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The earth orbiting around the sun has an angular momentum. But at one moment of time, each atom on earth is moving translationally, and the combined linear momenta of all the particles on earth would equal $MV$ where $M$ is the mass of earth and $V$ the velocity of earth tangential to the rotational axis.

Then how is angular momentum not a "simplification" of calculating the linear momentum of every atom? This is like the moment of inertia being a "simplification" of adding up the force it would take to move every atom in a object (with different velocities as an object farther from the axis would need more force to get to that speed) in the direction tangent to the axis.

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  • $\begingroup$ Adding forces acting on different objects (atoms in this example) do not make any sense. $\endgroup$ – user126422 Jan 15 '17 at 0:26
  • $\begingroup$ Why should it be a simplification? Why isn't it okay, that angular momentum is just something different than linear momentum? Just like the speed and any other property is different from angular momentum, then linear momentum is also something different. PS: to answer the title question, just look at the units. They aren't the same, so those two parameters aren't the same. $\endgroup$ – Steeven Jan 15 '17 at 2:21
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    $\begingroup$ Duplicate/related? physics.stackexchange.com/q/224545/104696 The problem with answering this question is deciding at which level the answer is to be given. $\endgroup$ – Farcher Jan 15 '17 at 5:00
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Angular momentum and linear momentum are fundamentally different quantities. At a basic level, they are generators/charges of two very different symmetries and are thus, conserved, according to Noether theorem, under two very different symmetries. In particular, angular momenta are the generators of the rotational symmetry, i.e. $SO(d)$ in a $d-$dimensional space while the linear momenta are the generators of the translational symmetry. The two symmetry groups are absolutely distinct, to wit, one (the group of rotations) is compact while the other (the group of translations) is not. To make the distinction even sharper, in $d$ dimensions, one would have $d$ linear momenta while $\frac{d(d-1)}{2}$ angular momenta. It is just a coincidence that in $d=3$, both those expressions yield $3$ which has the potential of creating confusion that they both might be representing the same quantity.

Coming to the specific situation OP raises, one cannot ascertain the linear momentum of the particle simply from knowing its angular momentum. Again, to wit, the angular momentum is conserved in a Kepler problem (actually, something even better is conserved in a Kepler problem but I won't go there for the moment) and thus, it simply never changes throughout the motion. But, the linear momentum continuously changes as the particle orbits around the focus of its orbit.

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Angular momentum is just linear momentum at a distance - i.e. a radius.

If a ball is thrown past you at a distance, with respect to you it has angular momentum. It only travels around you if it is attached to you by a rope or some gravity. But you don't need that constraint to have angular momentum.

Don't worry about all the particles making up a mass. Just consider the mass as a whole.

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    $\begingroup$ "Angular momentum is just linear momentum at a distance - i.e. a radius."--while this is technically true, it seems to boost the misunderstanding of the OP that angular momentum is not an individually important and distinct quantity than the linear momentum. In particular, one can be conserved independent of the other. Or, more fundamentally, they are charges of two completely distinct symmetries. To make the distinction sharp, in $d$ dimensions, one would have $d$ linear momenta while $\frac{d(d-1)}{2}$ angular momenta. It is just a coincidence that in $d=3$, both those expressions yield $3$. $\endgroup$ – Dvij Mankad Mar 6 at 10:09
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When an object moves in a circular fashion around the centre of force the linear momentum is directed as a tangent to the path and the angular momentum defined as perpendicular to both the radial vector and the linear momentum at any instant.

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  • $\begingroup$ -1: This doesn't seem to address the question that OP raises. $\endgroup$ – Dvij Mankad Mar 6 at 10:46
  • $\begingroup$ The answer is also unintelligible. $\endgroup$ – David White Aug 7 at 23:03

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