Of course you can define such a quantity, but the question is: does it mean anything physically?
Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great importance for the description of many phenomena in electromagnetism, is defined as the product of the magnitude of one of the charges and their displacement vector. It would be equal to the numerator of your expression if you put the negative charge to the origin, the only term in the sum would be the product of the charge and the position (displacement) vector. So far so good: but what about the statement that your expression would then be a "normalized dipole moment"? Let us look at the denominator: in case of a dipole, the sum of both charges amounts to zero, so we would have a division by zero. This does not give us something normalized, but rather something ill-defined. Hence, this concept does not make much sense.
This problem remains for any system where the sum of charges is equal to zero, i.e. for neutral systems. Thus, it is not defined for many physically important situations and even in cases where it is, it does not tell us anything about the properties of that system.
Your "charge momentum" is related to the current density, which is given by the product of charge density and velocity. Its time derivative would simply be the rate of change of a current, and meaningful for example in a system with time-dependent electrostatic potential: this can be found in electrotechnical application when dealing with alternate currents. But can this be compared to a force?
To answer this question, we have to examine the nature of the comparison of the quantity $q\vec{v}$ to momentum in mechanics. What makes momentum so special is the fact that it is conserved for closed systems, i.e. systems without any external forces. But what is required in order for a certain quantity to be conserved?
One of the key principles of classical mechanics, Noether's theorem, tells us that conserved quantities (also called "conserved charges" or "Noether charges") are related to continuous symmetries of a system (there are various sources on the internet and books which describe this principle in as much detail as one might imagine). The conservation of momentum is a consequence of translational invariance of a physical system. In order to be comparable to momentum, your "charge momentum" would have to correspond to a continuous symmetry of the underlying system, but it turns out that there is none.
However, it is not completely unrelated to that concept either. While your "charge momentum" is no conserved quantity, charge itself is indeed one, corresponding to the $U(1)$ symmetry of electromagnetism. Within the framework of Noether's theorem, there also exists the notion of a so-called conserved four-current $J^\mu$, which has to satisfy
$$\partial_\mu J^\mu=0,$$
i.e. its four-divergence vanishes. In the case of the $U(1)$ symmetry, splitting space and time components and writing out the equation explicitely gives
$$\frac{\partial\rho}{\partial t}+\partial_i j^i=0,$$
which is nothing but the continuity equation where $\rho$ is charge density and $j^i$ is three-dimensional current density, which is related to your "charge momentum", as was already pointed out in the comments. There is no indication that the latter is conserved by itself.
Since there is no conservation of "charge momentum", i.e. no analogue to Newton's first law, applicability of an analogue of Newton's second law, which states that force is defined as the time-derivative of conserved momentum is highly doubtful. Furthermore, there is no reason to assume that the principle of "actio=reactio" (Newton's third law) should hold.
This is indeed confusing. The confusion comes from this very peculiar hypothesis:
What if the person doesn’t apply a tangential friction force at his
feet?
It implies there is a radial contact force at the person's feet (I prefer "contact" to "friction", which refers to movement). And, indeed, for the person to move radially inwards, or even to stay immobile in the carousel, they need to at least counterbalance centrifugal acceleration.
So let's imagine how the person could be "frictionless" tangentially yet "frictionful" radially: suppose there are slippery concentric rails all over the carousel, on which the person can lean to move radially, but which prevent them to control rotational speed.
Suppose the person starts immobile with respect to the rail of radius $r$ on which they stand. When the person steps inwards, they undergo the said tangential Coriolis acceleration, which makes them start to glide counterclockwise along the inner rail of radius $r-δr$ on which they now stand, at $δω$ with respect to the carousel. Their rotational speed with respect to the lab is now $ω+δω$, and $δω$ is such that their angular moment has not changed: $rω=(r-δr)(ω+δω)$.
Best Answer
I am assuming by $G$ you actually mean $g$, local gravitational acceleration ($9.8 \,\mathrm{m/s^2}$ on Earth's surface) and not the universal gravitation constant, usually denoted $G$ ($6.7\cdot 10^{-11} \mathrm{N~m^2~kg^{-2}}$), which does not change.
If that's the case, your question is
There is a little bit of confusion here. Even a constant force will cause a constant acceleration; so a constant, uniform gravity field will cause continuous acceleration (constantly increasing speed v). Since momentum is mv, this means momentum is changing constantly even under a constant force (and accompanying constant acceleration).
The fact that the local value of $g$ changes in your scenario does not add anything significant, except that the acceleration and force will increase instead of being constant, and the momentum will increase more quickly than it would have in a constant gravity field.
Acceleration does not have to change in order for momentum to change. As discussed, a constant g will cause a continuous increase in speed and momentum.
The external force is gravity.