Consider the potential gravitational energy function
\begin{equation*}
P = - \frac{GmM}{r} = \frac{-GmM}{\sqrt{x^2 + y^2 + z^2}}
\end{equation*}
It would be nice to know how the potential energy changes while moving along a curve \(\gamma\text{.}\) For a particular example, consider a helical path out of the gravity well: \(\gamma(t) = (\sin t, \cos t,
t)\text{.}\) I differentiate along this path using the chain rule.
\begin{align*}
\frac{dP}{dt} \amp = \frac{\del P}{\del x}
\frac{dx}{dt} + \frac{\del P}{\del y} \frac{dy}{dt} +
\frac{\del P}{\del z} \frac{dz}{dt}\\
\amp = \frac{GmMx}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}
\frac{dx}{dt} + \frac{GmMy}{(x^2 + y^2 +
z^2)^{\frac{3}{2}}} \frac{dy}{dt} + \frac{GmMz}{(x^2 +
y^2 + z^2)^{\frac{3}{2}}} \frac{dz}{dt}\\
\amp = \frac{GmM\sin t}{(1 + t^2)^{\frac{3}{2}}} \cos t
+ \frac{GmM\cos t}{(1 + t^2)^{\frac{3}{2}}} (-\sin t) +
\frac{GmMt}{(1 + t^2)^{\frac{3}{2}}} 1\\
\amp = \frac{GmMt}{\sqrt{(1+t^2)^3}}
\end{align*}
Along this helical path, the potential energy is increases. The rate of increase, however, slows over time. This makes sense for leaving a potential energy well: the first steps are more difficult and later movement is not as difficult.
Notice that if I let \(|\gamma(t)| = \sqrt{1 + t^2}\) at the start, I could have written \(P(t) =
\frac{-GmM}{|\gamma(t)|} = \frac{-GmM}{\sqrt{1+t^2}}\) and the \(P^\prime(t) = \frac{GmMt}{\sqrt{(1+t^2)^3}}\) could have been calculated directly. That would have been easier, but its nice to get confirmation that working with the chain rule leads to the right result.
