Правка разделов «Statement», «Answer»

JMMA2006 правка от 3 июл. 2026 г., 19:34

правка #19605 предыдущая #19604 ← раньше

en/2.6.34.md +2 −2

		@@ -1,6 +1,6 @@
			### Statement

		−	$2.6.34.$ ~~[Insert the problem statement]~~
		+	$2.6.34.$ The kinetic energy of a satellite in a circular orbit is equal to K. What is its potential energy?

			### 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\
			$U=-2K$
		@@ -21,4 +21,4 @@Solution

			#### Answer

		−	[Insert a concise answer or boxed result]
		+	$U=-2K$

	@@ -1,6 +1,6 @@
	### Statement		### Statement

	$2.6.34.$ ~~[Insert the problem statement]~~		$2.6.34.$ The kinetic energy of a satellite in a circular orbit is equal to K. What is its potential energy?

	### Solution		### Solution

	For a circular orbit, the gravitational force provides the centripetal force\		For a circular orbit, the gravitational force provides the centripetal force\
	$\frac{GmM}{r^2}=\frac{mv^2}{r}$\		$\frac{GmM}{r^2}=\frac{mv^2}{r}$\
	Cancel m and multiply by r\		Cancel m and multiply by r\
	$\frac{GM}{r}=v^2$ (1)\		$\frac{GM}{r}=v^2$ (1)\
	The kinetic energy is\		The kinetic energy is\
	$K=\frac{mv^2}{2}$\		$K=\frac{mv^2}{2}$\
	$v^2=\frac{2K}{m}$\		$v^2=\frac{2K}{m}$\
	And subsituting into equation (1)\		And subsituting into equation (1)\
	$\frac{GM}{r}=\frac{2K}{m}$\		$\frac{GM}{r}=\frac{2K}{m}$\
	we get\		we get\
	$2K=\frac{GmM}{r}$\		$2K=\frac{GmM}{r}$\
	The potential energy is\		The potential energy is\
	$U=-\frac{GmM}{r}$ , so\		$U=-\frac{GmM}{r}$ , so\
	$U=-2K$		$U=-2K$
	@@ -21,4 +21,4 @@Solution

	#### Answer		#### Answer

	[Insert a concise answer or boxed result]		$U=-2K$