2.6.46 | revision #8669 | Savchenko Solutions

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.

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Solutions of Savchenko Problems in Physics
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    <h3> Statement </h3>
    <p>
        $2.6.46^*.$ 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.

    <h3>Solution</h3>
    <p>
        Kepler's Third Law

$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$
From where
$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$

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 $\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$

Almaskhan Arsen

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}}. \$

    <h4>Answer</h4>
    <p>
        $$T=2\pi\sqrt{\frac{a^3}{GM}}$$

$v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}$

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