I would suggest that you check the problem again, if neccesary with the teacher or person who gave it to you. In particular, See if they don't mean a "centre line" or other axis.
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
hi but I am sorry my English is not perfect so I sometimes will paraphrase in a way that make more sense in my understanding. but I realise its more helpful to you to read the question exactly. sorry again
does this help? can I have an advice to start please because the hardest part for me is knowing how to start an equation then I can use the normal rules to follow on to the next part and get the answer... so 2a/3 is a distance?
so what it means then?/? what is 'passes the centre' ther is masive emergency for solve this quick. and i dont have a speak to anyone for it or book. all have is wikipedias and etc. if it passes the centre how can it have a distance? because if it passes that mean there is no distance is a contridiction