Newtonian mechanics
Newtonian mechanics
Newtonian mechanics
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Problem 1.2<br />
8 OPGAVELØSNINGER 36<br />
Metode: Centripetalkraft leveres af snorekraften<br />
Løsning:<br />
S2 = M1ω 2 r2, S1 = M2ω 2 r1 + S2 = 3M1ω 2 r1<br />
Snor knækker ved M1 fordi S1 > S2. ωmax tilfredsstiller S1 = Smax, dvs.<br />
3M1ω 2 <br />
Smax 327N<br />
maxr1 = Smax ⇔ ωmax = =<br />
= 33.02s−1<br />
3M1r1 3 · 0.1kg1m<br />
2. Selv ved uendelig høj ω kan systemet ikke hænge helt vandret pga. gravitation.<br />
1. Givet: MPhobos = 9.4 · 106m, TPhobos = 27540s, jævn cirkelbevægelse<br />
Ukendt: MMars<br />
Metode: Centripetalkraft leveres af gravitationen<br />
Løsning:<br />
MPhobosω 2 rPhobos = G MPhobosMMars<br />
r 2 Phobos<br />
og idet ω 2 = 4π2<br />
T 2 f˚as<br />
MMars = 4π2r3 Phobos<br />
T 2 =<br />
PhobosG ⇔ MMars = ω2 r 3 Phobos<br />
G<br />
4π 2 (9.4 · 10 6 ) 3<br />
(27540s) 2 · 6.668 · 10 −11 Nm 2 kg−2 = 6.48 · 1023 kg<br />
2. Givet: rMars = 1.52rJord, jævn cirkelbevægelse<br />
Ukendt: ˚Ar p˚a Mars (dvs. TMars)<br />
Metode: Kepler<br />
Løsning:<br />
r3 T 2 = kst ⇔ r3 Jord<br />
T 2 =<br />
Jord<br />
r3 Mars<br />
T 2 ⇔<br />
Mars<br />
<br />
rMars<br />
3 TMars =<br />
TJord = (1.52)<br />
rJord<br />
3/2 1˚ar = 1.87˚ar
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