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I am self studying the two body problem and I'm stuck on the following:

I have given $$\ddot{\vec{x}}_1= - G m_2 \frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|^3}$$ and $$\ddot{\vec{x}}_2= - G m_1 \frac{\vec{x}_2-\vec{x}_1}{|\vec{x}_1-\vec{x}_2|^3}$$ with masses $m_1,m_2$ at the position $x_1,x_2$ and gravitational constant $G$.

How can you show only with these two equations that total momentum $m_1\dot{\vec x_1}+m_2\dot{\vec x_2}$ is constant in time?

And why is the total mechanical energy $$\frac12m_1|\dot{\vec x_1}|^2+\frac12m_2|\dot{\vec x_2}|^2-\frac{Gm_1m_2}{|\vec{x}_1-\vec{x}_2|}$$ conserved?

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2 Answers 2

OK, the first part: net force $\vec{F}=m_1 \ddot{\vec{x}}_1+m_2 \ddot{\vec{x}}_2=\vec{0}$, so by newton's second law total momentum is conserved(or force is rate of change of momentum, if total force is zero, total momentum doesn't change), and the second part: there is no sort of non conservative force acting, hence mechanical energy is conserved.

Non conservative-? That's a force which does non zero work in round trips. Say, take friction. You drag a block on a rough floor, then the total work is $-fl+(-fl)$ where $l$ is one way length. Try that with gravitation. Total work is zero then. That's why gravitation is conservative. Remember as a rule of thumb, only if NC forces act, then mechanical energy changes, otherwise it's constant.

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is there a way for the first part by just using the two equations without any other tools? –  peterson Apr 22 '13 at 20:45
    
@peterson: I don't understand what tools you are referring to. Whatever I said were Newtonian basics.$\vec{F}=m_1 \ddot{\vec{x}}_1+m_2 \ddot{\vec{x}}_2={\vec{dp} \over dt}=\vec{0}$. This implies $\vec{p}$ is constant. –  Ashish Gaurav Apr 24 '13 at 5:09
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Ashish already essentially said this, but using only the equations of motion we can show conservation of total momentum with the following calculation: \begin{align} \frac{d}{dt}(m_1\dot{\vec x_1} +m_2\dot{\vec x_2}) &= m_1\ddot{\vec x_1} +m_2\ddot{\vec x_2} \\ &= m_1\left(- G m_2 \frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|^3}\right) + m_2\left(- G m_1 \frac{\vec{x}_2-\vec{x}_1}{|\vec{x}_1-\vec{x}_2|^3}\right) \\ &= \frac{Gm_1m_2}{|\vec{x}_1-\vec{x}_2|^3}\Big[-(\vec x_1 - \vec x_2) - (\vec x_2-\vec x_1)\Big] \\ &=0 \end{align} and conservation of energy proceeds similarly. Just take the time-derivative of the total energy expression, do some manipulations using the equations of motion, and you'll get zero.

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