
central force
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

Re: central force
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.

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Re: central force
hi,
this wasn't set by any individual so there is no facility to check with anyone. I can screencap the question if it helps:

Re: central force
Oh! YOU miscopied! It doesn't say "crosses the centre", it says "passes the centre". That is, you are asked to show that shortest distance from the trajectory to the centre is 2a/3.

Re: central force
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?

Re: central force
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