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.

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    <h3 id="back-link"><a href="/#2.6">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.6.41^*.$ 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.
    </p>

    <h3>Solution</h3>
    <p>
        Let's consider two cases:</p><p>

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
$$\boxed{\frac{R_\text{1max}}{R_\text{2max}}=\frac{2u^2}{v^2}}$$

    <h4>Answer</h4>
    <p>
        $$R_1/R_2=2u^2/v^2$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Almaskhan Arsen<br>
    </p>


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