This is a supplementary answer with calculations (for 1) two bodies and 2) three bodies) problem which confirms that @MatthewWells answer is correct.
Would the Moon hit the Earth or not?
Yes, it would hit the Earth.
Do we have to do detailed numerical integration to find the answer, or can we use some simple equations that involve energy and/or eccentricity to get an answer?
To perform calculations for this answer I applied combination of the use of simple equations (force, acceleration, energy/eccentricity/velocity) and simple iterative approach (spreadsheet in OpenOffice Calc), i.e. no numerical integration of complex equations.
tl;dr
1. The simplest case of two initially static bodies: the Earth and the Moon (assuming there's no Sun, no other planets, and no galaxy):
Gravitational force that acts on each of the bodies can be found from Newton's law of universal gravitation:
Fem=Fme=G*(Mm*Me)/(r^2),
where:
Fem: Gravitational force acting on Earth, directed towards Moon;
Fme: Gravitational force acting on Moon, directed towards Earth;
G: universal gravitational constant;
Mm: mass of the Moon;
Me: mass of the Earth;
r: distance between centers of the Moon and the Earth.
Fem=Fme= 1.982*10^20 N
Acceleration of each body can be calculated using second Newton's law of motion: F=Ma , where F: force; M: mass; a: acceleration.
Earth's acceleration:
ae=Fem/Me= 3.31*10^-5 m/s^2
Moon's acceleration:
am=Fme/Mm= 2.79*10^-3 m/s^2
Then I applied a simple iteration process:
The above accelerations are calculated for time t=0. Velocity of both bodies Ve0, Vm0 and distance they traveled towards each other De0, Dm0 are equal to zero.
For the next step, after a time step dt: t1=0+dt, Earth's velocity grows from 0 to Ve1=0+ae0*dt, travelled distance is De1=0+Ve1*dt.
Moon's velocity grows from 0 to Vm1=0+am0*dt, travelled distance is Dm1=0+Vm1*dt.
Distance between the objects decreases:
r1=r0-De1-Dm1
Then we obtain "step-1" gravitational force Fem1=Fme1 as function of new r1 and new accelerations ae1 and am1 from the two formulas mentioned above.
For the next step t2=t1+dt, new velocities are: Ve2=Ve1+ae1*dt, Vm2=Vm1+am1*dt; travelled distance is De2=De1+Ve2*dt, Dm2=Dm1+Vm2*dt.
Distance between the bodies: r2=r1-De2-Dm2
and so on, until the distance between the centers of the bodies becomes equal to the sum of Earth and Moon radii.
how long would it take?
It would take 6898 minutes = 114.96 hours = 4.79 days for the Moon to hit the Earth.
At impact, Earth would have traveled 4573km km (or 72% of its radius) and would have velocity of 117 m/s; the Moon would have traveled 371,721 km and would have impact velocity of 9,544 m/s.
The Energy of the impact can be calculated from the formula for kinetic energy:
E=0.5*Me*(Ve)^2=0.5*Mm*(Vm)^2.
So, the energy of the impact would be E=3.506*10^26 J, or an equivalent of 1.67 billion Tsar Bomba's.
2. Including the Sun (and the fact that both Earth and Moon are orbiting the Sun) is a bit more tricky, but with some simplifying assumptions can be coarsely approximated by applying balance of the forces that act on the two orbiting bodies and simple iteration process as described above.
Let's assume Earth's trajectory around the Sun is perfect circle.
In absence of the Moon, and considering the local Earth-centered reference frame, there are two forces acting on Earth: gravitational (from the Sun, calculated as per formula above), directed from Earth towards the Sun and centrifugal force due to orbiting motion (directed radially, i.e. perpendicular to orbital velocity vector, outwards from the Earth and Sun). These two forces are equal in magnitude but opposite in direction, therefore cancelling each other, hence the Earth's radius vector is constant.
Although, apparently, the centrifugal force is a fictitious one for this case, let's assume it would be numerically equal to centripetal force:
Fce=(Me*(Vte)^2)/Re,
where Vte is tangential (orbital) velocity, Re is distance between Earth and Sun centers.
