Gravity vs. other fundamental forces Why is it that gravity is the weakest of the 4 fundamental forces? I know that from experimental data, we can see that it is the weakest, cf. e.g. this Phys.SE post, but is there any way to prove that it is? And if so, is it intuitive? Is it due to the inverse square law?
 A: It seems relevant in this context to mention the weak gravity conjecture (WGC) by Arkani-Hamed, Motl, Nicolis & Vafa (AHMNV). 
One of AHMNV's  arguments is that black holes (which satisfy an inequality of the form $M \geq |Q_i|\forall i$ in Planck units), should be able to completely evaporate without a remnant in order to save unitarity, see Fig. 2 on p. 6 in AHMNV. Here $Q_i$ denotes a charge of the $i$'th type of force/interaction. This requires "elementary particles" to obey the opposite inequality $M \leq |Q_i|\forall i$ in Planck units, i.e. gravity should be weaker than the $i$'th interaction.
A: $\def\ns#1#2{#1_{\mathrm{#2}}} \def\qy#1#2{#1\,\mathrm{#2}} 
 \def\10#1#2{#1\cdot10^{#2}}$
I strongly counter the use of term "force" in the present context. It's only meaningful for gravity and electromagnetism, but totally devoid of meaning in the other cases.
Let me explain. Gravity and electromagnetism belong - as to their
birth and development - to classical physics. Those fields of physics
describe a lot of phenomena of macroscopic scale, whose explanation
was initially given using ideas of Newtonian mechanics. First of all
that of force. You may measure gravitational, electric, magnetic
forces between bodies and give a law of force: gravitational
(Newton) electric (Cavendish) magnetic (Ampère). Nothing like that is
possible for the so-called nuclear, strong, weak "forces". From
whichever point of view, these are not at all forces. But more about
this afterwards.
The main reason for difference is that gravity and e.m. forces
are long range ones, a technical term describing their dependence on
distance. In both cases (Newton, Coulomb) it is $1/r^2$. The other
"forces" act only at very short distances and decay exponentially (I
will be more precise in the following).
Does this mean that a comparison between strengths of gravitational
and electric forces is easy? Not at all. Surely it is meaningless to
compare the respective constants. Though both force laws share the
same mathematical forms
$${G m_1 m_2 \over r^2} \qquad {k\,q_1 q_2 \over r^2}$$
a direct comparison of constants
$$G = \qy{\10{6.67}{-11}}{m^3 s^{-2} kg^{-1}}\qquad
  k = \qy{\10{8.98}9}{N\,m^2 C^{-2}}$$
is unphysical, in the first place because their values depend on units
system used. In other words, $G$ and $k$ have different physical
dimensions.
You may often find the comparison done using some charged particle,
e.g. electron or proton. In both cases you read that gravitational
force is largely smaller than electrostatic, for equal distances. For
electrons we have
$${\ns F{gr} \over \ns F{el}} = \10{2.4}{-43}$$
and for protons
$${\ns F{gr} \over \ns F{el}} = \10{8.4}{-37}.$$
This is OK to show the as far as systems of 2 or few more electrons,
protons, or other similar particles are concerned gravitational force is likely to be absolutely negligible wrt to electrical one (or other interactions, but it isn't trivial at this point). Actually it was a hard task to show that subatomic particles really feel gravitational force. As far as I can remember, the first direct proof was obtained through neutron interferometry (Staudenmann et al., 1980).
But in macroscopic experiments things are rather different. Force
ratios are much less unfavourable for gravity - otherwise Cavendish'
experiment would have been impossible. This is because for
macroscopic bodies the ratio $q/m$ is not so large as it is for
particles. E.g. it is $\qy{\10{9.4}7}{C/kg}$ for a proton, whereas it's largely out of question to give a charge of $\qy1C$ to a body of mass $\qy1{kg}$.

The above disposes of the only real forces in macroscopic world. But
when it comes to microscopic (quantum) world the concept of force
totally disappears. Just from the very beginning QM never spoke of
forces. Even in the simplest and historically first application of
Schrödinger equation, the hydrogen atom, electron-proton binding
is described in terms of potential energy, not of force. This could
be seen a secondary change of viewpoint - after all potential energy
already belongs to Newtonian physics. 
In fact the first attempts to understand the new "forces" were
conducted introducing a "nuclear force" between nucleons
(proton-proton, proton-neutron, neutron-neutron). The experimental fact
that these forces were of very short range (order $\qy1{fm}=\qy{10^{-15}}m$) explained why they produce no macroscopic effects.
But soon QFT come into play. Yukawa idea (1935) was that nuclear force
was mediated by a massive particle he named meson. A "force"
mediated by a massive particle has a range linked to the mediator's
mass. Yukawa introduced a (Yukawa) potential
$$V(r) \propto {e^{-kr} \over r}$$
where
$$k={m\,c \over \hbar}$$
if $m$ is mediator's mass. The range of such potential is $1/k$ and
equating it to $\qy1{fm}$ a value 
$$m = {\hbar\,k \over c} = \qy{200}{MeV}/c^2$$
results for meson's mass.
