Change In Total Energy:

Let us consider a two body system of masses \( m_1 \) and \( m_2 \) having position position vectors \( \vec{r_1} \) and \( \vec{r_2} \) respectively with respect to the origin O.

The distance between \( m_1 \) and \( m_2 \) is \( \vec{r}=r\hat{r} \), or, \( r=|\vec{r}|=\left|\vec{r_1}-\vec{r_2}\right| \).

Now the gravitational force of attraction between the two masses is

\( \displaystyle{F(r)\hat{r}=-G\frac{m_1m_2}{r^2}\hat{r}} \)
where, G is the gravitational constant.

If we take magnitude only then we can write the gravitational force is

\( \displaystyle{F(r)=-G\frac{m_1m_2}{r^2}}\tag{1} \)

Now the gravitational potential energy is given by

\( U=-\displaystyle{\int_{\infty}^{r}}F(r)dr \)

or, \( U=\displaystyle{\int_{\infty}^{r}}\frac{Gm_1m_2}{r^2}dr \)

or, \( U=-\left[\frac{Gm_1m_2}{r}\right]_{\infty}^{r} \)

or, \( \displaystyle{U=-G\frac{m_1m_2}{r}}\tag{2} \)

This equation (2) shows that the gravitational potential energy varies inversely with the distance between the two masses.

Now the total kinetic energy of the system = kinetic energy of mass \( m_1 \) + kinetic energy of mass \( m_2 \).

Therefore the total kinetic energy of the system is

\( \frac{1}{2}m_1{\left|\dot{\vec{r_1}}\right|}^2+\frac{1}{2}m_2{\left|\dot{\vec{r_2}}\right|}^2\tag{3} \)

where \( \dot{\vec{r_1}} \) and \( \dot{\vec{r_2}} \) are the velocities of the masses \( m_1 \) and \( m_2 \) respectively.

So the total energy of the system of the two masses is the sum of the potential enegy of interaction due to gravitational force between the two masses and the total kinetic energy of the two masses.

Therefore total energy of the system is given by,

\( E=\frac{1}{2}m_1{\left|\dot{\vec{r_1}}\right|}^2+\frac{1}{2}m_2{\left|\dot{\vec{r_2}}\right|}^2-G\frac{m_1m_2}{r}\tag{4} \)

Now this two body system is reduced into an equivalent one body system such that the reduced mass of the system is

\( \displaystyle{\mu=\frac{m_1m_2}{m_1+m_2}} \)

Putting the value of \( r=|\vec{r}|=\left|\vec{r_1}-\vec{r_2}\right| \) equation (4) becomes,

\( E=\frac{1}{2}m_1{\left|\dot{\vec{r_1}}\right|}^2+\frac{1}{2}m_2{\left|\dot{\vec{r_2}}\right|}^2-G\frac{m_1m_2}{\left|\vec{r_1}-\vec{r_2}\right|}\tag{5} \)

So the total energy of this reduced one body system is the sum of kinetic energy of the effective mass and the potential energy of interaction due to gravitational force.

Therefore,

\( \displaystyle{E’=\frac{1}{2}\mu{\left|\dot{\vec{r}}\right|}^2-\frac{Gm_1m_2}{r}} \)

or, \( \displaystyle{E’=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}{\left|\dot{\vec{r_1}}-\dot{\vec{r_2}}\right|}^2-\frac{Gm_1m_2}{\left|\vec{r_1}-\vec{r_2}\right|}} \)

or, \( \displaystyle{E’=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\left[{|\dot{\vec{r_1}}|}^2+{|\dot{\vec{r_2}}|}^2-2\cdot{|\dot{\vec{r_1}}|}\cdot{|\dot{\vec{r_2}}|}\right]-\frac{Gm_1m_2}{\left|\vec{r_1}-\vec{r_2}\right|}}\tag{6} \)

So the change in total energy is

\( E’-E\\=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\left[{|\dot{\vec{r_1}}|}^2+{|\dot{\vec{r_2}}|}^2-2\cdot{|\dot{\vec{r_1}}|}\cdot{|\dot{\vec{r_2}}|}\right]-\frac{Gm_1m_2}{\left|\vec{r_1}-\vec{r_2}\right|}\\-\frac{1}{2}m_1{\left|\dot{\vec{r_1}}\right|}^2-\frac{1}{2}m_2{\left|\dot{\vec{r_2}}\right|}^2+G\frac{m_1m_2}{\left|\vec{r_1}-\vec{r_2}\right|} \)

= \( -\frac{1}{2}\frac{1}{m_1+m_2}\left[{m_1}^2{|\dot{\vec{r_1}}|}^2+{m_2}^2{|\dot{\vec{r_2}}|}^2+2m_1m_2(\dot{\vec{r_1}}\cdot\dot{\vec{r_2}})\right] \)

= \( -\frac{1}{2}\frac{1}{m_1+m_2}{\left(m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}}\right)}^2 \)

If \( \vec{R} \) be the position vector of the centre of mass of the system of two masses \( m_1 \) and \( m_2 \), then

