Solution

From conservation of angular momentum, we have $m_v_pr_p=mv_ar_a$, where $v_p$ and $v_a$ are the speeds at the pericenter and the apocenter, respectively. Together with the conservation of energy

$$\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a},$$

we can solve for $v_p^2$ in terms of $r_p$ and $r_a$ and get

$$v_p^2=\frac{2GMr_a}{r_p(r_p+r_a)}.$$

Consequently, the energy of the space probe at the pericenter is

$$\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.$$

Therefore, the space probe requires a minimum energy of $GMm/(r_p+r_a)$ to escape from the planet (i.e. approaching infinity with very little kinetic energy left).