Why? The Rubik's cube group is a subgroup of S8 × S12 × Z38 × Z212; the four factors here arise from possible permutations of the edge and corner pieces, and the orientations of the edges and the corners. (This is in fact twelve times as large as the actual group.) More concretely, it's the subgroup of this group consisting of quadruples (σ, τ, v, w), where σ ∈ S8 and τ ∈ S12 have the same parity, v is an element of Z38 whose coordinates sum up to zero, and w is a similar element of Z212. The group operation in this group is given by working componentwise.
(If I have misunderstood what I've heard about Rubik's cubes, then please ignore the rest of this post.)
[edited, 8:48 pm: turns out I have, as the multiplication isn't exactly componentwise.]
So the order of (σ, τ, v, w) is the least common multiple of the orders of σ, τ, v, w in S8, S12, Z38, and Z212 respectively. Call these o(σ), o(τ), o(v), o(w). If σ has one 7-cycle and one 1-cycle, τ has one 11-cycle and one 1-cycle, and v and w are nontrivial, then o(σ) = 7, o(τ) = 11, o(v) = 3, o(w) = 2, and so the order of (σ, τ, v, w) is lcm(7, 11, 3, 2) = 462. (This corresponds to a sequence of moves, whatever it may be, that permutes seven of the corners and eleven of the edges cyclically, and disrupts the orientations of both the corners and the edges.) Note that σ and τ are both even permutations.
Furthermore, we can't do better. I don't think there's a particularly elegant proof, but it's not hard to check by brute-force computation that no choice of the cycle types for σ and τ which gives both the same parity allows us to do better. If you let σ have one cycle of order 8, and τ have one cycle of order 7 and one of order 5, and let v, w both be nontrivial, then you'd think the order of (σ, τ, v, w) was lcm(8, 35, 3, 2) = 840, but σ is odd and τ is even. In fact, that's how I got the number 462 -- take lcm(s, t, 3, 2) where s ran over possible orders of elements of S8 and t ran over possible orders of elements of S12; the largest number you get this way is 840, but it can only come about in the way I described. 462 is the next-largest.
In case you're wondering how I came to this question -- it's easy to see if you play around with the cube that if you do the same sequence of moves over and over again, you eventually get back where you started. (Theoretically, this is a consequence of the fact that any element of a group has finite order.) So I just started to wonder how many times you might have to do the same sequence of moves on a solved cube to get it back to the solved state.