A satellite of mass m moves in a circular orbit of radius 2R around the Earth of mass M and radius R. Air drag is absent and the Earth does not…

For a circular orbit of radius $2R$, $v_c=\sqrt{GM/2R}$, so (A) is false. Then $KE=\frac{GMm}{4R}$, $PE=-\frac{GMm}{2R}$ and $E=-\frac{GMm}{4R}$, so (B) is true. With speed $2v_c$ the energy is $\frac12m\cdot4\frac{GM}{2R}-\frac{GMm}{2R}=+\frac{GMm}{2R}>0$, so the satellite escapes (C true). With speed $v_c/2$ the satellite is at apogee $2R$ of an ellipse. Angular momentum gives $v_pr_p=\frac{v_c}{2}\cdot2R=v_cR$ and energy gives $\frac12v_p^2-\frac{GM}{r_p}=-\frac{7GM}{16R}$. Solving with $GM=R=1$ yields $7r_p^2-16r_p+4=0$, i.e. $r_p=2$ or $r_p=\frac{2R}{7}<R$. The perigee is inside the Earth, so it strikes the surface (D true). Note that the energy of the new orbit decides escape, while the speed reduction does not change the position, so the point of the change always remains the apogee of the new orbit.