# If angular momentum is conserved, what's wrong with this scenario?

I understand that, in a given system with no external torques applied, angular momentum is conserved. However, consider the following situation:

Let's say the earth revolves around the Sun in a circle. My reference point will be one point on the circle. When the earth is, say on the other side of the circle from my reference point, the angular momentum is non-zero. However, at the point when the earth crosses my reference point, the angular momentum is zero (since angular momentum is position cross momentum, and the position with respect to my reference point is zero).

But I thought it was conserved?

Is there some external torque I'm forgetting to include?

• What about the torque from the sun? – Prahar Nov 26 '14 at 23:46
• There is a torque from the sun (obviously) or we cant write a net torque equation. I just assumed that while writing my answer. Because the earth approaches the reference point, the reference point in this scenario is not fixed but instead keeps moving. – Sreram Nov 27 '14 at 16:59
• The torque around your arbitrary reference point is probably not zero. – Bryson S. Nov 27 '14 at 18:19

In your problem, "Earth" is not an isolated system. The combined "Sun-Earth" system, however, is, so we can know that the angular momentum of the Sun-Earth system is conserved. As the earth's mass is accelerating the sun, you have to take its angular momentum into account as well.

While the mass and size of the sun mean that we can ignore its motion with respect to the rest of the solar system, you can't do that for your calculation.

Alternatively, you can consider the sun's gravitational force on the earth to be a torque in your case because the force does not go through your reference point.

• -1. The Earth most certainly is a system because one can draw one's system boundary wherever one pleases. Some system boundaries are "better" (easier to work with) than are others (and the same goes for choice of reference frame). Change "system" to "isolated system" and I'll retract my downvote. – David Hammen Nov 27 '14 at 12:23
• I agree, that change would improve the answer. (I don't think it's worthy of a downvote for just that reason, but that's just me.) – David Z Nov 27 '14 at 13:37
• @DavidHammen I believe 10k rep is more than enough to edit a post, so why not just edit it and make it correct instead of downvoting? Even though it's one word, it's far from trivial since it changes the correctness of the answer. But it doesn't change the spirit of the answer, just makes it more correct. – corsiKa Nov 27 '14 at 15:45
• Because it is the answerer's answer and @DavidHammen respected their integrity. Edits are meant for editorial changes, not for correcting or guessing what the author meant. Comments are for pointing this out. Otherwise it should be a wiki answer. – luk32 Nov 27 '14 at 17:48
• Sorry luk32, but that isn't accurate. All the content posted to Stack Exchange is creative commons licensed (so it isn't "the answerer's answer") and is expected to change over time. Our real goal is to provide expert answers to real questions people to help people solve problems. High rep users are afforded the discretion to edit posts at will to that end. The real integrity to respect is giving people the best possible answer. =) If the edits are incorrect, they will either be further edited or rolled back. And wiki answer? I don't see how that plays in at all! – corsiKa Nov 27 '14 at 17:58

You are a point in the circle. The torque is: $$\mathbf\tau = \mathbf r\times\mathbf F$$

Where $\mathbf r$ is the position of Earth and $\mathbf F$ is the force (radially towards the sun). Notice when your reference point is somewhere in Earth's orbit, as you said, and your object is earth, the force will not be parallel to the position. Therefore, the cross product is non-null. And hence $\mathbf\tau\ne\mathbf 0$.

If torque is not null, there's no conservation of angular momentum.

Note: If your reference point is the sun, then $\mathbf r \parallel\mathbf F$ and therefore $\mathbf\tau = \mathbf r\times\mathbf F = \mathbf 0$. Conclusion: In this case angular momentum is conserved.

• This is true if you neglect the action-reaction law, but if you do take that into account you also have to include the torque from the Earth's gravitational force on the sun, and when you do that, the total torque is zero. So I think BowlOfRed's answer is more applicable to the real situation. – David Z Nov 27 '14 at 6:51
• @DavidZ - I disagree. IMO, this is a better answer than BowlofRed's because it directly answers the question. The question did not ask about the angular momentum of the Sun-Earth system. It asked about the angular momentum of the Earth, period. That latter quantity is not conserved from the perspective of the frame chosen by the OP for precisely the reasons given in this answer. +1. – David Hammen Nov 27 '14 at 12:19
• @DavidHammen Well, the question asked about a system with no external torques applied, which is not what this answer is describing. Basically, the question is internally inconsistent (not saying that's a bad thing) and there are two options for which premise to discard to make it consistent, corresponding to this answer and BowlOfRed's. I happen to think the other way gives a more useful explanation. – David Z Nov 27 '14 at 13:35
• @DavidZ From the perspective of an arbitrary inertial frame of reference, the Earth generally does have an external torque applied to it. The questioner set up a false premise. – David Hammen Nov 27 '14 at 15:37

before I answer your question, let me put forward another scenario. Imagine a mass less rod with two massive spherical objects (of mass $m$) is placed at either of its ends. Further imagine that the rod rotates about an axis. Let's define the axis to be the center of the rod. Let's choose the reference point at a distance $x$ from the axis and on the rod. now if we compute the torque of the nearest point mass from the reference point, we get $\mathbf t_1=\mathbf x\times \mathbf f_1$. but because the net torque is zero, the torque of the point mass away from the reference point (having distance $(l-x)$ if $l$ is the length of the rod) is $t_2(l-x)\mathbf r\times \mathbf f_2$(where $\mathbf r$ is the unit vector along the direction of the other mass).

we always have $sin(\pi/2) = 1$, so we can simply write $x\times f_1=(l-x)\times f_2$ as the angular momentum is conserved.

Now lets move further away from the axis towards the nearer point mass. we will find that the force $f_2$ keeps decreasing in magnitude and the force $\mathbf f_1$ keeps increasing (we can explain that by writing $f_1=((l-x)/x) \times f_2$ and $x$ keeps decreasing and tries to reach zero).

Now lets consider the limiting cases: $lim_{x\to0}f_1(x)=\infty$ To be precise $f_1\to\infty$ instead of actually being infinity. So hence, you have a finite value for the product $\mathbf t_1=\mathbf x\times \mathbf f_1$ even if $\mathbf x$ tends to zero (as a product of a infinitesimal quantity with a very large quantity gives a finite value).

Let's now consider the case where $\mathbf x$ is actually zero and not that its limit approaches zero. in this case, the force becomes plainly undefined in the set of real numbers and such torque does not exist (how can we say that there is some torque if the force acts on the reference point? it doesn't seem right!). This scenario contradicts our definition that torque is defined in two dimensional or three dimensional (more than one dimensions are required to compute torque) space (as many other physical quantities are); if the value of $x$ is zero the reference point and the position of the point mass would perfectly coincide, and that forms a 0 Dimentional system. We cannot compute torque in an zero dimensional system hence it is always assumed that the distance is only nearly zero and not perfectly zero.

The same case applies to you question; in your scenario, you have chosen the reference point away from the line joining the earth and the sun, on the earth's orbit. but in your case we won't always get $sin\theta=1$ that makes calculations "little" complicated, so hence I chose a similar but simpler scenario. note:- $\mathbf torque\times\mathbf dt=$ differential change in angular momentum, so without loss of generality we can say that this scenario is similar to yours.

About your reference point, torque acting on earth is not zero. But total torque of sun and earth is zero. That's why total angular momentum of earth and sun remains conserved.