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| <title>The plane of the satellite's orbit is divided into sectors with a common vertex in the center of the planet of mass M and the same small solution angles d \varphi. Find the change in the satellite's speed as it passes through each sector, if its speed is in the pericenter v_p, and the distance from the satellite to the center of the planet is in the pericenter r_p.</title> | | <title>The plane of the satellite's orbit is divided into sectors with a common vertex in the center of the planet of mass M and the same small solution angles d \varphi. Find the change in the satellite's speed as it passes through each sector, if its speed is in the pericenter v_p, and the distance from the satellite to the center of the planet is in the pericenter r_p.</title> |
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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> |
| $2.6.42^*.$ The plane of the satellite's orbit is divided into sectors with a common vertex in the center of the planet of mass $M$ and the same small solution angles $d \varphi$. Find the change in the satellite's speed as it passes through each sector, if its speed is in the pericenter $v_p$, and the distance from the satellite to the center of the planet is in the pericenter $r_p$. | | $2.6.42^*.$ The plane of the satellite's orbit is divided into sectors with a common vertex in the center of the planet of mass $M$ and the same small solution angles $d \varphi$. Find the change in the satellite's speed as it passes through each sector, if its speed is in the pericenter $v_p$, and the distance from the satellite to the center of the planet is in the pericenter $r_p$. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="2.6.42.png" | | <img src="2.6.42.png" |
| loading="lazy" width="120" /> | | loading="lazy" width="120" /> |
| <figcaption> | | <figcaption> |
| For problem $2.6.42^*$ | | For problem $2.6.42^*$ |
| </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="2.6.42_1.png" | | <img src="2.6.42_1.png" |
| loading="lazy" width="300" /> | | loading="lazy" width="300" /> |
| <figcaption> | | <figcaption> |
| Change in speed over a short period of time | | Change in speed over a short period of time |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| For a short time, $dt$, the body moves along the arc of the circle of arc $d\varphi$ with radius-curvature of the trajectory $R_n$.</p><p> | | For a short time, $dt$, the body moves along the arc of the circle of arc $d\varphi$ with radius-curvature of the trajectory $R_n$.</p><p> |
| Newton's second law for an external force, the gravitational force | | Newton's second law for an external force, the gravitational force |
| $$ma=\frac{GmM}{R_n^2}$$ | | $$ma=\frac{GmM}{R_n^2}$$ |
| $$\frac{dv}{dR}v_n=\frac{GM}{R_n^2}\Rightarrow \frac{dv}{d\varphi}v_n=\frac{GM}{R_n}$$ | | $$\frac{dv}{dR}v_n=\frac{GM}{R_n^2}\Rightarrow \frac{dv}{d\varphi}v_n=\frac{GM}{R_n}$$ |
| After the transformation, we can get a change in the satellite's speed | | After the transformation, we can get a change in the satellite's speed |
| $$\boxed{dv=\frac{GM\cdot d\varphi}{R_nv_n}}$$ | | $$\boxed{dv=\frac{GM\cdot d\varphi}{R_nv_n}}$$ |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Almaskhan Arsen<br> | | Almaskhan Arsen<br> |
| </p> | | </p> |
| <h3>Alternative solution</h3> | | <h3>Alternative solution</h3> |
| <p> | | <p> |
| Kepler's second law states that the speed of sweeping out a sector area is constant, i.e. | | Kepler's second law states that the speed of sweeping out a sector area is constant, i.e. |
| $$ | | $$ |
| \frac{dS}{dt}=\text{const} | | \frac{dS}{dt}=\text{const} |
| $$ | | $$ |
| We know the satellite's distance $r_n$ at the pericenter and its velocity $v_n$ at this point. At the pericenter, over a small interval $dt$, a very small sector is swept out, which can be considered an isosceles triangle with the thigh length $r_n$ and the base $v_n dt$. So at this point $dS_n=\frac{1}{2}v_ndt \cdot r_n$, or | | We know the satellite's distance $r_n$ at the pericenter and its velocity $v_n$ at this point. At the pericenter, over a small interval $dt$, a very small sector is swept out, which can be considered an isosceles triangle with the thigh length $r_n$ and the base $v_n dt$. So at this point $dS_n=\frac{1}{2}v_ndt \cdot r_n$, or |
| $$ | | $$ |
| \frac{dS}{dt}=\frac{v_nr_n}{2}. | | \frac{dS}{dt}=\frac{v_nr_n}{2}. |
| $$ | | $$ |
| Now let's move on to the question of the change in speed with each passage of the angle $d\varphi$. If at the moment the distance between the satellite and the planet is equal to $r$, then | | Now let's move on to the question of the change in speed with each passage of the angle $d\varphi$. If at the moment the distance between the satellite and the planet is equal to $r$, then |
| $$ | | $$ |
| dS=\frac{1}{2}r^2 d\varphi. | | dS=\frac{1}{2}r^2 d\varphi. |
| $$ | | $$ |
| Newton's second law is written as follows: | | Newton's second law is written as follows: |
| $$ | | $$ |
| \frac{dv_r}{dt}=\frac{GM}{r^2}=\frac{GM}{v_n r_n}\frac{d\varphi}{dt}, | | \frac{dv_r}{dt}=\frac{GM}{r^2}=\frac{GM}{v_n r_n}\frac{d\varphi}{dt}, |
| $$ | | $$ |
| where the index $r$ denotes the change in the radial component of the velocity in the direction toward the center of the planet. From here we obtain the final answer: | | where the index $r$ denotes the change in the radial component of the velocity in the direction toward the center of the planet. From here we obtain the final answer: |
| $$ | | $$ |
| dv=\frac{GM}{v_n r_n}d\varphi. | | dv=\frac{GM}{v_n r_n}d\varphi. |
| $$ | | $$ |
| | | |
| </p> | | </p> |
| <p style="text-align: right; font-style: italic; font-size: 14;"> | | <p style="text-align: right; font-style: italic; font-size: 14;"> |
| Alisher Yerkebayev<br> | | Alisher Yerkebayev<br> |
| </p> | | </p> |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $dv=\gamma Md\varphi/(v_{\mathrm{n}}r_{\mathrm{n}})$. The vector $dv$ is directed towards the centre of the planet | | $dv=\gamma Md\varphi/(v_{\mathrm{n}}r_{\mathrm{n}})$. The vector $dv$ is directed towards the centre of the planet |
| </p> | | </p> |
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