Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

Statement

$2.6.46^*.$ 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.

For problem $2.6.46^*$

Solution

Kepler's Third Law $$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$ From where $$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$

$v$ - sweep speed

Kepler's Second Law $$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$ $$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$ From here the velocity $v$ could be found as $$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$

Alternative solution

The radius of curvature of the orbit at the apex of the major axis of the ellipse \[ R = \frac{a}{k^2} = \frac{b^2}{a}. \] Therefore \[ \frac{v^2}{R} = \frac{v^2 a}{b^2} = \frac{G M}{r^2} \rightarrow vr = \sqrt{G M \frac{b^2}{a}}, \] \[ \frac{dS}{dt} = \frac{1}{2}vr = \frac{1}{2}b \sqrt{\frac{G M}{a}}. \] Satellite orbital period \[ T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}. \]

Answer

$$T=2\pi\sqrt{\frac{a^3}{GM}}$$ $$v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}$$

Almaskhan Arsen