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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?

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  • $\begingroup$ Yes, it is an approximation, for the case e<<r. What do you mean by GM=ar^2? What is r? GM is constant, isn't it? $\endgroup$
    – nasu
    Feb 22, 2017 at 21:59
  • $\begingroup$ My reasoning was that $F=ma=G\frac {Mm}{r^2}\Rightarrow ar^2=GM$ so I've substituted that into the equation and put $r$ for $e$ and $e+h$, I guess there's the flaw of my derivation? $\endgroup$ Feb 22, 2017 at 22:05
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    $\begingroup$ No acceleration is constant only if you assume r is constant. So your g here is not constant. The g you are talking about is obtained for situations near earths surface, when you can say Rearth=r. Than you get constant g. Your g is not constant...it is GM/e at one place and GM/(e+h) at another. mge-mge substraction is therefore illegal! $\endgroup$ Feb 22, 2017 at 22:11
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    $\begingroup$ But the acceleration (either you call it a or g) is not the same. You assume that g is constant and this is not the case. It it were, you don't need that integral. The formula mgh is valid for uniform gravitational field (constant g) which is an approximation of the real gravitational field of the Earth exactly for the case e<<r. $\endgroup$
    – nasu
    Feb 22, 2017 at 22:13

3 Answers 3

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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.)

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  • $\begingroup$ Nice addition regarding the tidal locking $\endgroup$ Feb 22, 2017 at 22:25
  • $\begingroup$ That fact about tidal locking is very interesting I'll have to read more about it! $\endgroup$ Feb 22, 2017 at 22:28
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Good question! You have committed the formal fallacy of "equivocation", which is a fancy word for "calling two different things (in this case, numbers) by the same name (in this case, $a$)."

So $a = GM/r^2$ is really a bunch of different numbers, one for each possible radius. When you pretend that you get the same $a$ for $GM/e^2$ that you got for $GM/(e + h)^2$, that is where you commit the fallacy. Let me call the first $a$ as $a_0$ and the second as $a_+$; you just wrote $m a_0 e^2/e - m a_+ (e + h)^2/(e+h) = m a_0 e - m a_+ (e + h),$ then you replaces both $a_0$ and $a_+$ with $g$, which is only approximately true if $h$ is small.

In fact if you want this as a series expansion, it is not very difficult. The first step is to factor out all of the common terms:$$\frac {GMm}{e} - \frac {GMm}{e + h} = \frac {GMm}e \left( 1 - \frac 1{1 + (h/e)}\right).$$It turns out that for $|r| < 1$ the geometric series $1 + r + r^2 + \dots = \frac 1{1-r}.$ To see this, consider the finite sum of the first $n$ terms, $S_n(r) = 1 + r + \dots + r^{n-2} + r^{n-1}.$ Now there are two different ways to get from $S_n(r)$ to $S_{n+1}(r)$ but they must both get you to the same number. The first is to imagine multiplying each term of $S_n(r)$ with $r$ and then adding 1; the second is to just add $r^n$ to the end: $$S_{n+1}(r) = 1 + r S_n(r) = S_n(r) + r^n.$$But that second equals sign is an equation that can be solved directly for $S_n(r) = (1 - r^n)/(1 - r).$ Then if $|r|<1$ the term $r^n$ will go to 0 as $n$ gets large, so we have $S_\infty(r) = 1/(1-r).$

In turn if $|h/e| < 1$ then the above series is$$1 - \frac1{1+h/e} = 1 - 1 + \frac he - \left(\frac he\right)^2 + \left(\frac he\right)^3 - \left(\frac he\right)^4 + \dots,$$with the alternating sign coming from a $(-1)^n$ term in the series that poses no theoretical problems so don't let it trouble you too much.

For the case where $GM/e^2 = g$ and $(h/e)^2 \ll (h/e)$ (which works out to $h \ll e$ only the $h/e$ term survives and you have simply $m g h.$

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The substitution you should make is that $mg = \frac{GM_{\rm E}m}{R^2}$ where $g$ is the value of the gravitational field strength at a distance $R$ from the centre of the Earth.
Note that the value of $g$ is not constant.

The change in potential energy in raising from distance $R$ from the centre of the Earth to a distance $R+h$ is $$ -\dfrac{GM_{\rm E}m}{R+h}+\dfrac{GM_{\rm E}m}{R} = \dfrac{GM_{\rm E}m}{R}\left ( \dfrac{-R+R+h}{R+h}\right )=\dfrac{GM_{\rm E}m}{R^2}\dfrac{h}{\left (1 +\frac h R \right )}=\dfrac{mgh}{\left (1 +\frac h R \right )}$$

So now you can make the approximation $h\ll R$ to get $mgh$ for the change in potential energy.

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