[tex]Dane:\\m_1 = m_2 = m = 70 \ kg\\r = 3 \ m\\G = 6,67\cdot10^{-11}\frac{Nm^{2}}{kg^{2}}\\Szukane:\\F = ?[/tex]
Rozwiązanie
Z prawa powszechnej grawitacji:
[tex]F = G\cdot\frac{m_1\cdot m_2}{r^{2}}\\\\m_1 = m_2 = m\\\\F = G\cdot \frac{m^{2}}{r^{2}}\\\\Podstawiamy \ wartosci \ liczbowe\\\\F = 6,67\cdot10^{-11}\frac{Nm^{2}}{kg^{2}} \cdot\frac{(70 \ kg)^{2}}{(3 \ m)^{2}}=6,67\cdot10^{-11}\frac{Nm^{2}}{kg^{2}}\cdot\frac{4900 \ kg^{2}}{9 \ m^{2}}\\\\\boxed{F\approx3,63\cdot10^{-8} \ N}[/tex]