Vte can be calculated from vis-viva equation with the special case of c=Re (I have substituted "a" in the formula from wikipedia with "c" to not confuse it with acceleration)
(Vte)^2=GMs(2/Re-1/c),
where Ms: mass of the Sun, c: semi-major axis of the orbital trajectory ellipse.
Now when we include the Moon, the third force would act on the Earth: gravitational force between the Earth and the Moon, directed outwards from Earth, towards the Moon.
It would depend where exactly around the Earth it has stopped. I've looked into two special cases of Sun, Earth and Moon aligned in one line, where the Moon is either "above" the Earth (i.e. on larger radius than Earth), or between the Earth and the Sun (i.e. on smaller radius then Earth).
2a) Moon "above" the Earth.
The same three forces would act on the Moon as described above for the Earth. The difference would be in the magnitude of centrifugal (calculated as centripetal) force: by constraints of the question, the Moon moves with the same speed as the Earth, but in a higher orbit (larger radius vector), which means it's going to orbit in ellipse and at t=0 it's in its periapsis.
Using the vis-viva equation,
(Vtm)^2=GMs(2/Rm-1/cm)
we can find semi-major axis cm, taking tangential (orbital) velocity Vtm equal to Vte (explained above to be calculated from circular case at Earth orbital radius) by definition of the question.
Rm is the Moon radius vector, i.e. the distance from the Moon center to the Sun center: Rm=Re+r.
By applying the iterative process as described above for the case 1) (i.e. the vector sum of the three acting forces is used to calculate the acceleration, then calculating traveled distance iteratively) the calculation continues until the distance between the Earth and the Moon centers becomes equal to the sum of their radii.
The result of the calculation is that collision will happen in this case after 6914 min=115.23 hrs, which is 15.8 minutes or 0.229% longer than calculated in the (two-body) case 1).
2b) Moon between the Earth and the Sun.
Everything else, described above for the case 2a) applies here, except the Moon is now in lower orbit with the orbital speed of the Earth, which means it would orbit the Sun in ellipse, and at t=0 it's in its apoapsis.
The time to collision in this case is 6914 min=115.23 hrs, which is 15.9 minutes or 0.23% longer than calculated in the (two-body) case 1) and practically identical to the same as in case 2a).
The Moon and the Earth velocities at impact are almost identical to case 1). Energy of collision is 3.505*10^26 J, a 0.04% deviation from that of case 1).
See below the plots of the {radius vector - 1au} (in meters) vs. time in minutes (my apologies for the quality):
Bonus calculation:
3) Let's assume the Earth and the Moon stopped altogether in the Solar system, and are aligned in one line, as described in cases 2a) and 2b).
By zero'ing orbiting velocity and centrifugal force in the calculations for the cases 2a) and 2b), we obtain the following plots (with comparison to the case 1)):
For the case 3a) (Moon is on larger radius than Earth), the time to collision is 6925 min=115.41 hrs, which is 26.9 minutes or 0.39% longer than calculated in the (two-body) case 1), but velocities at impact and impact energy differ significanly:
Earth velocity 2352 m/s (20 times higher), Moon velocity 12030 m/s (26% higher); impact energy: 4.420*10^26 J, a 26.1% increase compared to case 1)
For the case 3b) (the Moon is on the lower radius than the Earth), the time to collision is 6925 min=115 41 hrs, which is 27.0 minutes or 0.39% longer than calculated in the (two-body) case 1) and pretty much the same as for the case 3a), but velocities at impact and energy are significantly different again: Earth velocity 2587 m/s (22 times higher), Moon velocity 7093 m/s (26% lower). Energy of impact is 2.606*10^26 J, 25.7% lower than in case 1) and 1.7 times lower than case 3a).
So, to sum up, in the case of stopped Moon, the mutual gravitational force between the Moon and the Earth at the given distance between them prevails in comparison with the gravitational force of the Sun acting on these two bodies, resulting in relatively short time between the Moon stop and the impact of the two bodies, unless we consider stopping the Earth as well.
Does the answer depend on where the Moon is in its orbit?
It does very slightly and the Moon position can be neglected, unless we consider stopping the Earth as well, in which case, positioning of the moon around the Earth gives large effect on energy of the impact (but time to the impact stays very similar).
