$2.6.41^*.$ Two probes are launched from an orbital station moving at a speed of $u$ in a circular orbit around the planet. The initial velocity of the probes relative to the planet is $v$ ($\sqrt{2} u > v > u$). One probe moves in the direction of the planet's radius; the initial velocity of the other probe is perpendicular to its radius. Find the ratio of the maximum possible distances from the probes to the center of the planet.
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 formulae 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 $$\boxed{\frac{R_\text{1max}}{R_\text{2max}}=\frac{2u^2}{v^2}}$$
$$R_1/R_2=2u^2/v^2$$
Almaskhan Arsen