today i was curious about the potential energy, so, i started studying the Newton's Law of Universal Gravitation which its equation is \begin{eqnarray} U= -\frac{GMm}{r}.\end{eqnarray} Well, since i gotthis equation i really know how to reach at this point ( due some experiments, also some documents help me to get until here), i was very excited about reaching this point, and i told to myself : "Alright, good job." the excitement passed away when a friend told me that i can transform this equation into:\begin{eqnarray} U= m.g.h.\end{eqnarray} Sadly, i don't even know how to start with,i know that The potential should be an approximation for the general potential energy when r = r (of earth), but the problem is... Equation 1 Scales with radius and the other one with height.
1 Answers
Write $r=R+h$ where $r$ is the distance from the center of the Earth, $R$ is the radius of the Earth, and $h$ is height above the surface. Then the difference between the potential energy at height $h$ above the surface and the PE at the surface is
$$\begin{align} U(r)-U(R)&=-\frac{GMm}{r}+\frac{GMm}{R}\\ &=\frac{GMm}{R}-\frac{GMm}{R+h}\\ &=\frac{GMm}{R}\left[1-\left(1+\frac{h}{R}\right)^{-1}\right] \end{align}.$$
For $h<<R$, use the Taylor series
$$(1+x)^{-1}=1-x+O(x^2)$$
to get
$$\begin{align} U(r)-U(R)&=\frac{GMm}{R}\left[\frac{h}{R}+O\left(\frac{h^2}{R^2}\right)\right]\\ &\approx m\left(\frac{GM}{R^2}\right)h\\ &=mgh \end{align}.$$
So: Taking the PE outside the surface as $-GMm/r$, it is is zero at infinity and negative (namely $-GMm/R$) at the surface. Slightly above the surface, it is slightly less negative by $mgh$. (By the way, inside the Earth, $-GMm/r$ doesn’t apply.) In Newtonian physics it doesn’t really matter where you take the PE to be zero, so you can take it to be zero at the surface and say that it simply is $mgh$ for $h<<R$.
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