Why does an orbit become parabolic when total orbital energy is zero?

Question

I stumbled across while reading other stack posts where there was a discussiom of non calculus ways to find the trajectory under E negative and E positive conditions Feynmans technique was well used there, i was interested if that techinque can be applied for case of E=0 as well ? By oberving the velocity space diagram and then converting it to the space trajectory diagram . I know from general mechanics equations that E = 0 means v is escape speed . Like this link : https://physics.stackexchange.com/a/88252/297949

Answer 1

I'am answering the title. Referring to the attractive Coulombian interaction, the parabolic case is the first case of open orbit. It separates elliptic (closed) orbits from hyperbolic (open) orbits. The energy is constant along each given orbit, so that it can be computed at infinity since open orbits reach the infinity. Along every open orbit at infinity, the potential energy is zero end the energy is completely kinetic. The first case of open orbit corresponds to the case where the particle reaches the infinity with the least possible kinetic energy. That is zero kinetic energy. In summary, the total energy is $0+0=0$. A grater (kinetic) energy at infinity corresponds to some hyperbolic orbit.