Answer
Since ${a_A} > {a_B}$, so by Kepler's Third Law: ${T_A} > {T_B}$. Hence, $A$'s orbital period ${T_A}$ will increase and she will fall farther and farther behind $B$.
Work Step by Step
From Exercise 20, we obtain the total energy: $E = - \frac{{GMm}}{{2a}}$. Since $A$ and $B$ orbit the earth along the same solid trajectory, their total energies are the same. Notice that the larger the value of $a$, the larger is the total energy. Thus, shuttle $A$ will move off into a larger orbit if its pilot increases the shuttle's kinetic energy.
Let ${a_B} = a$. Let ${a_A}$ denote the semi-major axis of $A$ after it moves off into a larger orbit. So, ${a_A} > {a_B}$.
By Kepler's Third Law, we have
${T_A}^2 = \left( {\frac{{4{\pi ^2}}}{{GM}}} \right){a_A}^3$ ${\ \ \ }$ and ${\ \ \ }$ ${T_B}^2 = \left( {\frac{{4{\pi ^2}}}{{GM}}} \right){a_B}^3$
Since ${a_A} > {a_B}$, so ${T_A} > {T_B}$. Hence, $A$'s orbital period ${T_A}$ will increase and she will fall farther and farther behind $B$.
