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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> |
| $2.6.37.$ The Molniya-1 communications satellite has a perigee over the southern hemisphere of the Earth at an altitude of about $500$ km, and an apogee at an altitude of about $40,000$ km over the northern hemisphere. What is the ratio of the angular velocities of rotation of this satellite at perigee and apogee? | | $2.6.37.$ The Molniya-1 communications satellite has a perigee over the southern hemisphere of the Earth at an altitude of about $500$ km, and an apogee at an altitude of about $40,000$ km over the northern hemisphere. What is the ratio of the angular velocities of rotation of this satellite at perigee and apogee? |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.6.37.png" | | <img src="2.6.37.png" |
| loading="lazy" width="140" /> | | loading="lazy" width="140" /> |
| <figcaption> | | <figcaption> |
| For problem $2.6.37$ | | For problem $2.6.37$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.6.37_1.png" | | <img src="2.6.37_1.png" |
| loading="lazy" width="300" /> | | loading="lazy" width="300" /> |
| <figcaption> | | <figcaption> |
| Sattelite at apogee and perigee | | Sattelite at apogee and perigee |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| Law of energy preservation | | Law of energy preservation |
| $$\frac{mv^2_a}{2}-G\frac{mM}{R_a}=\frac{mv^2_p}{2}-G\frac{mM}{R_p}\quad (1)$$ | | $$\frac{mv^2_a}{2}-G\frac{mM}{R_a}=\frac{mv^2_p}{2}-G\frac{mM}{R_p}\quad (1)$$ |
| Conservation of angular momentum | | Conservation of angular momentum |
| $$mv_aR_a=mv_pR_p$$ | | $$mv_aR_a=mv_pR_p$$ |
| From where velocity at closest position to the planet, at perigee | | From where velocity at closest position to the planet, at perigee |
| $$v_p=v_a\frac{R_a}{R_p}\quad (2)$$ | | $$v_p=v_a\frac{R_a}{R_p}\quad (2)$$ |
| Let's substitute value of $v_p$ to $(1)$: | | Let's substitute value of $v_p$ to $(1)$: |
| $$\frac{v_a^2}{2} = \frac{GMR_p}{R_a(R_a+R_p)}$$ | | $$\frac{v_a^2}{2} = \frac{GMR_p}{R_a(R_a+R_p)}$$ |
| From where, we could value of the velocity at farthest position to the planet, at apogee | | From where, we could value of the velocity at farthest position to the planet, at apogee |
| $$v_a=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a}}$$ | | $$v_a=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a}}$$ |
| Considering dependence $v_p(v_a)$ from $(2)$, we obtain | | Considering dependence $v_p(v_a)$ from $(2)$, we obtain |
| $$v_p=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p}}$$ | | $$v_p=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p}}$$ |
| By definition of angular velocity | | By definition of angular velocity |
| $$\omega_a = \frac{v_a}{R_a}=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a^3}}$$ | | $$\omega_a = \frac{v_a}{R_a}=\sqrt{\frac{2GMR_p}{(R_a+R_p)R_a^3}}$$ |
| $$\omega_p = \frac{v_p}{R_p}=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p^3}}$$ | | $$\omega_p = \frac{v_p}{R_p}=\sqrt{\frac{2GMR_a}{(R_a+R_p)R_p^3}}$$ |
| The desired relationship could be found by dividing one equation on another | | The desired relationship could be found by dividing one equation on another |
| $$\xi = \frac{\omega_p}{\omega_a} = \frac{R_a^2}{R_p^2}\quad (3)$$ | | $$\xi = \frac{\omega_p}{\omega_a} = \frac{R_a^2}{R_p^2}\quad (3)$$ |
| Considering the radius of the Earth | | Considering the radius of the Earth |
