[tex]Dane:\\Obw = 40 \ 000 \ km\\G = 6,67\cdot10^{-11}\fracPNm^{2}}{kg^{2}}\\g = 9,81\frac{m}{s^{2}}\\Szukane:\\M = ?[/tex]
Rozwiązanie
Promień Ziemi:
[tex]Obw = 40 \ 000 \ km\\\\2\pi R = 40 \ 000 \ km \ /:2\pi\\\\R = \frac{40000 \ km}{2\pi} = \frac{40000 \ km}{2\cdot3,14} = 6 \ 370 \ km = 6 \ 370 \ 000 \ m[/tex]
1. Prawo powszecnego ciążenia:
[tex]F = G\frac{Mm}{R^{2}}[/tex]
2. II zasada dynamiki:
[tex]F = mg[/tex]
[tex]F = \frac{GMm}{R^{2}} \ \ / \cdot\frac{R^{2}}{Gm}\\\\M = \frac{FR^{2}}{Gm}\\\\F = mg\\\\M = \frac{mgR^{2}}{Gm}\\\\M = \frac{gR^{2}}{G}[/tex]
[tex]M = \frac{9,81\cdot(6370000)^{2}}{6,67\cdot10^{-11}} = 5,97\cdot10^{24} \ kg\\\\\underline{M \approx6\cdot10^{24 \ kg}}[/tex]
Wyprowadzenie jednostki:
[tex][M] =[ {\frac{gR^{2}}{G}]= \frac{\frac{m}{s^{2}}\cdot m^{2}}{N\cdot\frac{m^{2}}{kg^{2}}} = \frac{\frac{m^{3}}{s^{2}}}{\frac{kg\cdot m}{s^{2}}\cdot\frac{m^{2}}{kg^{2}}}=\frac{\frac{m^{3}}{s^{2}}}{\frac{m^{3}}{s^{2}\cdot kg}} = kg[/tex]
