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| ### Statement |
| ### Statement |
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| $2.6.9.$ [Insert the problem statement] |
| $2.6.9.$ Find the force of gravitational attraction acting on you from the Earth, Moon, and Sun. |
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| ### Solution |
| ### Solution |
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| By Newton's law gravitation:\ |
| By Newton's law gravitation:\ |
| $F=G\frac{m_1 m_2}{r^2}$\ |
| $F=G\frac{m_1 m_2}{r^2}$\ |
| 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 $G=6.674×10^{-11}\frac{Nm^2}{kg^2}$ |
| Using $G=6.674×10^{-11}\frac{Nm^2}{kg^2}$ |
| For me and Earth:\ |
| For me and Earth:\ |
| $m_1=70kg$\ |
| $m_1=70kg$\ |
| $m_earth=5.972×10^{24}kg$\ |
| $m_{Earth}=5.972×10^{24}kg$\ |
| $r_earth=6.371×10^6m (assuming that I'm at sea level)\ |
| $r_{Earth}=6.371×10^6m$ (that's Earth's radius, assuming that I'm at sea level)\ |
| and calculating\ |
| 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:\ |
| For me and the Moon:\ |
| $m_Moon=7.342×10^22kg$\ |
| $m_{Moon}=7.342×10^{22}kg$\ |
| Average Earth–Moon distance: r=3.844×10^8m\ |
| Average Earth–Moon distance: $r=3.844×10^8m$\ |
| and calculating\ |
| and calculating\ |
| $F_Moon\approx 2.4×10^{-3}N$\ |
| $F_{Moon}\approx 2.4×10^{-3}N$\ |
| For me and the Sun:\ |
| For me and the Sun:\ |
| $m_Sun=1.989×10^{30}kg$\ |
| $m_{Sun}=1.989×10^{30}kg$\ |
| Average Earth–Sun distance: r=1.496×10^{11}m\ |
| Average Earth–Sun distance: $r=1.496×10^{11}m$\ |
| and calculating\ |
| and calculating\ |
| $F_Sun\approx 0.414N |
| $F_{Sun}\approx 0.414N$ |
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| #### Answer |
| #### Answer |
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| $F_{Earth}\approx 686N$ |
| [Insert a concise answer or boxed result] |
| $F_{Moon}\approx 2.4×10^{-3}N$ |
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| $F_{Sun}\approx 0.414N$ |
