See subject. I like to have the wind at my back, so to speak. In a thread asking if can you travel through time while standing still (spoiler: you can’t, even if you could) poster mlees at #22 mentions
With all this whipping around, in which direction should I walk to expend the least amount of energy? I’d like to visit some sort of hill or valley as well–I’m not that lazy. Also, so I can adjust my new sports watch-pedometer-calorie thingie, how much energy would I save over a trip in the other direction?
The only force you could take advantage of is the centrifugal force, it’ll make it slightly easier to walk up a hill at the equator, and work against you when you walk back down. The wikipedia article on the Gravity of Earth says the rotation gives a decrease of 0.3%, with an additional 0.2% resulting from the increased distance from the center due to the equatorial bulge.
So if we just look at the gain and loss of potential energy:
The force involved is F = m*v^2 / r
For the spinning Earth this works out to an apparent accelleration of (456 m/s)^2 / 6,378,713 m = 0.0326 m/s^2 (Which unsurprisingly is about 0.3% of 10 m/s^2)
For the orbit of the Earth, if you make sure to take your hike at midnight at perihelion, we get:
F = (30,000 m/s)^2 / 147,095,000,000 m = 0.006118 m/s^2
Which works out to a 0.12 % decrease compared to doing the same hike at high noon at the same point in the orbit.
Thnx. Actually, I phrased the last part wrong: what’s the hardest trip from any location.
I know, sort of, in real life rocket guys (and airplane fuel budgeting–or is that only a weather thing) figure this stuff out before breakfast.
I’m basically trying, as a goof, to learn about the minuscule forces of all these non-overlapping rotational force. And what word is the right one instead of “overlapping but not equally centered centripetal (centrifugal?) forces.”
If you meant hardest you shouldn’t have put “easiest” in the title and written “least amount” in the question itself.
In any case, the largest influence of these forces on your motion will be when you move directly towards or away from the center of rotation, whether that is the axis of the Earth, the Sun or Sagittarius A. And the magnitude of the forces will be the accelerations mentioned, multiplied by your mass.
I’m thinking of myself as that spider on the flying frisbee, expending energy with each step. I’m also understanding, I think, that the tide moves and the moon, so shouldn’t…
Wait, I just thought: the tide doesn’t “expend” more energy depending where the moon is–precisely by shifting depending on where the moon is, it “spends” exactly the same energy as always.
Is this right relative to my OP? Ditto a tide on a frisbee, hence your answer?
Which is also to say, perhaps you could educate me on what I thought I was thinking and where I went so wrong.
The usual error in this sort of thing is to regard the ground you’re walking on as unchanging. You think that the Earth is speeding along at X m/s, and you’re walking along the Earth at Y m/s, and if X is large enough, then the energy to go from X m/s to X+Y m/s can be huge. Except that most of that energy doesn’t actually come from you: When you walk relative to the Earth in such a way that you’re benefiting from the Earth’s motion, you’re also slowing the Earth down. You’re only slowing it by a minuscule amount, of course, since the Earth has so much more mass than you, but then, also because it has so much mass, that small change in velocity can result in a large change in energy. If you conserve momentum in the you-Earth system and do the calculations, then you’ll find that walking at a certain speed relative to the Earth always requires the same energy input from you, regardless of what speed the Earth is traveling at. Which is as it should be, with the Principle of Relativity and all.
