Yeah one is for measuring potential energy between the objects of two masses $M$ and $m$
We recently started studying about gravitation and I'm really confused when swtiching back and forth, or can we derive $mgh$ from $-\frac{GMm}{r}$? we derived $mgh$ through work done formula, he recently referred "Gravitational work" as $mgh$. It's really confusing.
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Qmechanic
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1Possible duplicates: https://physics.stackexchange.com/q/521412/2451 , https://physics.stackexchange.com/q/286360/2451 , https://physics.stackexchange.com/q/35878/2451 and links therein. – Qmechanic Aug 17 '21 at 09:28
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The expression mgh assumes that the displacement is small enough that g can be considered a constant. – R.W. Bird Aug 17 '21 at 13:50