a particle moves under an attracting central force, with magnitude (at distance r from fixed point O) F(r) = -m^2/r^2 where m is a constant. if the particle starts at distance a from O with radial velocity -m(root(3/2a)) and transverse velocity m(root(2/a)), show that the particle moves on a path which crosses the centre at a distance 2a/3. ^^^^^ ok my concern is that, if it crosses the centre, surely the distance means is 0? because that is the centre. I don't know really what I have to work out. I can do all calculations and integrations easily but setting up an equation here I don't know what to do. if anyone here is a scientist because this is science math can you kick me into the right direction