Gps global positioning system satellites orbit at an altitude of

The Navstar Global Positioning System (GPS) utilizes a group of 24 satellites orbiting the Earth. Using “triangulation” and signals transmitted by these satellites, the position of a receiver on the Earth can be determined to within an accuracy of a few centimeters. The satellite orbits are distributed evenly around the Earth, with four satellites in each of six orbits, allowing continuous navigational “fixes.” The satellites orbit at an altitude of approximately 11,000 nautical miles [1 nautical mile = 1.852 km = 6076 ft].

(a) Determine the speed of each satellite.
m/s___________
(b) Determine the period of each satellite.
s ____________

Answer:

a)3.86 ×10³m/s

b)4.83× 10⁷ s

Explanation:

The speed that a satellite at a given radius must travel so as to orbit in the presence of gravity is given by;

$v=\sqrt{\frac{Gm_2}{r} }$

where;

G is the universal gravitational constant given as 6.67 × 10⁻¹¹ N.m²/kg²

m is mass of Earth given as 5.972×10²⁴kg

r is the radius at which the satellites orbit

v is the speed of satellite

r is obtained by the sum of distance from center of Earth and height of satellite from earth

Height of satellite from earth= 11000 nautical miles

1 nautical mile=1.852 km

11000 nautical miles =?

11000*1.852 = 20372 km

1 km = 1000 m

20372 km=?

=20372000m

=2.04×10⁷ m

r= 6.38×10⁶ m +2.04×10⁷ m

r=2.68×10⁷ m

Substitute values in equation;

$v^2=\frac{(6.67*10^{-11})*(5.972*10^{24}) }{2.68*10^{7} } \\\\\\v^2=1.49*10^{7}\\\\\\v=\sqrt{1.49*10^7} \\\\\\v=3.86*10^3 m/s$

b)The formula for period is

$T=2\pi *\sqrt{\frac{r^3}{Gm} }$

where T is the period of the satellite

Substituting values

$T=2*3.14*\sqrt{\frac{(2.68*10^7)^3}{(6.67*10^-11)*(5.972*10^24)} } \\\\\\T=4.83*10^7 s$