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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.

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Welcome to physics.SE. You will be much more likely to get responses to your questions if you format your math properly: physics.stackexchange.com/help/notation –  Ben Crowell Oct 1 '13 at 22:39
    
Does this SE answer give you what you want? physics.stackexchange.com/q/29737 –  BMS Oct 1 '13 at 22:41
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The two-body problem reduces to a single-body problem, because the center of mass doesn't accelerate. In the two-body problem, the two angular momenta aren't equal, and they don't cancel. They're separately conserved, since each body is acted on by a central force that produces no torque. This sounds like homework, so I'm going to put the homework tag on it. If it's not homework, please remove the tag. If I'm right and it is homework, then please add the tag yourself in the future. –  Ben Crowell Oct 1 '13 at 23:24
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Hi user2836596. It seems that @Ben Crowell is correct in adding the homework tag. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. –  Qmechanic Oct 2 '13 at 7:54
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I don't want to get in an edit war over this homework tag, but user2836596, rather than just reverting my addition of the tag without explanation, it would have been better to explain what's up. Phrases like "I gather I must now show that ..." sound like this is homework. –  Ben Crowell Oct 2 '13 at 15:43

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