Dear all, Suppose we have a uniform spherical planet, of density p and radius R, and we drill a straight hole from one point to its antipode (neglecting any changes in the gravitational force that this will entail). Then, we drop a point mass from one of the ends, and it will travel through the hole. It is simple enough to show that the particle will oscillate with SHM, and the half-period of this motion will be sqrt(3pi/4Gp). It turns out that, even if the two ends of the hole are not antipodal, the particle will oscillate with the same half-period, and proving this is not much harder than in the previous case. I talk about the half-period because that is the time taken by the particle to cross the planet. In the case in which the points are not antipodal, this might not be the shortest time our particle could take to travel between the two points. For example, if the path travels towards the centre and then away, in a "curvy" shape, then the particle will be travelling faster for the journey, and this might compensate for the fact that the particle has a longer distance to travel. So the problem is to decide what the path is which will lead to the shortest time of travel. We'll assume air resistance and friction don't exist, and that no time is wasted turning (if that were possible). Regarding my own efforts on this problem, I've managed to prove some ostensibly wrong results, so am interested to see what you think. -- James