Новое решение

JMMA2006 правка от 2 июл. 2026 г., 17:39

правка #19516 позже →

en/2.6.9.md +30 −0

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		+	### Statement
		+
		+	$2.6.9.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	By Newton's law gravitation:\
		+	$F=G\frac{m_1 m_2}{r^2}$\
		+	where $m_1$ is my mass, and $m_2$ is the mass of the celestial body, and $r$ is the distance between the center of the body and me\
		+	Using $G=6.674×10^{-11}\frac{Nm^2}{kg^2}$
		+	For me and Earth:\
		+	$m_1=70kg$\
		+	$m_earth=5.972×10^{24}kg$\
		+	$r_earth=6.371×10^6m (assuming that I'm at sea level)\
		+	and calculating\
		+	$F_earth\approx 686N$ (this is equeal to my weight!, $mg=70kg×9.8\frac{m}{s^2}$)
		+	For me and the Moon:\
		+	$m_Moon=7.342×10^22kg$\
		+	Average Earth–Moon distance: r=3.844×10^8m\
		+	and calculating\
		+	$F_Moon\approx 2.4×10^{-3}N$\
		+	For me and the Sun:\
		+	$m_Sun=1.989×10^{30}kg$\
		+	Average Earth–Sun distance: r=1.496×10^{11}m\
		+	and calculating\
		+	$F_Sun\approx 0.414N
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.6.9.$ [Insert the problem statement]

			### Solution

			By Newton's law gravitation:\
			$F=G\frac{m_1 m_2}{r^2}$\
			where $m_1$ is my mass, and $m_2$ is the mass of the celestial body, and $r$ is the distance between the center of the body and me\
			Using $G=6.674×10^{-11}\frac{Nm^2}{kg^2}$
			For me and Earth:\
			$m_1=70kg$\
			$m_earth=5.972×10^{24}kg$\
			$r_earth=6.371×10^6m (assuming that I'm at sea level)\
			and calculating\
			$F_earth\approx 686N$ (this is equeal to my weight!, $mg=70kg×9.8\frac{m}{s^2}$)
			For me and the Moon:\
			$m_Moon=7.342×10^22kg$\
			Average Earth–Moon distance: r=3.844×10^8m\
			and calculating\
			$F_Moon\approx 2.4×10^{-3}N$\
			For me and the Sun:\
			$m_Sun=1.989×10^{30}kg$\
			Average Earth–Sun distance: r=1.496×10^{11}m\
			and calculating\
			$F_Sun\approx 0.414N

			#### Answer

			[Insert a concise answer or boxed result]