Although I continued to use the word "force" in QFT this idea doesn't
exist and even that of a potential energy is a by-product. The basic
idea is an interaction term added to free-field lagrangian. The
potential energy applies only to a limited subset of situations, of
scarce interest: the very low energy interaction between two particles
(in present case, two nucleons).
I won't recall ensuing developments which brought us still farther
from forces and potentials: QCD, gauge theories. All this leads me to
state that present status of fundamental interactions has no place for
the concept of force, and the very word should be ruled out, as an
inevitable source of confusion for laypersons. The only correct term
IMHO is interaction.

I still have to write about "weak force" and its supposed weakness.
This was never seen as a force in classical sense (as nuclear force
was at the beginning and still is at a phenomenological level in
nuclear physics). There are no particles held together or acted on in
some other sense by a weak force. It only makes itself felt in some
decays - first of all in historical sequence nuclear $\beta$ decay,
then neutron decay, muon decay, pion decay, and so on.
But before of all that, when only nuclear $\beta$ decay was known,
neutrino hypothesis was born to explain continuous spectrum of
electrons emitted and the spin puzzle (Pauli 1930). Fermi (1933)
coined the first QFT model of $\beta$ decay, as a 4-line vertex
(interaction). The original process was
$$n \to p + e^- + \nu$$
later replaced by
$$d \to u + e^- + \bar \nu_e \tag1$$
and in electroweak unification (Glashow, Salam, Weinberg, late
'60s) by
$$d \to u + W^- \to u + e^- + \bar \nu_e.\tag2$$
The reason why Fermi 4-field interaction works at low energies is
the heavy mass of $W$ boson, about $\qy{80}{GeV}/c^2$. The $W$ propagator in eq. (2) has a $M_W^2+q^2$ denominator ($q$ momentum transfer). If $q^2\ll M_W^2$ it's almost constant and allows reducing (2) to (1), with a constant factor absorbed in the coupling constant.
Fermi theory predicts a decay rate increasing as the square of excess
energy. So only at small energies it is correct to see weak
interaction as really weak - in fact it would increase without bounds for
increasing energy. Such increase is bounded in electroweak theory
because the $W$ propagator at high $q$ begins to decrease.
A: It is true that gravity is the weakest force and Nuclear Force is the strongest.
Ordered from strongest to weakest, the forces are...
1. the strong nuclear force, 2. the electromagnetic force, 3. the weak nuclear force, and 4. gravity.
But If we consider about There ranges , Gravity has Longest effective range ,and that of nuclear force is smallest. Each force dies off as the two objects experiencing the force become more separated. The rate at which the forces die off is different for each force. The strong and weak nuclear forces are very short ranged, meaning that outside of the tiny nuclei of atoms, these forces quickly drop to zero. The tiny size of the nuclei of atoms is a direct result of the extreme short range of the nuclear forces.
The low value of Gravitational constant might be due to the low coupling factor between matter and the gravitational field.Every force works differently and they have different constants. When you put a large amount of charge with a small opposite charge , the force will be strong.
But if you put a small amount of mass , you will need very much mass to attract effectively
A: I think that the only meaningful way to state something about gravity being the weakest of the fundamental forces is to fully qualify such a statement by adding that it refers to the corresponding interactions between any pairs of the known elementary particles. Missing that, it is quite obvious that the gravitational force between two neutral planets dominates over their electrostatic interaction.
A: Gravity is not always the weakest "force".  In some circumstances (if there are lots of matter/energy content), it may be the strongest force of all, able to crush an entire neutron star against the internal nuclear forces and thus producing a black hole after the gravitational collapse.
The "strenght" of a force depends on the energy scale involved.  On the human scale, gravity is weak compared to all other interactions because if it was otherwise, we wouldn't be there to measure it! If gravity was stronger, life wouldn't be possible in our universe.  Gravity had to be the weakest so there could be observers at some scale.  Our own existence implies that gravity is a "weak" phenomenon at our scale, and that life is dominated by the other interactions (electromagnetism, in our case).
The usual forces couplings are defined with some dimensionless parameter.  The fine structure constant of electrodynamics is a good example :
\begin{equation}\tag{1}
\alpha = \frac{e^2}{4 \pi \varepsilon_0 \hbar c} \approx \frac{1}{137}.
\end{equation}
In the case of gravity, the coupling parameter is $G$, which has dimensions of a squared length (in natural units) :
\begin{equation}\tag{2}
G \equiv \frac{\hbar \, G_{\text{N}}}{c^3} \equiv \ell_{\text{Planck}}^2 \approx 2.61 \times 10^{-70} \, \text{m}^2.
\end{equation}
There's a lot to say about all this, but I'm lacking time to elaborate, and I think this answer is a good start to understand why gravity is "weak" in the usual human scale.
A: OP is not asking the right question. It's "not even wrong"! since comparing a dimension-full interaction (gravity) with a dimensionless interaction (standard model interactions) is an ultimate sin if no circumstance is provided. 
Given that the "charge" of the gravitational force is mass (energy tensor), the correct question is: why are the masses of the elementary particles so small compared with the Planck scale? As of yet, mortal physicists are still scratching their heads and fretting about this nasty "naturalness/hierarchy/fine-tuning problem". The Nobel laureate Frank Wilczek actually devoted a whole book about it: The Lightness of Being.
