Added 2.6.43

astrosander edited
revision #6378 parent #6375 GitHub e888bb6 ← older newer →
@@ -89,13 +89,39 @@
After the transformation, we can get a change in the satellite's speed
$$\boxed{dv=\frac{GM\cdot d\varphi}{R_nv_n}}$$
</p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
+ </p>
+ <h3>Alternative solution</h3>
+ <p>
+ 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>
− <p style="text-align: right; font-style: italic; font-size: 14;">
− Almaskhan Arsen<br>
</p>
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