Why is gravitational potential energy negative, and what does that mean?
About negative energies: they set no problem:
On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.
However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:
let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:
$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$
as expected: we lose $PE$ and win $KE$.
Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.
Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $ r = r_{earth}$.
The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).
By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:
$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.
As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$
Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:
$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.
Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.
So, from what I understood, your logic is totally correct, apart from two key points:
energy is defined apart of a constant value.
in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.
I will first (1) summarize the differences between the definitions of PE1 and PE2 and then I will (2) equate the two.
(1) First, as this answer to "Why is gravitational energy negative?" says, PE1 defines the potential energy of a body of mass m in the gravitational field of a mass M as the energy (work) required to take it from its current position $r$ to infinity. PE1 assumes $r=\infty$ is $PE=0$ $$PE1=\frac{−GMm}{r}$$
PE2, on the other hand, is defined as the negative of the work done by gravity to lift a body of mass m from the surface of a planet to a height h above the planet.
$$PE2=-W=-Fdcos\theta =mgh$$
PE2 has a different frame of reference than PE1, as it assumes $PE=0$ at $r=R$, or on the surface of the planet. Also, and very importantly, PE2 is only used when an object is close to the surface of a planet, when $h<<<R$ (R is radius of planet), and g can be assumed to be constant:
$$g=\frac{GM}{(R+h)^2} \approx \frac {GM}{R^2}$$
(2) OK, now on to equating the two. Though the frames of references for PE1 and PE2 are different, $|\Delta PE|$ between two points should surely be the same. For the sake of example, let's say the two points are the surface of the planet and height h above the planet.
PE1 says $|\Delta PE|=mgh-mg(0)=mgh$
PE2 says $|\Delta PE|=\frac {-GMm}{R+h}- \frac {-GMm}{R}=GMm\left(\frac{1}{R}-\frac{1}{R+h}\right)=GMm\left(\frac{h+R-R}{(R)(h+R)}\right)=\frac{GMmh}{(R)(R+h)}$
and because $h<<<R$, $\frac{GMmh}{(R)(R+h)} \approx \frac{GMmh}{R^2} =mgh$
And thus, PE1 and PE2 both represent the same form of energy, but we must keep in mind the frames of reference and the conditions of use when we use them.
Hope this helps!! Peace.
It is because gravitational force is attractive and work is done by gravitational force itself. When system does work itself energy is taken as negative and when work is done by external agency on system energy is take as positive.