Fixed bags in 2.6.42, 2.6.46, 2.8.11, 4.1.12 sol.

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			<h2>Solutions of Savchenko Problems in Physics</h2>
			<p class="author">
			Aliaksandr Melnichenka <br/>
			October 2023
			</p>
			</header>

			<h3 id="back-link"><a href="../../#2.6">$\leftarrow$Back</a></h3>

			<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.
			</p>
			<center>
			<figure>
			<img src="2.6.46.png"
			loading="lazy" width="230" />
			<figcaption>
			For problem $2.6.46^*$
			</figcaption>
			</figure>
			</center>
			<p>
			</p>

			<h3>Solution</h3>
			<p>
			Kepler's Third Law
			$$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$
			From where
			$$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$

			</p>
			<center>
			<figure>
			<img src="2.6.46_1.png"
			loading="lazy" width="290" />
			<figcaption>
			$v$ - sweep speed
			</figcaption>
			</figure>
			</center>
			<p>
			Kepler's Second Law
			$$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$
			$$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$
		@@ -84,6 +84,9 @@
			From here the velocity $v$ could be found as
			$$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$
			</p>
		+	<p style="text-align: right; font-style: italic; font-size: 14;">
		+	Almaskhan Arsen<br>
		+	</p>


			<h3>Alternative solution</h3>
			<p>
			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
			\[
			T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}.
			\]
			</p>

			<h4>Answer</h4>
		@@ -109,9 +112,6 @@
			<p>
			$$T=2\pi\sqrt{\frac{a^3}{GM}}$$
			$$v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}$$
		−	</p>
		−	<p style="text-align: right; font-style: italic; font-size: 14;">
		−	Almaskhan Arsen<br>
			</p>


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	<header style="text-align:center;">		<header style="text-align:center;">
	<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>
	$2.6.46^*.$ 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.		$2.6.46^*.$ 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.
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="2.6.46.png"		<img src="2.6.46.png"
	loading="lazy" width="230" />		loading="lazy" width="230" />
	<figcaption>		<figcaption>
	For problem $2.6.46^*$		For problem $2.6.46^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	Kepler's Third Law		Kepler's Third Law
	$$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$		$$\frac{T^2}{a^3}=\frac{4\pi ^2}{GM}$$
	From where		From where
	$$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$		$$\boxed{T=2\pi\sqrt{\frac{a^3}{GM}}}$$

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="2.6.46_1.png"		<img src="2.6.46_1.png"
	loading="lazy" width="290" />		loading="lazy" width="290" />
	<figcaption>		<figcaption>
	$v$ - sweep speed		$v$ - sweep speed
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	Kepler's Second Law		Kepler's Second Law
	$$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$		$$\frac{dS}{dt}=\text{const};\quad \frac{dS}{dt}=v$$
	$$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$		$$v=\frac{S}{T}=\frac{\pi ab}{2\pi}\cdot\sqrt{\frac{GM}{a^3}}$$
	@@ -84,6 +84,9 @@
	From here the velocity $v$ could be found as		From here the velocity $v$ could be found as
	$$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$		$$\boxed{v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}}$$
	</p>		</p>
			<p style="text-align: right; font-style: italic; font-size: 14;">
			Almaskhan Arsen<br>
			</p>


	<h3>Alternative solution</h3>		<h3>Alternative solution</h3>
	<p>		<p>
	The radius of curvature of the orbit at the apex of the major axis of the ellipse		The radius of curvature of the orbit at the apex of the major axis of the ellipse
	\[		\[
	R = \frac{a}{k^2} = \frac{b^2}{a}.		R = \frac{a}{k^2} = \frac{b^2}{a}.
	\]		\]
	Therefore		Therefore
	\[		\[
	\frac{v^2}{R} = \frac{v^2 a}{b^2} = \frac{G M}{r^2} \rightarrow vr = \sqrt{G M \frac{b^2}{a}},		\frac{v^2}{R} = \frac{v^2 a}{b^2} = \frac{G M}{r^2} \rightarrow vr = \sqrt{G M \frac{b^2}{a}},
	\]		\]
	\[		\[
	\frac{dS}{dt} = \frac{1}{2}vr = \frac{1}{2}b \sqrt{\frac{G M}{a}}.		\frac{dS}{dt} = \frac{1}{2}vr = \frac{1}{2}b \sqrt{\frac{G M}{a}}.
	\]		\]
	Satellite orbital period		Satellite orbital period
	\[		\[
	T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}.		T = 2\pi \frac{ab}{\frac{dS}{dt}} = 2\pi \frac{a^{3/2}}{\sqrt{G M}}.
	\]		\]
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	@@ -109,9 +112,6 @@
	<p>		<p>
	$$T=2\pi\sqrt{\frac{a^3}{GM}}$$		$$T=2\pi\sqrt{\frac{a^3}{GM}}$$
	$$v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}$$		$$v=\frac{1}{2}b\cdot\sqrt{\frac{GM}{a}}$$
	</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
	Almaskhan Arsen<br>
	</p>		</p>


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