From guns to the Moon

The path of the shell lay between the Earth and the Moon. As the shell moved away from the Earth, the Earth's gravitational pull decreased, while the Moon's gravitational pull increased. At some point along the way, both gravitational forces—lunar and terrestrial—would become balanced, and the shell would lose all weight. At this equilibrium point, any object without speed or engine would remain forever motionless, because both celestial bodies would attract it with equal force, and nothing could make it move in one direction or another.

If the force of the initial impulse was calculated correctly, the shell would reach this point with zero speed, having lost any sign of weight along with all objects inside it. What would happen then? Three possibilities were considered.

Either the shell, retaining some speed and passing the point of equal gravity, would fall to the Moon, as lunar gravity would prevail over the Earth's.

Or the shell would not have enough speed to reach the neutral point, and it would fall back to Earth, as Earth's gravity would prevail over the Moon's.

Or finally, moving with enough speed to reach the neutral point, but not enough to pass beyond it, it would remain forever stationary, floating along the line as the legendary Mohammed's coffin, between zenith and nadir.

Jules Verne, "Around the Moon"

Write a program that determines whether the Earth's or the Moon's gravitational force affects a shell more at the point $(x,y)$. The center of the Earth is at $(0,0)$, and the center of the Moon is at $(384000,0)$. The mass of the Earth is $81$ times the mass of the Moon. The force of mutual attraction between two bodies of mass m_1 and m_2, separated by distance $r$, can be calculated by the formula F = g \cdot m_1 \cdot m_2 / r^2, where $g$ is the gravitational constant, equal to 6.673 \cdot 10^{-11}.