homework and exercises - Is my derivation of the potential energy formula $m*g*h$ correct?

I've just wondered where the formula $E_{pot} = mgh$ you learn at school comes from so I've tried to work it out - is my reasoning correct? The change in energy is given by

$$\Delta E=\int_{e}^{e+h}G\frac{mM}{r^2}dr,$$ where $e$ is the radius of earth. The integral is equal to $$\Delta E=\left [-G\frac{mM}{r} \right ]_e^{e+h}.$$ Because $GM=ar^2$, $$\Delta E = \frac{mae^2}{e}-\frac{ma(e+h)^2}{e+h}.$$ On earth the acceleration is $g$ and because of that $$\Delta E=mge-mge+mgh=mgh.$$

But I've heard that $mgh$ is only an approximation if the change in height is approximatly constant - so why does my derivation equal to $mgh$ no matter how big the change in height is? Do I need to integrate with respect to acceleration from the start?

Answer

It looks like you figured out your mistake in the comments; here's the correct derivation. Let's start with your expression

$$\Delta E = \left[-\frac{GMm}{r}\right]^{e+h}_e = GMm \left(\frac{1}{e} - \frac{1}{e+h}\right),$$

which is the correct general expression for the change in gravitational potential energy outside a spherically symmetric body. If $h \ll e$, then we may expand

$$\frac{1}{e} - \frac{1}{e+h} = \frac{1}{e}\left[1-\frac{1}{1+h/e}\right] \approx \frac{1}{e}\left[1 - \left(1-\frac{h}{e} + \cdots\right)\right] = \frac{h}{e^2} + \cdots,$$

where we used the binomial expansion on $1/(1+h/e) = (1+h/e)^{-1} = 1-h/e + \cdots$, where the dots are terms that go like $(h/e)^2$. Plugging this back in, we get

$$\Delta E \approx \frac{GMmh}{e^2} = \frac{GM}{e^2} mh = mgh,$$

where we recognize that the gravitational acceleration at the surface of the Earth is $g = GM/e^2$. This is the expression you're looking for.

(For extra fun: if you keep the next term in the binomial expansion, i.e. the term that goes like $(h/e)^2$, you can show that if an oblong object like a rod is free to pivot about its center of mass, it will ever so slightly prefer to hang up-and-down rather than horizontally. This effect is partly responsible for the phenomenon of tidal locking, wherein, say, the rotation of the moon gets synced with that of the Earth, so that we only ever see the same side of the moon.)