Gravitational Potential and Center of Mass

\begin{equation} \vec{F_g} = - \frac{GM_1M_2}{r^2} \hat{r} \end{equation}

where, \(\vec{F_g}\) is the force of gravity on \(M_2\); \(M_1\) and \(M_2\) are two point masses; \(G\) the universal gravitation constant; \(r\) the magnitude of the vector \(\vec{r}\) from \(M_1\) to \(M_2\) and \(\hat{r}\) the unit vector in the \(\vec{r}\) direction.

To deduct the escape velocity of the Earth, our object \(M_2\) must do enough work such that the work Earth exerts upon it — as deducted above — is canceled out.

The net force of a system is described as:

\begin{equation} \sum^n_{i=1} \vec{F_{net,i}} = (\sum^n_{i=1} m_i) \ddot{\vec{r_{CM}}} \end{equation}

where, \(\vec{F_{net,i}}\) is the net force on \(m_i\), \(\vec{r_{CM}}\) the position vector of the center of mass \(CM\).

In order to isolate \(\vec{r_{CM}}\), the equation above must be integrated twice w.r.t. time, namely:

\begin{equation} \int \sum^n_{i=1} \vec{F_{net,i}} dt = \int (\sum^n_{i=1} m_i) \ddot{\vec{r_{CM}}} dt \end{equation}

with the integrals performed twice.

For this evaluation, we will also set \(\sum^n_{i=1} m_i = M\), define \(\vec{a_i}\) as acceleration of \(m_i\), \(\vec{r_i}\) as position of \(m_i\), and apply Newton's second law:

We first perform the first integration:

\begin{align} \int \sum^n_{i=1} \vec{F_{net,i}} dt &= \int (\sum^n_{i=1} m_i) \ddot{\vec{r_{CM}}} dt \\ \sum^n_{i=1} m_i \int \vec{a_i} dt &= M \int \ddot{\vec{r_{CM}}} dt \\ \sum^n_{i=1} m_i \int \frac{d^2\vec{r_i}}{dt^2} dt &= M \int \frac{d^2\vec{r_{CM}}}{dt^2} dt \\ \sum^n_{i=1} m_i \frac{d\vec{r_i}}{dt} &= M \frac{d\vec{r_{CM}}}{dt} + C_1 \\ \end{align}

Given the statement for velocity, we need to integrate again to figure position.

\begin{align} \int (\sum^n_{i=1} m_i \frac{d\vec{r_i}}{dt}) dt &= \int M \frac{d\vec{r_{CM}}}{dt} + C_1 dt \\ \sum^n_{i=1} m_i \int \frac{d\vec{r_i}}{dt} dt &= M \int \frac{d\vec{r_{CM}}}{dt} + C_1 dt \\ \sum^n_{i=1} m_i \vec{r_i} + C_2 &= M \vec{r_{CM}} \\ \frac{1}{M} \sum^n_{i=1} m_i \vec{r_i} + C_2 &= \vec{r_{CM}} \end{align}

In the case of the first integral application, the constant of integration — \(C_1\) — is defined to be zero due to the fact that, were it to be nonzero, the center of mass would be traveling at a faster rate than that of the object. As the object increases in speed, the center of mass would therefore increase in speed as well more than the object: leading it to travel away from the object, which would be absurd.

In the latter case, the constant of integration — \(C_2\) — is equally defined as zero due to the fact that, were the value to be nonzero, the center of mass would perhaps be consistently ahead or behind from the object, which would equally be absurd.

Collecting all constants, as both constants of integration is defined to be at zero, we could set \(C=0\). As such, the position \(\vec{r_{CM}}\) of the center of mass \(CM\) is therefore:

\begin{equation} \vec{r_{CM}} = \frac{1}{M} \sum^n_{i=1} m_i \vec{r_i} \end{equation}

Where, \(M\) is the total mass of the system, \(m_i\) the mass of component \(i\) of the system, and \(\vec{r_i}\) the position of \(m_i\).

Center of mass can also be defined for a more general system, as follows:

\begin{equation} \vec{r_{CM}} = \frac{\int y dm}{\int dm} \end{equation}

that, the center of mass is the position at which the differential mass-weighted sum is the same.

Because of Newton's Third Law, if forces acted internal to a system, the target of the force — given its inside the system — will preform an equal and opposite reaction. Hence, the net internal force will be \(0\), meaning that we could ignore internal forces.

Because of the fact that \(\vec{r_{CM}} = \frac{1}{M} \sum^n_{i=1} m_i \vec{r_i}\) as derived above, \(\vec{r_{CM}}\) changes when the system as a whole moves. However, internal forces does not do this, meaning the existence of internal forces does not change \(\vec{r_{CM}}\).

This fact allows for a simplification of the equation:

\begin{equation} \sum^n_{i=1} \vec{F_{net,i}} = (\sum^n_{i=1} m_i) \ddot{\vec{r_{CM}}} \end{equation}

to:

\begin{equation} \sum^m_{j=1} \vec{F_{ext,j}} = M \ddot{\vec{r_{CM}}} \end{equation}

by applying \(\sum^n_{i=1} m_i = M\) as per aforementioned and the external forces argument above.