Mechanics11/Seite15/Loesung5.10

a)

$\frac{a_1^3}{T_1^2} =\frac{a_2^3}{T_2^2}$

$T 1 = 1 a a 2 = 3,0 A E a 1 = 1 A E T 2 = ?$

$\frac{T_1^2}{a_1^3}*a_2^3= T_2^2$

$(\frac{1a^2}{1AE^3}*3,0AE^2)^\frac{1}{2} = T_2$

$T 2 = 5,19 a$

b)

$a = 10,5 a$

$c= \frac{T^2}{a^3}$

$\frac{T^2}{c}=a^3$

einsetzen a= 4,87AE

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--Max 08:24, 10 July 2008 (UTC)

zweite Schülerlösung

a)

1 AE = astronomische Einheit = a 1 AE = 150 Millionen km = 1,5*10^11 m

$(\frac{T}{T_E})^2 = (\frac{a}{a_E})^3$

$T^2 = T_E^2 (\frac{a}{a_E})^3$

$T = T_E (\frac{a}{a_E})^3/2$

$T = 1a (\frac{3,0 AE}{1 AE})^3/2$

$T = 5,2 a\$

b)

$a = 4,8 AE\$

--FatF 08:15, 10 July 2008 (UTC)