gravity - Angular Momentum Conservation in Gravitational Interaction

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.