# Conservation of angular momentum on Earth

If I start spinning while standing on the floor, conservation of angular momentum says something needs to start spinning in the opposite direction but what's that?

• Please, do not ask two potentially unrelated questions in one. Ask a separate question instead. – FGSUZ Oct 17 '18 at 13:26
• I'm performing a rollback to the previous version to remove the unrelated question (cc @FGSUZ). Can you ask that one separately? It's really awesome, and I'd like to answer that! Try searching for "co-axial rotors" in helicopters for some alternatives for that vertical rotor. – user191954 Oct 17 '18 at 13:40
• Some things to think about: imagine you are standing on an extremely slippery flat surface that is on the earth -- say, a wet, smooth ice rink, and you have slippery shoes. Now what happens when you try to spin, and what does that tell you about the premise of your question? – Eric Lippert Oct 17 '18 at 21:48

If I want to start spinning, I have to push on something (try imagining starting to spin in space - you won't be able to because there will be nothing to push off of).

This will most likely be the ground, or if you like the Earth itself, and so by changing my angular momentum, I also have to change the Earth's angular momentum.

However, because the Earth is so big and heavy compared to myself, me pushing on the ground has a negligible effect on the motion of the Earth.

A very simple example - say I gain angular momentum \begin{align} L&=mvr\\&\sim100\text{ kg }\times1\text{ m s}^{-1}\text{ }\times1\text{ m}\sim100\text{ kg m}^{2}\text{ s}^{-1} \end{align} then the change in angular momentum that must occur on the Earth is \begin{align} \Delta L&\sim100\text{ kg m}^{2}\text{ s}^{-1}\\&=M_\text{Earth}\Delta v_{\text{Earth}}R_\text{Earth}\\&\sim6\times10^{24}\text{ kg }\times\Delta v_\text{Earth}\times6\times10^{6}\text{ m} \end{align} and solving for $$\Delta v$$, we find the change in the Earth's velocity at the surface of the Earth is $$\Delta v \sim 3\times10^{-30}\text{ m s}^{-1}$$ I think it's safe to say you can spin freely without worrying about disrupting the planet's spin.

• Does this mean Earth will spin on the axis connecting me and center of Earth? – kapsi Oct 17 '18 at 13:12
• Yes I think in this "toy" example, the Earth would spin (or rather the spin vector adding to the already existing spin) will be oriented in such a way to exactly cancel yours. – Garf Oct 17 '18 at 13:14
• Though (as I've just added to my answer) the amount by which you spin the Earth is incredibly negligible - other effects would swamp your spinning actions. – Garf Oct 17 '18 at 13:15
• mandatory xkcd: xkcd.com/162 – Oxy Oct 17 '18 at 15:17
• @Oxy mandatory xkcd is a redundant title – Aaron Stevens Oct 17 '18 at 16:51

Angular momentum is conserved only in closed systems. Since you're applying an external torque to start the rotation (perhaps by pushing something else to start off), it isn't a valid principle any more. When you consider a larger system, for example the combination of the planet and the person, angular momentum is conserved: if you push the ground to start the rotation, the planet's angular momentum changes (although this isn't very visible; Garf's answer provides an excellent numerical estimate), or if you push a different object next to you, it starts rotating. The same explanation applies even when you're on a different planet.

Conservation laws generally need to be applied with caution. When there's an external force, you can't always say that linear momentum is conserved, can you?

• @Mohan Running is quite inefficient: though a person maintains a constant-ish kinetic energy, they expend a lot of energy. Does that help? ;) – user191954 Oct 17 '18 at 13:16
• @Mohan Okee, so the person isn't a closed system. They loose energy, but the kinetic energy appears constant. That's because energy is lost as heat (friction with the ground), kinetic energy of surrounding air, kinetic energy of some particles on the ground, and making the person accelerate upwards slightly (off the top of my head). Momentum appears to not be conserved, because the momentum change experienced by the ground/earth is not apparent due to the huge mass of the earth . – user191954 Oct 17 '18 at 13:34
• @Mohan Yes, you can consider the person and the Earth as one closed system. Essentially you can always make your system large enough so that your system is closed (you could even make your closed system the entire universe, unless there is some unknown entity outside of the universe that can interact with our universe). – Aaron Stevens Oct 17 '18 at 14:59
• The important difference between the comments by @AaronStevens and me is that his comment talks about the system of the human and the earth, while I talked about just the human. His phrase "Essentially you can always make your system large enough so that your system is closed" is a very good way to describe this (within reason): he considers a larger system and hence energy is conserved there. – user191954 Oct 17 '18 at 15:04
• The question clearly asks what other body angular momentum is being conserved with. Responding "There's something else" isn't an answer. That would be like responding to a question "Where does the power to run headlights on a car come from" with "Headlights aren't a closed system". On top of that, the title of the question says "on Earth", they're clearly considering the Earth to be the system under consideration. – Acccumulation Oct 17 '18 at 21:35