If atoms were held by gravitational (instead of electrical) forces 
"If atoms were held together by gravitational (instead of electrical) forces, a single hydrogen atom would be much larger than the known universe."
  - from Grifffith's Introduction to Electrodynamics

I don't understand this. If the atoms were held together by gravitational forces, then the atoms will have to be extremely close together, but why extremely big?
 A: To understand this, you need to consider the semi-classical description of an atom (so-called Bohr Model). For the sake of simplicity let's only talk about a hydrogen atom, that is, a proton with a positive charge ($q_p$) somewhere in space and an electron with a negative charge ($q_e$) orbiting in a circular path around the proton.
In this case, the electron is kept in the circular orbit by only electrostatic attraction, that is, the centripetal force is equal to the Coulomb force.
$$\frac{m_ev^2}{r}=k\frac{q_{p}q_{e}}{r^2},$$
where $k$ is Coulomb's constant, $m_e$ is mass of electron, $v$ is the speed of electron and $r$ is the radius of the orbit (or size of an atom). The above equation can be re-organised as 
$$v=\sqrt{\frac{kq_{p}q_{e}}{m_{e}r}}.$$
Now, the quantum mechanics says that the angular momentum must be an integer multiple of the Planck's constant $\hbar$, that is,
$$L=mvr=n\hbar.$$
If we plug in the expression that we found above for speed into this equation we get
$$m_e\sqrt{\frac{kq_{p}q_{e}}{m_{e}r}}r=n\hbar.$$
By re-organisation we reach
$$r_n=\frac{n^2\hbar^2}{kq_{p}q_{e}m_e}.$$
If you plug in the numbers for the ground state ($n=1$) you would get
$$r_1\approx5.29\times10^{-11}m.$$
Now, if you assume that only the gravitational force keeps the proton and electron together the new equation for the centripetal force would be
$$\frac{m_ev^2}{r}=G\frac{m_{p}m_{e}}{r^2},$$
where $G$ is Gravitational constant and $m_p$ is mass of proton. In this case the speed would be
$$v=\sqrt{\frac{Gm_{p}}{r}}.$$
Applying the the constraint on angular momentum we get
$$m_e\sqrt{\frac{Gm_{p}}{r}}r=n\hbar,$$
which leads to
$$r_n=\frac{n^2\hbar^2}{Gm_{p}m_{e}^{2}}.$$
Please plug in the constants yourself and see what value you get for the radius of an atom in ground state (I got a number in the order of $10^{29} m$).
A: As two answers have discussed the semi-classical model, I'll mention that if you were to solve the Schrodinger equation for a radial hydrogen atom held together by gravity alone, with no EM forces, you'd find that the expectation value ("the average") value of the radius would be :
$$<r>= \frac 3 2 \frac {\hbar^2} {Gm_p {m_e}^2}$$
Which is of the same order of magnitude as the semi-classical model.
So even "proper" quantum theory predicts a huge gravitational atom. 
I think your intuitive difficulty is that you're thinking in terms of having to get close to match the forces that an EM field would produce, which would require the electron and proton to be closer.  But that's not needed - there's no reason the forces in the gravitational atom should equal those in the EM atom - they can be quite different.
A: To find the radius of the orbit, we first need to consider the centripetal force on the electron ($\frac{m_ev^2}{r}$) which is equal to the gravitational force between the proton and the electron in this case. ($F_g=\frac{Gm_em_p}{r^2}$)
Using this, and the fact that the angular momentum of an orbit is quantised, (i.e. $m_evr=\frac{nh}{2\pi}$),
the radius ($r$) of the orbit turns out to be: $$\frac{n^2h^2}{4\pi^2Gm_pm_e^2}$$
For $n=1$ (i.e. the principal orbit), this has a value of $1.2\times10^{29}m$, which is ridiculously large, as Griffith states.
A more intuitive way to understand this would to notice that the force needed to keep an electron in orbit is inversely proportional to $r$. The force of gravitation between an electron and a proton is practically negligible ($10^{42}$ times lesser) when compared to the electromagnetic force between them. This is why $r$ has to be very large (compared to the electromagnetic case), as this reduces the amount of force needed by the electron to be kept in orbit, i.e. achieving balance in some way.
A: At first sight, you would indeed expect that the electron has to be much closer to the nucleus, so the force needed to keep the electron in a circular orbit will be the same as that in the case of the electric force. 
But for the centripetal force (directed inwards and provided by the gravitational force or the electric force), proportional to $\frac 1 {r^2}$ (for conservative forces), to be in balance with the centrifugal force, proportional to $\frac 1 r$ (which is the condition for a circular orbit), the distance where this balance will occur will be the bigger the smaller the force which binds the electron and the nucleus. 
So essentially it's because of this difference in the $\frac 1 {r^2}$-dependence of the force which delivers the centripetal force and the $\frac 1 r$-dependence of the centrifugal force which will increase the radius of a circular orbit if the centripetal force gets smaller (as is the case for the gravitational force).
