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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $2.6.46^*.$ A satellite moves around a planet of mass $M$ in an ellipse with semi-major and minor axes $a$ and $b$. Determine the area that the radius vector drawn from the center of the planet to the satellite "sweeps" per unit time. Find the period of rotation of the satellite. | | $2.6.46^*.$ A satellite moves around a planet of mass $M$ in an ellipse with semi-major and minor axes $a$ and $b$. Determine the area that the radius vector drawn from the center of the planet to the satellite "sweeps" per unit time. Find the period of rotation of the satellite. |
| </p> | | </p> |
| <center> | | <center> |
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| <img src="2.6.46.png" | | <img src="2.6.46.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $2.6.46^*$ | | For problem $2.6.46^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Kepler's Third Law | | Kepler's Third Law |
| $$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$ | | $$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$ |
| From where | | From where |
| $$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$ | | $$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$ |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.6.46_1.png" | | <img src="2.6.46_1.png" |
| loading="lazy" width="290" /> | | loading="lazy" width="290" /> |
| <figcaption> | | <figcaption> |
| $v$ - sweep speed | | $v$ - sweep speed |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Kepler's Second Law | | Kepler's Second Law |
| $$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$ | | $$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$ |
| $$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$ | | $$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$ |
| From here the velocity $v$ could be found as | | From here the velocity $v$ could be found as |
| $$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$ | | $$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$ |
| </p> | | </p> |
| | | |
| | | |
| <h3>Alternative solution</h3> | | <h3>Alternative solution</h3> |
| <p> | | <p> |
| The radius of curvature of the orbit at the apex of the major axis of the ellipse | | The radius of curvature of the orbit at the apex of the major axis of the ellipse |
| \[ | | \[ |
| R = \frac{a}{k^2} = \frac{b^2}{a}. | | R = \frac{a}{k^2} = \frac{b^2}{a}. |
| \] | | \] |
| Therefore | | Therefore |
| \[ | | \[ |
| \frac{v^2}{R} = \frac{v^2 a}{b^2} = \frac{G M}{r^2} \rightarrow vr = \sqrt{G M \frac{b^2}{a}}, | | \frac{v^2}{R} = \frac{v^2 a}{b^2} = \frac{G M}{r^2} \rightarrow vr = \sqrt{G M \frac{b^2}{a}}, |
| \] | | \] |
| \[ | | \[ |
| \frac{dS}{dt} = \frac{1}{2}vr = \frac{1}{2}b \sqrt{\frac{G M}{a}}. | | \frac{dS}{dt} = \frac{1}{2}vr = \frac{1}{2}b \sqrt{\frac{G M}{a}}. |
| \] | | \] |
| Satellite orbital period | | Satellite orbital period |
| \[ | | \[ |
| T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}. | | T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}. |
| \] | | \] |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
