Question

Energy of a hyperbolic orbit

Answer 1

Total energy is given as \[E=\frac{ GM }{ 2c }(k^2-2)\] where c is distance from the center of the Earth to the spacecraft. If I know that \[k>\sqrt{2}\] and that trajectory is hyperbolic, why does it hold that \[\frac{ GM }{ 2c }(k^2-2)=\frac{ GM }{ 2a }\] where a is standard hyperbola parameter?

Answer 2

Maybe this will be clearer :)

Answer 3

I don't understand equation for angular momentum too