For typical solar neutrino energies, and given Fermilab and South Dakota are about 750 miles apart, around one in 10^15 neutrinos would interact with the ground

For typical solar neutrino energies, and given Fermilab and South Dakota are about 750 miles apart, around one in 10^15 neutrinos would interact with the ground (more specifically, the nucleus of an atom in the ground). It would take about 10^14 miles (17 light years) of solid lead to block 50% of neutrinos. Imagine that you are a tiny, tiny, tiny neutrino. You are a million times smaller than an electron. You are drifting through space, and the only other component of space that you can interact with is the nucleus of atoms. But atoms also have electron shells that repel each other, so the nuclei have to be centered inside electron clouds that refuse to go anywhere near each other. The nucleus of each atom is only 1/25,000 the width of the electron cloud, or the "full width" of the atom. In other words, travelling through solid lead, for neutrinos, is like flying through mostly empty space. From the perspective of a neutrino, in solid lead, if each lead atom was centered on a star in the Milky Way galaxy, each lead nucleus would be the diameter of Jupiter's orbit. Everything else would be pure emptiness. Continuing with this blown-up scale, you would need to fly your spaceship for 10^27 lightyears before you have a 50% chance of flying within the Jupiter orbit distance of a star. The universe is only 10^11 light years wide, so you would need to fly through 10^16 universes before you have a 50% chance of passing within the Jupiter orbit of a star. And this is assuming the entire universe is as dense as the milky way's disk around us.