Edits to “Statement”, “Solution”, “Answer”

JMMA2006 edited Jul 3, 2026, 7:09 PM

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revision #19600 parent #19599 ← older

en/2.6.12.md +6 −6

		@@ -1,17 +1,17 @@
			### Statement

		−	$2.6.12.$ ~~[Insert the problem statement]~~
		+	$2.6.12.$ Determine the radius of the asteroid's circular orbit if the angular velocity of its revolution around the Sun is w, and the mass of the Sun is m.

			### Solution

			For a circular orbit, the gravitational force provides the centripetal force:\
		−	$\frac{GmM}{r^2}=mw^2 r$\
		−	where $G$ is the gravitational constant, $M$ is the mass of the Sun, $m$ the asteroid's mass, $r$ is the orbital radius and $w$ is the angular velocity\
		+	$\frac{Gmm_1}{r^2}=m_1w^2 r$\
		+	where $G$ is the gravitational constant, $m$ is the mass of the Sun, $m_1$ the asteroid's mass, $r$ is the orbital radius and $w$ is the angular velocity\
			we get\
		−	$\frac{GM}{r^2}=w^2 r$\
		+	$\frac{Gm}{r^2}=w^2 r$\
			and from that\
		−	$r=(\frac{GM}{w^2})^1/3$
		+	$r=(\frac{Gm}{w^2})^1/3$

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$r=(\frac{Gm}{w^2})^1/3$

	@@ -1,17 +1,17 @@
	### Statement		### Statement

	$2.6.12.$ ~~[Insert the problem statement]~~		$2.6.12.$ Determine the radius of the asteroid's circular orbit if the angular velocity of its revolution around the Sun is w, and the mass of the Sun is m.

	### Solution		### Solution

	For a circular orbit, the gravitational force provides the centripetal force:\		For a circular orbit, the gravitational force provides the centripetal force:\
	$\frac{GmM}{r^2}=mw^2 r$\		$\frac{Gmm_1}{r^2}=m_1w^2 r$\
	where $G$ is the gravitational constant, $M$ is the mass of the Sun, $m$ the asteroid's mass, $r$ is the orbital radius and $w$ is the angular velocity\		where $G$ is the gravitational constant, $m$ is the mass of the Sun, $m_1$ the asteroid's mass, $r$ is the orbital radius and $w$ is the angular velocity\
	we get\		we get\
	$\frac{GM}{r^2}=w^2 r$\		$\frac{Gm}{r^2}=w^2 r$\
	and from that\		and from that\
	$r=(\frac{GM}{w^2})^1/3$		$r=(\frac{Gm}{w^2})^1/3$

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$r=(\frac{Gm}{w^2})^1/3$