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