Правка разделов «Statement», «Solution», «Answer»

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правка #19517 предыдущая #19516 ← раньше
@@ -1,30 +1,31 @@
### Statement
−$2.6.9.$ [Insert the problem statement]
+$2.6.9.$ Find the force of gravitational attraction acting on you from the Earth, Moon, and Sun.
### 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\
+where $m_1$ is my mass, $m_2$ 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:\
$m_1=70kg$\
−$m_earth=5.972×10^{24}kg$\
−$r_earth=6.371×10^6m (assuming that I'm at sea level)\
+$m_{Earth}=5.972×10^{24}kg$\
+$r_{Earth}=6.371×10^6m$ (that's Earth's radius, assuming that I'm at sea level)\
and calculating\
−$F_earth\approx 686N$ (this is equeal to my weight!, $mg=70kg×9.8\frac{m}{s^2}$)
+$F_{Earth}\approx 686N$ (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\
+$m_{Moon}=7.342×10^{22}kg$\
+Average Earth–Moon distance: $r=3.844×10^8m$\
and calculating\
−$F_Moon\approx 2.4×10^{-3}N$\
+$F_{Moon}\approx 2.4×10^{-3}N$\
For me and the Sun:\
−$m_Sun=1.989×10^{30}kg$\
−Average Earth–Sun distance: r=1.496×10^{11}m\
+$m_{Sun}=1.989×10^{30}kg$\
+Average Earth–Sun distance: $r=1.496×10^{11}m$\
and calculating\
−$F_Sun\approx 0.414N
+$F_{Sun}\approx 0.414N$
#### Answer
−
−[Insert a concise answer or boxed result]
+$F_{Earth}\approx 686N$
+$F_{Moon}\approx 2.4×10^{-3}N$
+$F_{Sun}\approx 0.414N$