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+ <meta name="description" content="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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+ <title>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.</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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:</p><p>
+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</p><p>
+From Kepler's first law, the orbit will be an ellipse
+</p>
+<center>
+ <figure>
+ <img src="2.6.41_1.png"
+ loading="lazy" width="300" />
+ <figcaption>
+ Elliptical orbit
+ </figcaption>
+ </figure>
+</center>
+<p>
+ <i>$R_0$ - radius of circular orbit<br>
+$R_1$ - maximum orbit at radial velocity<br>
+$R_2$ - maximum orbit at perpendicular velocity</i></p><p>
+
+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:</p><p>
+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
+$$\boxed{\frac{R_\text{1max}}{R_\text{2max}}=\frac{2u^2}{v^2}}$$
+ </p>
+
+ <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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