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

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The equations U=mgh and U=Gm1m/R yield similar results because the former is a limiting case of the latter when considering gravitational potential energy near the Earth's surface. The equation U=mgh approximates gravitational potential energy by assuming height (h) is small compared to the Earth's radius (Re). By applying a Taylor expansion to the universal law of gravity, it simplifies to U=-mgh, where g is derived from the gravitational constant and Earth's radius. This shows that both equations converge to similar values under the specified conditions. Thus, the similarity arises from the approximation used in the context of Earth's gravitational field.
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 R=R_e+h where Re is the radius of the Earth, and h is your height above the Earth.

Now,

U=GM_em\frac{1}{R_e+h}

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:

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

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,U=-GM_em\frac{1}{R_e^2}(h)=-mgh

with g=\frac{GM_e}{R_e^2}

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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