If an object is taken from earth's surface to infinity, it's gravitational potential energy becomes zero (always taken as zero), but it doesn't make any sense as energy can never be destroyed so where did that kinetic energy go?
[Physics] Why the gravitational potential energy at infinity is zero
conventionsenergy-conservationnewtonian-gravitynewtonian-mechanicspotential energy
Related Solutions
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 $PE_2=mgh$.
You say:
Therefore, the potential energy of an object has to be absolute, a constant in some sense.
The (gravitational) potential energy is a function of position so it varies as objects move around in space. The PE at some fixed point is sort of constant in the sense that all observers will agree on it's value, but note that PE has a global gauge symmetry - you can add a constant value to it without changing the physics because only changes of PE appear in the equations of motion.
We normally take the gravitational PE to be zero in flat spacetime far from any other masses. Therefore the PE near Earth's surface is negative, and the PE near whatever passes for Jupiter's surface is considerably more negative. The (positive) kinetic energy is therefore a lot greater near Jupiter's surface to keep the total energy constant.
Best Answer
Potential energy doesn't have any physical meaning. Only potential energy difference has physical meaning. You can add a constant to the potential energy at all points, and it will not make any difference. Hence, we have the liberty to choose any point in space as the point with potential energy as 0. Choosing potential energy at infinite as 0 is most convenient for calculations. You may choose any other point to have 0 potential energy, as long as you are consistent about it.
The kinetic energy (assuming, it becomes 0 when it reaches infinity, which means the velocity is equal to escape velocity), when added to the negative potential energy near the Earth, will be equal to zero. At infinity, both the energies are zero. That is, the sum of potential and kinetic energies remain constant.