Новое решение

JMMA2006 правка от 3 июл. 2026 г., 19:31

правка #19604 позже →

en/2.6.34.md +24 −0

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		+	### Statement
		+
		+	$2.6.34.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	For a circular orbit, the gravitational force provides the centripetal force\
		+	$\frac{GmM}{r^2}=\frac{mv^2}{r}$\
		+	Cancel m and multiply by r\
		+	$\frac{GM}{r}=v^2$ (1)\
		+	The kinetic energy is\
		+	$K=\frac{mv^2}{2}$\
		+	$v^2=\frac{2K}{m}$\
		+	And subsituting into equation (1)\
		+	$\frac{GM}{r}=\frac{2K}{m}$\
		+	we get\
		+	$2K=\frac{GmM}{r}$\
		+	The potential energy is\
		+	$U=-\frac{GmM}{r}$ , so\
		+	$U=-2K$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.6.34.$ [Insert the problem statement]

			### Solution

			For a circular orbit, the gravitational force provides the centripetal force\
			$\frac{GmM}{r^2}=\frac{mv^2}{r}$\
			Cancel m and multiply by r\
			$\frac{GM}{r}=v^2$ (1)\
			The kinetic energy is\
			$K=\frac{mv^2}{2}$\
			$v^2=\frac{2K}{m}$\
			And subsituting into equation (1)\
			$\frac{GM}{r}=\frac{2K}{m}$\
			we get\
			$2K=\frac{GmM}{r}$\
			The potential energy is\
			$U=-\frac{GmM}{r}$ , so\
			$U=-2K$

			#### Answer

			[Insert a concise answer or boxed result]