# Eccentricity of projectile in Newtonian gravitational model

1. Sep 25, 2012

### taninamdar

When we were taught gravitation, we were taught that if a projectile (with mass very less compared to that of the Earth) is projected from the Earth, its path is:
1. Ellipse, if the velocity is less than critical (orbital) velocity.
2. Circle, if the velocity is exactly equal to critical velocity.
3. Ellipse, if the velocity is more than critical velocity, but less than escape velocity.
4. Parabola, if the velocity is exactly equal to escape velocity.
5. Hyperbola, if the velocity is more than escape velocity.
So for remembering these conditions, I had intuitively come up with a formula,
e = abs[(v - vc)/(ve - vc], where v is initial velocity of the body, vc is critical velocity, ve is the escape velocity and e is the eccentricity of the projectile. abs is absolute value.
This formula seems to work with all these cases (gives corresponding condition of eccentricity for given condition of velocity) and gives exact eccentricities for cases of circle and parabola. However, I am not sure whether it will give exact value of eccentricity of the projectile.
So, my question is, whether the formula really works and gives exact value of eccentricity?
Also, I would like to see the derivation, if it works.
Thanks. :)