Mechanics11/Seite15/Loesung5.13

Lösung n°1

a)

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

m₁ = m_E (Erde)

m₂ = m_M (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 = m_E (Erde)

m₂ = m_S (Sonne)

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

$(8 L i c h t m i n u t e n = 1,44 * 10 11)$

$F g = 3,86 * 10 22 N$

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