2.6.42 | revision #8667 | Savchenko Solutions

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.

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    <h3> Statement </h3>
    <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 $r_p$.

    <h3>Solution</h3>
    <p>

Change in speed over a short period of time

For a short time, $$dt$$ , the body moves along the arc of the circle of arc $d\varphi$ with radius-curvature of the trajectory $$R_n$$ .

Newton's second law for an external force, the gravitational force
$ma=\frac{GmM}{R_n^2}$
By definition of acceleration
$a=\frac{dv}{dt}\Rightarrow \frac{dv}{dt}=\frac{GM}{R_n^2}\quad (1)$
In a small amount of time
$dR=v_ndt\quad (2)$
By definition the arc length of the circle corresponding to the angle $d\varphi$
$dR=d\varphi \cdot R_n\quad (3)$
After substituting $$(2)$$ and $$(3)$$ into $$(1)$$
$\frac{dv}{dR}v_n=\frac{GM}{R_n^2}\Rightarrow \frac{dv}{d\varphi}v_n=\frac{GM}{R_n}$
After the transformation, we can get a change in the satellite's speed
$\boxed{dv=\frac{GM\cdot d\varphi}{R_nv_n}}$

Almaskhan Arsen

Alternative solution

Kepler's second law states that the speed of sweeping out a sector area is constant, i.e.
$\frac{dS}{dt}=\text{const}$
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
$\frac{dS}{dt}=\frac{v_nr_n}{2}.$
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
$dS=\frac{1}{2}r^2 d\varphi.$
Newton's second law is written as follows:
$\frac{dv_r}{dt}=\frac{GM}{r^2}=\frac{GM}{v_n r_n}\frac{d\varphi}{dt},$
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:
$dv=\frac{GM}{v_n r_n}d\varphi.$

    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Alisher Yerkebayev<br>
    </p>
    <h4>Answer</h4>
    <p>
        $dv=\gamma Md\varphi/(v_{\mathrm{n}}r_{\mathrm{n}})$. The vector $dv$ is directed towards the centre of the planet
    </p>


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