Mechanics11/Seite15/Loesung5.13

Lösung n°1
a)

$$F_g = G * \frac {m_1 * m_2} {r^2} $$

m1 = mE (Erde)

m2 = mM (Mond)

$$F_g = G * \frac {m_E * m_M} {r^2} $$

$$F_g = 6,67 * 10^{-11} * \frac {(6,0 * 10^{24}) * (0,0123 * 6,0 * 10^{24})} {(384.000.000 m)^2} $$

$$F_g = 2,00 * 10^{20} N $$

b)

$$F_g = G * \frac {m_1 * m_2} {r^2} $$

m1 = mE (Erde)

m2 = mS (Sonne)

$$F_g = G * \frac {m_E * m_S} {r^2} $$ $$F_g = 6,67 * 10^{-11} * \frac {(6,0 * 10^{24}) * (2,0 * 10^{30})} {(8 Lichtminuten)^2} $$

$$( 8 Lichtminuten = 1,44 * 10^{11})$$

$$F_g = 3,86 * 10^{22} N $$

Lösung von: Kenny und Marc am 26 June 2008, um 10:15