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

Question

3 years ago

15

mass of the Earth to be 5.97 1024 kg and its radius 6.38 106 m. (a) What is the orbital period, in hours, of this GPS satellite

1 answer:

Korolek [52] · Answer 1 · 2022-04-01T05:29:35+03:00

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

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

a) Orbital period

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

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

T= 2 $\pi$ r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs