Two final velocities for a mass launched from Earth

Question

i'm trying to find the final velocity of a mass that is launched vertically from the earth, only considering the gravitational force. My first attempt was with angular momentum: $r$ is the distance from the center of earth to the final position, $v_0$ is the velocity of launch. $\vec l= \vec r \times\vec p \Rightarrow \dot{\vec{l}} = \vec r \times \frac{-GmM}{r^3} \vec r =0 $ then $l_o=R_tv_0m=l_f=rmv \Rightarrow v=\frac{R_t v_0 }{r}$ But then i tried with energy and i got other velocity:

$T_2-T_1=GmM(\frac{1}{r}-\frac{1}{R_t}) \Rightarrow v^2=2GM(\frac{1}{r}-\frac{1}{R_t})+v_0^2$

Why this velocities are different?

Answer 1

If the mass is launched directly away from the Earth, then its angular momentum is zero, since the displacement from the center of the earth $\mathbf{r}$ is parallel to the momentum of the mass $\mathbf{p}$ (due to the cross product in the definition of angular momentum, $\mathbf{L}=\mathbf{r}\times\mathbf{p}$).

Conservation of angular momentum then tells you that the mass always travels along a line directed away from the center of the Earth, but it doesn't tell you the magnitude of the velocity.