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Okay so for an object of mass m at a distance r from the Earth's centre, the GPE is $$U(r) = {{ - GMm} \over r}$$ For an object at height z above the surface (which obviously means at radius $r = {r_{Earth}} + z$) I am supposed to show the GPE near the surface can be approximated $U(z) = {U_0} + mgz$ where ${U_0}$ & g are constants which can be expressed in terms of G, M, m & ${r_{Earth}}$.

My friend says he thinks you should use the approximation $${(1 + \alpha )^{ - 1}} \approx 1 - \alpha $$ which is of course only valid when alpha is much less than 1. I have no idea how to answer this, it's part of a question set we're supposed to be able to answer before the exam in a couple of months so I want to be able to answer it.

Ahmed
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    Your friend is right. The radius of the Earth is $\sim 6500$km. $\frac{1}{r_{\rm Earth} + z} = \frac{1}{r_{\rm Earth}\left(1 + \frac{z}{r_{\rm Earth}}\right)} \approx \frac{1}{r_{\rm Earth}}\left(1-\frac{z}{r_{\rm Earth}}\right)$. In your notation $\alpha = \frac{z}{r_{\rm Earth}}$ which means that the approximation is good as long as $z \ll 6500$ km. – Winther May 12 '16 at 14:34

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For $r = r_\text{earth} + z$ where $z \ll r_\text{earth}$ we have

$$ U = \frac{-GMm}{r_\text{earth} + z} = \frac{-GMm}{r_\text{earth}} \cdot \left (1 + \frac{z}{r_\text{earth}} \right)^{-1} $$

where $z/r_\text{earth} \ll 1$. Now use your friend's approximation.