I have problems with this equation: $$U_G ~=~ G\frac{m_1m_2}{r}.$$
It's for potential energy of say something placed on Earth.
But it intrigues me. $r$ is the distance from the very center of the Earth to its surface. But anything placed on the surface of Earth will never fall any deeper into the earth. So I think there will not be any potential energy for anything placed at the very lowest point possible on Earth. Though it's no where close to the core of the Earth.
The equation you cited makes use of "big G", the universal gravitational constant. Generally, this equation is used if you want to calculate the attractive force between two bodies, such the moon and the Earth, or a satellite and the Earth, or the Earth and the Sun, or if you want to calculate escape velocity from he Earth's gravitational field. The center of the Earth then is appropriate for the measurement.
But if you place a small object on the surface of the Earth, or somewhat above the surface, you would use "little g", acceleration due to the local gravitational field, which is 9.8 m/sec^2 at the surface of the Earth, and you wouldn't multiply the mass of the object by the mass of the Earth. Rather, you would use this equation:
Gravitational potential energy = weight * height = mass * g * height
"Little g" is the acceleration of an object toward the Earth during free fall from a height relatively near the surface.
Here is a better explanation, and a calculator for computing gravitational potential energy: http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html. If you scroll down the page, you also will see how the universal gravitational constant is used.
We treat a mass producing gravity as if all the mass is concentrated at the centre. Does that make sense to you? Also potential energy values are relative, it is only the difference in two values of P.E. that counts, not any absolute value.
But anything placed on the surface of earth will never fall any deeper into the earth.
We can assign a value of 0 to the the surface of the earth, it's an arbititrary value. But then what happens if you fall down a mineshaft?
I think of P.E. as a bookkeeping exercise, (that's my personal working definition of it) to make sure total energy, kinetic and potential, is conserved, which is a law of nature.
Couple of things to point out here, and I'm not sure whether either of those answer your question, so here are some ideas to help you think about the problem and guide you towards the full answer:
The lowest possible value of Earth's gravitational potential is met:
a)if the object is at Earth's exact center (r=0, namely there is no mass "below" him (m1=0), and the pull of all the mass around him cancels out perfectly, so U tends to a minimum) or
b)very far away (r tends to infinity, so from the equation U tends to a minimum). We're obviously focusing on the first case here.
The only important thing about potential energy is its difference between two points, so we can add an arbitrary constant without changing any physics behind it. In this case the arbitrary constant is the minimum at r=0 and r->oo, and we may define it as U=0. Another confusion may arise because we can look at the two different definitions of potential energy (the general, which you have stated, and the near-surface approximation, U=mgz, the equation itself and difference between G and g is explained in one of the previous answers). Having negative potential energy arising from the second equation in the case of z<0 (below the ground) is not a problem, since we can always define that position as a new "point 0".
Now you have to come to terms with what you meant by the "lowest possible point on Earth". My intuition would tell me it's its center, where you have the lowest possible potential energy anyway so your problem is solved (by moving the object away from the center, assuming it has the room to move, it would just oscillate around Earth's center, according to the gravitational application of Gauss's law (https://en.wikipedia.org/wiki/Gauss%27s_law_for_gravity), one of its consequences being that the effects of gravity would be the same whether we're dealing with an Earth sized ball, or a tennis ball of the same mass (let's pretend spacetime curving from general relativity doesn't happen here).
If by "lowest possible point" you mean somewhere on Earth's surface, then your question can be formulated into "Why doesn't the object fall through the surface into Earth's center?". The shortest answer for that is that Earth's crust is sturdy enough not to collapse under all the weight on the surface, and that Earth's gravity is weak compared to that of Jupiter, the Sun or a typical black hole. Earth's gravity just confines us to the surface of the planet (we don't fall of the edge of the globe, nor fly away), but at least it doesn't limit us in moving anywhere on the sphere's surface (excluding the effects of friction).
