Eccentricity of projectile in Newtonian gravitational model

[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. :)

 

[gtoguy97]

A:The formula you present is a good approximation to the eccentricity of an elliptic orbit, but in general is not exact. The eccentricity of an elliptic orbit is given by$$e = \sqrt{1 - \frac{b^2}{a^2}}$$where $a$ and $b$ are the semi-major and semi-minor axes of the ellipse, respectively. This can be derived from the equation for the ellipse$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$by taking the derivative with respect to $x$ and solving for the slope of the tangent line at any point on the ellipse.If you make the approximation that the escape velocity is much greater than the orbital velocity, then you can approximate the semi-major axis as the ratio of the escape velocity to the orbital velocity times the orbital radius, i.e. $a \approx r\frac{v_e}{v_c}$, and the semi-minor axis as the orbital radius, i.e. $b \approx r$. Substituting these into the equation for eccentricity gives$$e \approx \sqrt{1 - \left(\frac{r}{r\frac{v_e}{v_c}}\right)^2} = \sqrt{1 - \left(\frac{v_c}{v_e}\right)^2}$$which is approximately equal to the expression you presented, i.e.$$e \approx \left|\frac{v - v_c}{v_e - v_c}\right|$$as long as $v_e \gg v_c$.