Explanation: often steps are skipped in conveying a derivation. To facilitate formal validation, the implicit steps need to be documented. Who does this work: human content creator Motive for doing this work: clearer explanation of mathematical steps. Facilitates student learning and reproducibility in a research context. 
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source: "Derivation of Gravitational Potential Energy" by Rhett Allain
Suppose an object starts an infinite distance from a moon and is dropped, falling towards the moon due to gravitational acceleration. What is the speed of the object when it is distance \(r\) from the moon?
The initial conditions are \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} The work is calculated using W_{\rm to\ system} = \(\Delta E\) since the force changes. To find the cumulative work done on the object, integrate over all positions between \(\infty\) and \(r\) \begin{equation} W_{\rm to\ system} = \int_{\infty}^r \vec{F}\cdot d\vec{r} \label{eq:work as function of force} \end{equation} Substituting the gravitational force into Eq. \ref{eq:work as function of force}, \begin{equation} W_{\rm to\ system} = \int_{\infty}^r \frac{G m_1 m_2}{x^2} dx \end{equation} Factor out the constants, \begin{equation} W_{\rm to\ system} = G m_1 m_2\int_{\infty}^r \frac{1}{x^2} dx \end{equation} \begin{equation} W = \frac{G m_1 m_2}{r} \end{equation} Another definition of work is that it is the change in energy for a system: \(W_{\rm by\ system} = \Delta KE \). Because the initial velocity was zero, the work here is \begin{equation} W_{\rm by\ system} = \frac{1}{2} m_1 v^2 \end{equation} Thus we can combine the two definitions of work to get \begin{equation} \frac{1}{2} m_1 v^2 = \frac{G m_1 m_2}{r} \end{equation} The \(m_1\) cancels, leaving \begin{equation} v(r) = \sqrt{\frac{2Gm_2}{r}} \end{equation}