\( \displaystyle{\vec{R}=\frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}} \)

differentiating both side with respect to time we get,

\( \displaystyle{\vec{V}=\dot{\vec{R}}=\frac{m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}}}{m_1+m_2}} \)

or, \( {\left|m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}}\right|}^2={m_1+m_2}^2{\left|\dot{\vec{R}}\right|}^2 \)

Therefore,

\( \displaystyle{E’-E=-\frac{1}{2}(m_1+m_2){\left|\dot{\vec{R}}\right|}^2} \)

Change In Angular Momentum:

Let \( p_1=m_1\dot{\vec{r_1}} \) and \( p_2=m_2\dot{\vec{r_2}} \) be the linear momentum of the masses \( m_1 \) and \( m_2 \) respectively, then the total angular momentum of the system is

\( \vec{J}=\vec{r_1}\times{p_1}+\vec{r_2}\times{p_2} \)

or, \( \vec{J}=\vec{r_1}\times{m_1\dot{\vec{r_1}}}+\vec{r_2}\times{m_2\dot{\vec{r_2}}} \)

Now the angular momentum of the equivalent one body system is given by,

\( \vec{J’}=\vec{r}\times(\mu\dot{\vec{r}}) \)

or, \( \vec{J’}=(\vec{r_1}-\vec{r_2})\times\frac{m_1m_2}{m_1+m_2}\left(\dot{\vec{r_1}}-\dot{\vec{r_2}}\right) \)

or, \( \vec{J’}=\frac{m_1m_2}{m_1+m_2}\left(\vec{r_1}\times\dot{\vec{r_1}}-\vec{r_1}\times\dot{\vec{r_2}}-\vec{r_2}\times\dot{\vec{r_1}}+\vec{r_2}\times\dot{\vec{r_2}}\right) \)

Therefore the change in angular momentum is given by,

\( \vec{J’}-\vec{J} \)

= \( \frac{m_1m_2}{m_1+m_2}\left(\vec{r_1}\times\dot{\vec{r_1}}-\vec{r_1}\times\dot{\vec{r_2}}-\vec{r_2}\times\dot{\vec{r_1}}+\vec{r_2}\times\dot{\vec{r_2}}\right)-\left(\vec{r_1}\times{m_1\dot{\vec{r_1}}}+\vec{r_2}\times{m_2\dot{\vec{r_2}}}\right) \)

= \( \frac{m_1m_2}{m_1+m_2}\left(\vec{r_1}\times\dot{\vec{r_1}}-\vec{r_1}\times\dot{\vec{r_2}}-\vec{r_2}\times\dot{\vec{r_1}}+\vec{r_2}\times\dot{\vec{r_2}}\right)-\left(m_1\vec{r_1}\times{\dot{\vec{r_1}}}+m_2\vec{r_2}\times{\dot{\vec{r_2}}}\right) \)

= \( -\frac{m_1m_2}{m_1+m_2}\left(\vec{r_1}\times\dot{\vec{r_2}}-\vec{r_2}\times\dot{\vec{r_1}}\right)\\+\left(\frac{m_1m_2}{m_1+m_2}-m_1\right)\vec{r_1}\times\dot{\vec{r_1}}\\+\left(\frac{m_1m_2}{m_1+m_2}-m_2\right)\vec{r_2}\times\dot{\vec{r_2}} \)

= \( \frac{1}{m_1+m_2}\left[-m_1m_2\vec{r_1}\times\dot{\vec{r_2}}-m_1m_2\vec{r_2}\times\dot{\vec{r_1}}\\+(m_1m_2-m_1^2-m_1m_2)\vec{r_1}\times\dot{\vec{r_1}}\\+(m_1m_2-m_1m_2-m_2^2)\vec{r_2}\times\dot{\vec{r_2}}\right] \)

= \( \frac{1}{m_1+m_2}\left[-m_1m_2\vec{r_1}\times\dot{\vec{r_2}}-m_1m_2\vec{r_2}\times\dot{\vec{r_1}}-m_1^2\vec{r_1}\times\dot{\vec{r_1}}m_2^2\vec{r_2}\times\dot{\vec{r_2}}\right] \)

= \( -\frac{1}{m_1+m_2}\left[m_1\vec{r_1}\times{m_2}\dot{\vec{r_2}}+m_2\vec{r_2}\times{m_1}\dot{\vec{r_1}}+m_1\vec{r_1}\times{m_1}\dot{\vec{r_1}}+m_2\vec{r_2}\times{m_2}\dot{\vec{r_2}}\right] \)

= \( -\frac{1}{m_1+m_2}\left[(m_1\vec{r_1}+m_2\vec{r_2})\times(m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}})\right] \)

Now the angular momentum of the centre of mass is given by,

\( \vec{R}\times(m_1+m_2)\dot{\vec{R}} \)

= \( \frac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}\times(m_1+m_2)\frac{m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}}}{m_1+m_2} \)

= \( \frac{1}{m_1+m_2}\left[(m_1\vec{r_1}+m_2\vec{r_2})\times(m_1\dot{\vec{r_1}}+m_2\dot{\vec{r_2}})\right] \)

So the angular momentum of the system of two masses \( m_1 \) and \( m_2 \) decreases by an amount equal to the angular momentum of the centre of mass.