Aufgaben zum Gravitationsgesetz
Aufgaben zum Gravitationsgesetz
Aufgaben zum Gravitationsgesetz
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Physik * Jahrgangsstufe 10 * <strong>Aufgaben</strong> <strong>zum</strong> <strong>Gravitationsgesetz</strong> * Lösungen<br />
1. a) Für die Gewichtskraft einer Masse m auf der Erdoberfläche gilt<br />
m<br />
6 2<br />
2 9,8 ⋅(6,37 ⋅10 m)<br />
2<br />
* m ⋅MErde<br />
g ⋅R Erde<br />
24<br />
m ⋅ g = F s<br />
G<br />
= Fgrav = G ⋅ ⇒ M<br />
2 Erde<br />
= = = 6,0⋅10 kg<br />
*<br />
3<br />
R<br />
Erde<br />
G<br />
−11<br />
m<br />
6,67 ⋅10<br />
2<br />
kg ⋅s<br />
2 * mMond ⋅MErde<br />
b) FZentripetal = Fgrav ⇔ mMond ⋅ω ⋅ r = G ⋅ ⇔<br />
2<br />
r<br />
2 3 2 3<br />
2 6 3<br />
ω ⋅ r 4 π ⋅(60,3⋅<br />
R<br />
E<br />
) 4 π ⋅(60,3⋅6,370 ⋅10 m)<br />
24<br />
MErde = = = = 6,03⋅10 kg<br />
* 2 * 2 −11 3 −1 −2<br />
G T ⋅G (27,1⋅ 24⋅3600s) ⋅6,67 ⋅10 m kg s<br />
2.<br />
2 * mErde ⋅MSonne<br />
FZentripetal = Fgrav ⇔ mErde ⋅ω ⋅ d = G ⋅ ⇔<br />
2<br />
d<br />
2 3 2 3 2 11 3<br />
ω ⋅d 4π ⋅d 4 π ⋅(1,496 ⋅10 m)<br />
30<br />
MSonne = = = = 1,99 ⋅10 kg<br />
* 2 * 2 −11 3 −1 −2<br />
G T ⋅G (365, 26⋅ 24⋅3600s) ⋅6,67 ⋅10 m kg s<br />
3. Bestimme zunächst die Masse des Mars<br />
2 * mPhobos ⋅MMars<br />
FZentripetal = Fgrav ⇔ mPhobos ⋅ω ⋅ d = G ⋅ ⇔<br />
2<br />
d<br />
2 3 2 3 2 6 3<br />
ω ⋅d 4π ⋅d 4 π ⋅(9,3⋅10 m)<br />
23<br />
MMars = = = = 6, 23⋅10 kg<br />
* 2 * 2 −11 3 −1 −2<br />
G T ⋅G (0,32 ⋅24⋅3600s) ⋅6,67 ⋅10 m kg s<br />
⋅ ⋅ ⋅<br />
F = m ⋅ g = G ⋅ = 6,67 ⋅10 ⋅ = 36 N<br />
3 23<br />
* m MMars<br />
11 m 10kg 6, 23 10 kg<br />
G,grünesMännchen Mars 2 2 6 2<br />
R<br />
Mars<br />
kg ⋅s (6,76 ⋅10 m : 2)<br />
4. a)<br />
b)<br />
2<br />
mDeimos ⋅ v * mDeimos ⋅MMars<br />
= G ⋅ ⇔<br />
2<br />
r<br />
r<br />
* −11 3 −1 −2 23<br />
G ⋅ MMars<br />
6,67 ⋅10 m kg s ⋅6, 40⋅10 kg km<br />
v = = = 1,35 ⋅<br />
6<br />
r 23,5 ⋅10 m s<br />
3<br />
2π⋅ r 2π⋅r 2π⋅23,5 ⋅10 km<br />
v = ⇒ T = = = 30,4h<br />
1<br />
T v 1,35kms −<br />
5. Die Umlaufdauer des Satelliten muss genau 24,0 Stunden betragen.<br />
2 * mSat ⋅MErde 3 * MErde G ⋅ MErde<br />
⋅T<br />
FZentripetal = Fgrav ⇔ mSat ⋅ω ⋅ r = G ⋅ ⇔ r = G ⋅ =<br />
2 2 2<br />
r ω 4π<br />
* 2<br />
* 2 −11 3 −1 −2 24 2<br />
G ⋅M 3 Erde<br />
⋅T 6,67 ⋅10 m kg s ⋅5,977 ⋅10 kg ⋅(24⋅3600s)<br />
r = = 3<br />
= ⋅<br />
2 2<br />
4π<br />
4π<br />
6 6 6<br />
h = r − R<br />
Erde<br />
= 42,24 ⋅10 m − 6,37 ⋅ 10 m = 36⋅<br />
10 km<br />
6<br />
42, 24 10 m<br />
6. a)<br />
b)<br />
*<br />
* m ⋅MErde G ⋅ MErde<br />
1<br />
m ⋅ g(r) = G ⋅ ⇒ g(r) = ∼ also<br />
2 2 2<br />
r r r<br />
m R<br />
Erde<br />
m 6370km m<br />
g(500km über der Erde) = 9,8 ⋅ = 9,8 ⋅ = 9,1<br />
2 2 2<br />
s r s (6370 + 500)km s<br />
1 g(R + h) R 5,0 m / s R<br />
g(r) ∼ also = also<br />
= ⇒<br />
r g(R ) (R h) 9,8 m / s (R h)<br />
2 2<br />
2<br />
Erde Erde Erde<br />
2 2 2 2<br />
Erde Erde<br />
+<br />
Erde<br />
+<br />
9,8 ⎛ 9,8<br />
5,0 ⎜<br />
⎝ 5,0<br />
3<br />
R<br />
Erde<br />
+ h = R<br />
Erde<br />
⋅ ⇒ h = R<br />
Erde<br />
⋅⎜<br />
− 1⎟<br />
= 2,55⋅10 km<br />
⎞<br />
⎟<br />
⎠
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