Answer:
42.96 km/s
Explanation:
From the conservation of Energy
[tex](PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2[/tex]
Mass gets cancelled
[tex]-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}[/tex]
[tex]v_e=\sqrt{\frac{2Gm}{R}}[/tex] = Escape velocity of Earth = 11.2 km/s
[tex]v_i[/tex] = Velocity of projectile = 44.4 km/s
[tex]v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s[/tex]
The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s
