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ivann1987 [24]
3 years ago
13

Please help, please show work if possible.

Physics
1 answer:
aalyn [17]3 years ago
7 0
Hey So your equation would be ma=load/effort so we need one more part to this
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A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the
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Answer:Orbital period =21.22hrs

Explanation:

given that

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radius of a satellite's orbit, R=  earth's radius + height of the satellite

6.38X 10^6 +  3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= \sqrt GM/R

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

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\sqrt GM/R = \frac{2\pi r}{T}

V= \frac{2\pi r}{T}

T= 2 \pi r/ V

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=76,385.1 s

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76,385.1s =hr

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3 years ago
Please help on this one?
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