Oh no it won't! That's what Aristotle thought, but Galileo proved him wrong. In air, particle size and shape will affect how quickly the dust settles, but particle density is irrelevant. Talcum powder and U238 dust of the same shape and dimensions would settle at the same speed.
http://galileo.phys.virginia.edu/classes/109N/lectures/gal_accn96.htm
If an object is allowed to fall on Earth it accellerates in obeyance to Newton's formula a=F/m.
But this will happen only until the force due to air resistance equals that of gravity. The drag force is (Cd*rho*V^2 )/2
Cd= coefficient of drag
rho=density of air
V=velocity
A=area in the plane perpendicular to the direction of travel
So when ma=(Cd*rho*A*V^2)/2 the object will be at terminal velocity.
We can see that denser objects will have a larger terminal velocity that less dense ones
We also know that a dust particle or a feather falling horizontally will reach it's , slower , terminal velocity faster than say a hammer. (BTW drop the feather with the point of the qill aimed straight down and for a short distance(several feet) both the feather and hammer will drop pretty close to the same rate).
In an atmosphere dust drops slower than in no atmosphere obviously since a dust particle has very little mass and therefore reaches it's terminal velocity very quickly.
First, let's determine the fall time of a particle in a vacuum. This is an easy calculation. Let's say the particle is dropped from a height of 2 meters, we have:
on Earth,
t = (2d/a)^0.5 = (2*2/9.807)^0.5 = 0.64 s
on the Moon,
t = (2*2/1.62)^0.5 = 1.57 s
However, because of drag, soil-sized particles may actually fall slower on Earth, which is exactly the opposite of what the HBs claims. A small particle will reach its terminal velocity very quickly (in only a fraction of a second), thus we can assume the particle will fall at a constant velocity (the terminal velocity) over the entire distance. From this we can determine the fall time.
The lunar soils are mainly classified as sandy-silt or silty-sand. Therefore, let's assume a particle with a diameter of 10 microns, which falls into the size range of silt. If we assume a density of 2700 kg/m^3, which is fairly typical of rocky material, we can calculate the mass of the particle,
r = 10 / 1E+6 / 2 = 0.000005 m
volume = 4/3*pi*r^3 = 4/3*pi*0.000005^3 = 5.236E-16 m^3
m = vol*density = 5.236E-16*2700 = 1.414E-12 kg
The cross-sectional area of the particle is,
A = pi*r^2 = pi*0.000005^2 = 7.854E-11 m^2
The drag force on a particle is given by,
Fd = Cd*rho*A*V^2/2
where Cd is the drag coefficient, rho is the air density, A is the area normal to the flow, and V is the velocity. We know that F=ma, thus we can write,
ma = Cd*rho*A*V^2/2
Terminal velocity occurs when the acceleration caused by the drag force is equal to the acceleration of gravity. In this condition we have a net acceleration of zero and the velocity remains constant. We therefore replace the acceleration, a, in the above equation with the acceleration of gravity, g, giving us,
mg = Cd*rho*A*V^2/2
Rearranging variables we get,
V = [ 2mg/(Cd*rho*A) ]^0.5
which is the terminal velocity.
We know that g equals 9.807 m/s^2, and the standard density of air at sea level is 1.225 kg/m^3. In my previous calculations I used a value of 0.5 for Cd, which is the normal coefficient of friction for a sphere. However, this number is generally higher for very small particles. Furthermore, a soil particle is not a perfect sphere. Nonetheless, let's use Cd=0.5 knowing that our calculation is conservative, i.e. the time we calculate will be the *fastest* time in which the particle will fall. We have,
V = [ 2*1.414E-12*9.807 / (0.5*1.225*7.854E-11) ] ^0.5 = 0.759 m/s
Thus the fall time is,
t = d/v = 2/0.759 = 2.64 s
Don't forget that if we use a higher value of Cd, the fall time increases. For instance, if Cd=1 then fall time becomes 3.73 seconds.
Now that you have an example, you can vary the particle size, the soil density, and the drag coefficient to calculate other examples. I think these computations show quite convincingly that small soil particles will fall more slowly on Earth than they will on the Moon.
One more thing, these calculations assume the air is completely still. In reality, very small particles will usually remain in suspension longer because of air currents.
Thus, if one drives down a dirt road when it is dry you need not to be going very fast at all in order to kick up a great cloud of dust behind the vehicle. This dust cloud remains in the air for quite some time due to it's particles reaching their small terminal velocity very quickly,,, AND because the turbulent air following the vehicle keeps lifting the dust.
IF the lunar vehicle was on Earth , in an atmosphere, the dust would behave exactly the same way. The lack of a following dust cloud proves that this was not on Earth and or in an atmosphere.
In contrast to this an astronaut jumping into the air on Earth would fall about 2.5 times faster than that same astronauts jumping up to the same height on the surface of the Moon. This is because the much greater mass of the astronaut requires a much much greater velocity before drag equals gravitational forces on Earth but gravitational force on the Moon is so much les than that on Earth.