Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question is an exact duplicate of:

thanks for any help.

I'm trying to show that in a 2body problem, angular momentum is conserved given that $\dfrac{dp}{dt}=\dfrac{-GMm(rv)}{r³}$, where p is momentum, t time, G gravitational constant, M mass of 1 object, m mass of the other, (rv) is the vector between them and r is the magnitude of the vector between them.

I've had a lot of attempts but I don't seem to get very far.

Where L is angular momentum, $r_1$ is position vector of mass M and $r_2$ is position vector of mass m, $p_1$ is momentum of mass M and $p_2$ is momentum of mass m.

L=$r_1Xp_1+r_2Xp_2$

If $\dfrac{dL}{dt}=0$ then dL will be conserved over time, applying product rule $L=\dfrac{dr_1}{dt}Xp_1+r_1X\dfrac{dp_1}{dt}+\dfrac{dr_2}{dt}Xp_2+r_2X\dfrac{dp_2}{dt}$

noticing that $p_1=m\dfrac{dr_1}{dt}$ and $p_2=m\dfrac{dr_2}{dt}$ clearly $p_1$ is parallel to$ \dfrac{dr_1}{dt}$ and $p_2$ is parallel to $\dfrac{dr_2}{dt}$ and hence when cross producted with them, the result is 0.

Giving $L=r_1X\dfrac{dp_1}{dt}+r_2X\dfrac{dp_2}{dt}$

I gather I must now show that either these two terms are 0, or they are equal and opposite. I can't see any reason why they would be 0, and (slightly guessing) it seems to make sense to me that they are terms related to how the angular momentum of one body changes with the other body, and I assume they must therefore be equal and opposite. However when I do

$r_1X\dfrac{dp_1}{dt}=-r_2X\dfrac{dp_2}{dt}$ I notice $\dfrac{dp_1}{dt}=\dfrac{-GMm(rv)}{r³}$ and $\dfrac{dp_2}{dt}=\dfrac{GMm(rv)}{r³}$ (since (rv)'s direction is flipped) Therefore $r_1X\dfrac{-GMm(rv)}{r³}=-\dfrac{r_2XGMm(rv)}{r³}$ Defining $w=\dfrac{GMm}{r³}$:

$r_1X(rv)(-w)=r_2X(rv)(-w)$

Scalar products are commutative so the (-w)'s easily cancel giving

$r_1X(rv)=r_2X(rv)$ Which then the only solutions are if all components of $r_1$ are equal to all components of $r_2$, however angular momentum should clearly be conserved in all positions, not just when the masses are ontop of each other.

Thanks again for any help.

share|improve this question

migrated from stackoverflow.com Oct 4 '13 at 3:12

This question came from our site for professional and enthusiast programmers.

marked as duplicate by Qmechanic Oct 7 '13 at 18:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
... but, I didn't read your entire question but gravity simply conserves angular momentum because it's a radial force. That is, break p into radial and tangential components: L is only determined by the tangential p and gravity only changes the radial p. –  tom10 Oct 2 '13 at 3:50
1  
Since $dp1/dt = -dp2/dt$, if you simply replace $dp2/dt$ with $-dp1/dt$ in the expression for $dL/dt$, you end up with $dL/dt = const * (r1 - r2) x rv = 0$. Q.E.D. –  Hristo Iliev Oct 2 '13 at 14:18
add comment

1 Answer 1

This equation:

$$\mathbf{r}_1\times \mathbf{rv}=\mathbf{r}_1\times \mathbf{rv}$$

has solutions other than $\mathbf{r}_1=\mathbf{r_2}$. It is sufficient that (as in this case) $\mathbf{r}_1=\mathbf{r_2}\pm \mathbf{rv}$.

Also, look at your own strategy: you assume what you're trying to prove, then see where it leads. If it leads to a contradiction, you have a proof that the premise is false. (If your conclusion $-$ that $L$ is conserved only if the masses are collocated $-$ had been correct, that would have been a lovely disproof of conservation of $L$.) But if the conclusion is true, all you have is the hope that all of your steps are invertible, so that you can work your way back along the chain and prove the premise.

share|improve this answer
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.