How Do the Equations U=mgh and U=Gm1m/R Yield Similar Results?

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SUMMARY

The equations U=mgh and U=Gm1m/R yield similar results because U=mgh is a limiting case of U=Gm1m/R when considering gravitational potential energy near the Earth's surface. The first equation approximates gravitational potential energy, while the second is derived from Newton's universal law of gravity. By substituting R with Re+h, where Re is the Earth's radius and h is the height above the surface, and applying a Taylor expansion, the two equations converge, demonstrating that both yield comparable values under specific conditions.

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Loppyfoot
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Hi, I was wondering this question:

Why does U=mgh and U=Gm1m/R come out to similar values even when their equations are completely different?

Thanks
 
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The short answer is that this is because the first equation is just a limiting case of the second.

The first equation is and approximation for the potential energy due to gravity near the Earth's surface. This equation is a limiting case of of Newton's universal law of gravity(the first equation), if we assume we are near the Earth's surface.

Let [itex]R=R_e+h[/itex] where Re is the radius of the Earth, and h is your height above the Earth.

Now,

[tex]U=GM_em\frac{1}{R_e+h}[/tex]

Since h is really small compared to R_e(we are close to the Earth's surface), we can Taylor expand the fraction and keep only the first few terms:

[tex]U=GM_em\frac{1}{1+h/R_e}=GM_em\frac{1}{R_e}(1-h/R_e)[/tex]

We get two terms, but we can drop the first, as it is a constant, and constants of potential energy don't matter. Let's drop that constant. (This is equivalent to setting the Earth's surface as 0 potential.)

So,[tex]U=-GM_em\frac{1}{R_e^2}(h)=-mgh[/tex]

with [itex]g=\frac{GM_e}{R_e^2}[/itex]

So, near the Earth's surface, both equations must give similar values, because the second is just an approximate limiting case of the first.
 

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