# Why are the trajectories velocity specific? [duplicate]

Consider projecting a particle from or above the earth surface (at a distance $$r$$ from center of earth) with velocity $$v$$. My teacher told me that if $$v particle will follow elliptical path, $$v=v_0$$ circle, $$v_e>v>v_0$$ ellipse, $$v=v_e$$ parabola, $$v>v_e$$ hyperbola.

$$v_e=$$ escape velocity in that orbit
$$v_o=$$ orbital velocity in that orbit

I asked my teacher for the reasons for the above miracles. He told me that it is experimentally observed.

But I can't believe how exactly an ellipse could be predicted experimentally without knowing the equation of a conic.

Could someone help me knowing the reasons?

• Does this answer your question? Why does an orbit become hyperbolic when total orbital energy is positive?, in particular the third answer Jun 19, 2023 at 19:42
• @DanDan0101 I am unable to understand that language of physics. (As per now) Jun 19, 2023 at 20:10
• I know you are confused, but as the current wording of the question, I would have to point out that it is obvious that trajectories are velocity specific. Just by changing the initial velocity of a projectile, you can convert between something moving in a curve to just a dropping straight down a line. Of course, Claudio below properly answers the equation of conic part, so I will omit that here. Jun 20, 2023 at 4:42
• These trajectories follow theoretically from Newtonian mechanics and Newton’s law of gravity. The point of these theories was of course to agree with observation, but once you know the theory you don’t have to observe anything other than the initial conditions to determine trajectories using math. You solve some differential equations and… bingo!… you get the various conic sections. What your teacher told you ignores the glorious success of having this theoretical understanding, and that’s just sad. Observation without theory is as boring as theory without observation. Jun 20, 2023 at 6:06