 Solution

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.