A satellite moves in a circular orbit of radius r about a planet of mass M (speed v0, period T0). By a short rocket burn its speed is suddenly raised…

With $v_0^2=GM/r$, the new specific energy is $\varepsilon=\tfrac12(1.44)v_0^2-v_0^2=-0.28\,GM/r$. For an ellipse $\varepsilon=-GM/2a$, so $a=r/0.56=25r/14$. The burn point is the nearest point (velocity is perpendicular to the radius), so the farthest distance is $r_a=2a-r=18r/7$: (A) is true. Angular momentum conservation gives $v_ar_a=1.2v_0r$, so $v_a=1.2\times7/18\,v_0=7v_0/15$: (B) is true. Kepler's third law gives $T/T_0=(a/r)^{3/2}=(25/14)^{1.5}\approx2.39$: (C) is true. The energy is negative ($-0.28\,GM/r$, since $1.2<\sqrt2$ the escape speed ratio), so the orbit is bound: (D) is false. Kepler's second law is consistent with this: the satellite moves slowest at the farthest point, where the speed is only $7/15\approx0.47$ of the original circular speed, and fastest at the burn point where it began.