The acceleration due to gravity for an object on the surface of the Earth is g. The distance from the Earth to the Moon is roughly 60 RE, where RE is the radius of the Earth. What is the centripetal acceleration of the moon during its (roughly circular) revolution around the Earth

The centripetal acceleration that the moon experiences will be almost equal to the gravitational force that the Earth does in the moon,

Now, remember these two things:

F = m*a

and Fg = G*M1*M2/r^2

the first equation says that the force applied to something is equal to the mass of the object times the acceleration.

The second equation is for the gravitational force, where G is a constant, M1 and M2 are the masses of both objects, in this case, the Earth and the moon, and r is the distance.

We know that the acceleration in the surface of the Earth is:

a = Fg/M2 = g = G*M1 / (RE) ^2

now, for the moon we will have:

a = G*M1 / (60RE) ^2 = (G*M1 / (RE) ^2) * (1/60^2)

Here the term in the left is equal to g, so we have:

(G*M1 / (RE) ^2) * (1/60^2) = g * (1/60^2)

So the centripetal acceleration of the moon is 60^2 = 3600 times smaller than g.