We consider an object of mass m that orbits a much more massive object of mass M. Initially, m is at r = (Subscript[r, 0]) and with a velocity v= (Subscript[v, 0]) in the transverse direction (i.e. (OverVector[(Subscript[v, 0])])⊥ (OverVector[(Subscript[r, 0])])).

The mechanical energy is a constant of the motion and is given by: 
(Subscript[E, mechanical]) = (1/2)m (v^2) - ((G M m)/r) = constant

The constant value of (Subscript[E, mechanical]) can be computed at the initial location
(Subscript[E, mechanical]) = (1/2)m ((Subscript[v, 0])^2) - ((G M m)/(Subscript[r, 0]))
Or, using the constant of motion h equal to  h = (r^2)(OverDot[θ]) = (Subscript[v, 0])(Subscript[r, 0]), 
(Subscript[E, mechanical]) = (1/2)m ((h^2)/((Subscript[r, 0])^2)) - ((G M m)/(Subscript[r, 0]))

In polar coordinates, (v^2)= ((OverDot[r])^2) + (((r (OverDot[θ])))^2) = ((OverDot[r])^2) + ((h^2)/(r^2)), so the general form of (Subscript[E, mechanical]) may be written as
(Subscript[E, mechanical]) = (1/2)m (((OverDot[r])^2) + ((h^2)/(r^2))) - ((G M m)/r)
If we write 
 (Subscript[V, eff]) = (1/2)m  ((h^2)/(r^2))- ((G M m)/r)
sometimes called the effective potential (http://en.wikipedia.org/wiki/Effective_potential). 
the mechanical energy becomes
(Subscript[E, mechanical]) = (1/2)m ((OverDot[r])^2) + (Subscript[V, eff])
Since (1/2)m((OverDot[r])^2) ≥ 0, we must have (Subscript[V, eff])≤ (Subscript[E, mechanical])
At the closest and farthest points (perigee and apogee when orbiting the Earth), (Subscript[V, eff])= (Subscript[E, mechanical]). 

To simplify the plots, we can introduce unitless quantities. The variables of the problem are r and h. They can be made unitless by dividing r by (Subscript[r, 0]) and  h by (Sqrt[G M (Subscript[r, 0])]).
If we write h' = (h/(Sqrt[G M (Subscript[r, 0])]))and r' = (r/(Subscript[r, 0])), we get 
(Subscript[E, mechanical])= ((G  M m )/(Subscript[r, 0]))((1/2)(((h'))^2) - 1)
(Subscript[V, eff]) = ((G  M m )/(Subscript[r, 0]))((1/2)((((h'))^2)/(r')) - (1/(r'))) 
In the animation, h’ varies from .5 to 2. At h’=0.5, (Subscript[r, 0]) is the r coordinate of the apogee of the ellipse ((Subscript[r, a])'= 1, (Subscript[r, p])'= (((h')^2)/(2-((h')^2)))= 0.14). As h’ increases, (Subscript[v, 0]) increases and (Subscript[r, apogee]) gets closer to (Subscript[r, perigee]). The trajectory becomes a circle for h’= 1.  For 1≤h'≤ (Sqrt[2]), r’=1 is the perigee, and (Subscript[r, a])'= (((h')^2)/(2-((h')^2))) the apogee.  For h = (Sqrt[2]), (Subscript[E, mechanical]) = 0, and the trajectory is a parabola. If h > (Sqrt[2]), the trajectory is a hyperbola.
Using the polar coordinates (r’, θ), the equation of the trajectory is given by
r' = (((h')^2)/(1 + e Cos[θ]))
The eccentricity e can be found from r’= 1 for θ = 0, that is 1 = ((((h'))^2)/(1+e))or e = (((h'))^2)-1.

Emech[h_] = (1/2) (h^2) - 1; (* energy in units of ((G M m)/(Subscript[r, 0])), and h in units of (Sqrt[G M (Subscript[r, 0])]) *)
Veff[r_, h_] =(1/2) ((h^2)/(r^2))- (1/r); (* r in units of (Subscript[r, 0]) *)
r[θ_,h_] = ((h^2)/(1 + ((h^2)-1) Cos[θ])); (* polar equation of the trajectory *)
raOrp[h_] = If[2-(h^2)>0, ((h^2)/(2-(h^2))), +Infinity]; (* closest and farthest distances on the trajectory *)
(* Print the energy, eccentricity, apogee, and perigee to 2 decimal places*)
message[h_] := "E = "<>ToString[((Round[Emech[h]100])/100.0)]<>", e = " <>ToString[((Round[Abs[(h^2)-1]100])/100.0)]<>", rp = " <>ToString[((Round[Min[1,raOrp[h]]100])/100.0)]<>", ra = " <>ToString[((Round[Max[1,raOrp[h]]100])/100.0)];
Manipulate[Grid[{{Show[Plot[{Veff[r,h], Emech[h]}, {r,Min[((h^2)/3),1], 4(h^2)}, AxesOrigin->{0, 0}, ImageSize->Large, PlotRange->All],Graphics[Text[ Style[message[h], 18],Scaled[{0.5,0.9}]]]]},{PolarPlot[r[θ,h], {θ,0, 2π}, ImageSize->Large]}}] ,  {h, .5,2}]