, moving 15° every hour, while the spoon takes off in the (originally) due-east straight-line red path, moving one tick every hour.

However, since my daughter will be moving in a curved-eastwards path, the spoon appears to be moving up-and-west, relative to my daughter: in the (hat u, hat e) coordinate system moving with my daughter and the earth, normalizing out the r = R_E * cos(LAT), then vec(up) = (\theta*sin(\theta\) + cos(\theta) – 1)*hat u, and vec(east) = (\theta*cos(\theta) – sin(\theta))*hat e, where \theta = \pi/(12hr) * time. (It’s graphed with east pointing right, and up pointing up.) After about 9 hours, the spoon would drop below my daughter’s western horizon (as seen when the grey lines start intersecting the circle).

However, this is technically only the first-order approximation, where the antigravity device only cancels the earth’s gravity, keeping the spoon affected by the sun’s gravity. Without that, I’m not sure (yet), but I think the direction of the spoon will depend on the time of day and time of year. (And beyond that, we’d have to consider galactic and universal centers of gravity as well. But since we’re not changing much with respect to those centers in the realm of watching a spoon fly away, I’ll ignore those.) The solar analysis, including the tilt of the earth relative to it’s orbital plane, will come later.

If you want the mathy-details, see my derivations.