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source: "Derivation of Gravitational Potential Energy" by Rhett Allain

scenario description Suppose an object starts an infinite distance from a moon and is dropped, falling towards the moon due to gravitational acceleration.
question What is the speed of the object when it is distance \(r\) from the moon?
figure of scenario
Figure 1: small mass falling towards a moon from initial position at infinity.

initial condition The initial condition is \begin{equation} v(x=\infty) =0 \label{eq:initial_velocity} \end{equation} The force acting on the object is \begin{equation} \vec{F} = \frac{-G m_1 m_2}{x^2} \hat{x} \label{eq:gravitational force} \end{equation}
step The work is calculated using W = \(\Delta E\) since the force changes. To find the cumulative work done on the object, integrate over all positions between \(\infty\) and \(r\) \begin{equation} W = \int_{\infty}^r \vec{F}\cdot d\vec{r} \label{eq:work as function of force} \end{equation}
step Substituting the gravitational force into Eq. \ref{eq:work as function of force}, \begin{equation} W = \int_{\infty}^r \frac{-G m_1 m_2}{x^2} dx \end{equation}
step Factor out the constants, \begin{equation} W = -G m_1 m_2\int_{\infty}^r \frac{1}{x^2} dx \end{equation}
step which leads to \begin{equation} W = \frac{G m_1 m_2}{r} \end{equation}
step Another definition of work is that it is the change in energy for a system: \(W = \Delta E \) Because the initial velocity was zero, the work here is \begin{equation} W = \Delta KE \end{equation} Thus we can combine the two definitions of work to get \begin{equation} W = \frac{1}{2} m_1 v^2 = \frac{G m_1 m_2}{r} \end{equation}
step The \(m_1\) cancels, leaving \begin{equation} v(r) = \sqrt{\frac{2Gm_2}{r}} \end{equation}