Question

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 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

Answer 1

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