Got to thinking that the moon's influence on ICBM and other trajectories is better thought of as a straight forward tidal force, given we are looking at the moon's gravitational gradient from 2 sides of the planet Earth. Tidal forces drop off as the cube of the distance. So instead of a gravitational force being 6.8% stronger on the side of the earth facing the moon, we have instead 240 X 240 X 240 (earth-moon distance in thousands of miles)= 13,824,000. On the far side of the Earth, the side away from the moon, we'll cube the distance, but add the Earth's diameter of 8,000 miles and get; 248 X 248 X 248 = 15,252,992. From this latter figure we will subtract the former to get the cubed difference, and we get 1,428,992. We divide this by 13,824,000 to get a percentage in the drop off of the tidal force a rocket driven ICBM or free falling warhead would realize floating in the sea of space-time about the Earth. We get 10.3%. The moon's tidal force will be 10.3% weaker on the far side of the Earth. The side without the moon. Big difference when it comes to the precise targeting of ICBMs and so forth. And we can clearly see, the "tidal solution" as opposed to the simpler gravitational one is appropriate with our needing to evaluate a gravitational gradient, tidal force differential across the Earth's diameter.