Solution

Let's consider two cases:

1. Velocity $\vec{v}$ in radial direction:

For this case, let's consider preservation of energy $$\frac{mv^2}{2}-\frac{GmM}{R_0}=\frac{m{v'}^2}{2}-\frac{GmM}{R_1}$$ Analyzing the components, when $v'\to 0,$ $R_1$ tends to the maximum value $R_1 \to R_\text{1max}$ $$\frac{mv^2}{2} - \frac{GmM}{R_0} = - \frac{GmM}{R_\text{1max}}$$ $$\frac{1}{2}v^2 = GM \left( \frac{1}{R_0} - \frac{1}{R_\text{1max}} \right)$$ 2) The velocity $v$ is perpendicular to the radius

From Kepler's first law, the orbit will be an ellipse

_$R_0$ - radius of circular orbit $R_1$ - maximum orbit at radial velocity $R_2$ - maximum orbit at perpendicular velocity_

Similarly, the preservation of energy $$\frac{mv^2}{2}-\frac{GmM}{R_0}=\frac{m{v'}^2}{2}-\frac{GmM}{R_\text{2max}}\quad(1)$$ Conservation of angular momentum $$mvR_0=m{v'}R_\text{2max}$$ From where $$v'=v\frac{R_0}{R_\text{2max}}\quad(2)$$ Substituting $(2)$ into $(1)$ $$\frac{v^2}{2}\left(1+\frac{R_0}{R_\text{2max}}\right) = GM \frac{1}{R_0}$$ Now, let's consider circular motion:

Given the centripetal acceleration $a=\frac{u^2}{R_0}$, the presence of rotational motion of which causes centrifugal force, compensating gravitational force, in the equilibrium condition. $$\frac{mu^2}{R_0}=\frac{GmM}{R_0^2} \Leftrightarrow \frac{GM}{R_0}=u^2$$ Substituting into the final formula of motion $$\frac{1}{2}v^2=u^2\left(1-\frac{R_0}{R_\text{1max}}\right)$$ $$\frac{1}{2}v^2\left(1+\frac{R_0}{R_\text{2max}}\right)=u^2$$ Transforming the obtained expressions $$R_0=R_\text{1max}\cdot\left(1-\frac{1}{2}\frac{v^2}{u^2}\right)=R_\text{2max}\cdot\left(2\frac{u^2}{v^2}-1\right)$$ $$R_\text{1max}\cdot \frac{2u^2-v^2}{2u^2} = R_\text{2max}\cdot \frac{2u^2-v^2}{v^2}$$ From where we express the ratio of the maximum possible distances from the probes to the center of the planet $$\boxed{\frac{R_\text{1max}}{R_\text{2max}}=\frac{2u^2}{v^2}}$$