I like math, and I strongly suspect Bret Sheppard does not, so I thought I'd do Bret's homework (assigned by myself) for him.
In a two-body system in which the smaller body is in tidal lock with the larger body, the radius of a synchronous orbit around the smaller body will always be greater than the distance to the L1 libration point.
Consider a two-body system separated by distance D.
Let M be the mass of the larger body (kg)
Let m be the mass of the smaller body (kg)
Let T be the orbital period of the smaller body about the larger body (sec)
Let r be the radius of a synchronous satellite orbit round the smaller body (m)
Let d be the distance from the smaller body to the L1 libration point (m)
For the smaller body's orbit, by Kepler's third law:
T = 2π √ (D3/GM) where G is the gravitational constant
Since the smaller body is in tidal lock with the larger body, the rotational period of the smaller body must also be T, and so must be the orbital period of a synchronous satellite.
So for the synchronous orbit:
T = 2π √(r3/Gm)
r3/Gm = T2/4π2
Substituting for T
r3 = 4π2GmD3/4π2GM
r = 3√(D3Gm/GM) = D 3√(m/M)
By a simplified formula for d, the distance of the L1 libration point
d = D 3√(m/3M) reference
Therefore d < r, Q.E.D.
The significance of this is that, for a tidally-locked moon, a synchronous orbit is impossible since a satellite would prefer to orbit the parent body. Therefore, anyone who claims to have found a satellite dish on the moon is mathematically full of shit. Thank you.
For the Earth-Moon system, m/M = 0.012 r = 88,465 km d = 61,135 km