Question

The escape velocity from the surface of the earth of radius R is v. What is the escape velocity from a height of 2R?

Answer 1

$mv_0^2/2-GMm/R=0\to v_0=\sqrt{2GM/R}$

$mv^2/2-GMm/(2R)=0\to v=\sqrt{GM/R}=v_0/\sqrt{2}$ . Answer