2019-10-31T07:17:10-04:00
It takes a certain satellite exactly 1.9 hours to orbit the Earth. Find:
a) the period of the orbit in seconds
b) the orbital radius
c) the centripetal acceleration
d) the angular velocity
e) the orbital velocity
f) the altitude of the satellite above Earth’s surface
1
2019-11-01T11:24:28-0400
a.
T = 1.9 ( 3600 ) = 6840 s T=1.9(3600)=6840\ s T = 1.9 ( 3600 ) = 6840 s
b.
r 3 = G M 4 π 2 T 2 r^3=\frac{GM}{4\pi^2}T^2 r 3 = 4 π 2 GM T 2
r 3 = ( 6.76 ⋅ 1 0 − 11 ) ( 5.97 ⋅ 1 0 24 ) 4 π 2 684 0 2 r^3=\frac{(6.76\cdot 10^{-11})(5.97\cdot 10^{24})}{4\pi^2}6840^2 r 3 = 4 π 2 ( 6.76 ⋅ 1 0 − 11 ) ( 5.97 ⋅ 1 0 24 ) 684 0 2
r = 7.82 ⋅ 1 0 6 m r=7.82\cdot 10^{6}m r = 7.82 ⋅ 1 0 6 m
c.
g = G M r 2 g=\frac{GM}{r^2} g = r 2 GM
g = ( 6.76 ⋅ 1 0 − 11 ) ( 5.97 ⋅ 1 0 24 ) ( 7.82 ⋅ 1 0 6 ) 2 = 6.60 m s 2 g=\frac{(6.76\cdot 10^{-11})(5.97\cdot 10^{24})}{(7.82\cdot 10^{6})^2}=6.60\frac{m}{s^2} g = ( 7.82 ⋅ 1 0 6 ) 2 ( 6.76 ⋅ 1 0 − 11 ) ( 5.97 ⋅ 1 0 24 ) = 6.60 s 2 m
d.
ω = g r = 6.6 7.82 ⋅ 1 0 6 = 0.000919 r a d s \omega=\sqrt{\frac{g}{r}}=\sqrt{\frac{6.6}{7.82\cdot 10^{6}}}=0.000919\frac{rad}{s} ω = r g = 7.82 ⋅ 1 0 6 6.6 = 0.000919 s r a d
e.
v = ω r = ( 0.000919 ) ( 7.82 ⋅ 1 0 6 ) = 7190 m s v=\omega r=(0.000919)(7.82\cdot 10^{6})=7190\frac{m}{s} v = ω r = ( 0.000919 ) ( 7.82 ⋅ 1 0 6 ) = 7190 s m
f.
h = r − R = 7.82 ⋅ 1 0 6 − 6.37 ⋅ 1 0 6 = 1.45 ⋅ 1 0 6 m h=r-R=7.82\cdot 10^{6}-6.37\cdot 10^{6}=1.45\cdot 10^{6}m h = r − R = 7.82 ⋅ 1 0 6 − 6.37 ⋅ 1 0 6 = 1.45 ⋅ 1 0 6 m
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