The Earth's mass is 81 times the Moon's, and the distance between their centres is 60RE, where RE =…

Step 1. The balance point. At a distance $x$ from the Earth's centre: $\dfrac{GM_E}{x^2} = \dfrac{GM_M}{(60R_E - x)^2}$. Take square roots: $\dfrac{60R_E - x}{x} = \dfrac19$, so $x = 54R_E = 345600\ \text{km}$.

Step 2. The least speed. Before the balance point the Earth pulls the body back; after it, the Moon pulls it on. So the body must just reach the balance point. The potential (per kilogram) at a point is $-\dfrac{GM_E}{r_E} - \dfrac{GM_M}{r_M}$. With $GM_E = gR_E^2$ and $gR_E = 64\ \text{MJ/kg}$:

• At the Earth's surface: $-gR_E\left(1 + \dfrac{1}{81 \times 59}\right) = -64.013\ \text{MJ/kg}$.
• At the balance point: $-gR_E\left(\dfrac{1}{54} + \dfrac{1}{81 \times 6}\right) = -1.317\ \text{MJ/kg}$.

So $\tfrac12 v^2 = -1.317 - (-64.013)\ \text{MJ/kg}$, which gives $v = 11.20\ \text{km/s}$. That is a little less than the escape speed from the Earth alone, $\sqrt{2gR_E} = 11.31\ \text{km/s}$: the Moon helps a little.

Step 3. Arrival at the Moon. At the Moon's surface the body is $59.73R_E$ from the Earth's centre and $0.27R_E$ from the Moon's:

From the balance point, where the speed is nearly zero: $\tfrac12 v^2 = -1.317 - (-3.998)\ \text{MJ/kg}$, so $v = 2.32\ \text{km/s}$.