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 centrimeters. The satellite orbits are distributed around the Earth, 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. (b) Determine the period of each satellite.

The speed of a satellite is proportional to the square root of the ratio of the mass of satellite and the distance of satellite from the centre of Earth.

The time period to revolve the satellite around the Earth depends upon the velocity of Satelite.

The distance of satellite from the earth surface () \({r_u} = 11,000\;{\rm{nautical miles}}\).

The mass of Earth is \({M_{{\rm{Earth}}}} = 5.97 \times {10^{24}}\;{\rm{kg}}\).

The radius of the Earth is \({R_{{\rm{Earth}}}} = 6.38 \times {10^3}\;{\rm{km}}\).

(a)

The distance between the centre of Earth and the GPS satellite can be calculated as,

\(\begin{aligned}r &= {R_{{\rm{Earth}}}} + \left( {{r_u}} \right)\\ &= \left( {6.38 \times {{10}^3}\;{\rm{km}}} \right) + \left( {11,000\;{\rm{nautical miles}}} \right)\left( {\frac{{1.852\;{\rm{km}}}}{{1\;{\rm{nautical miles}}}}} \right)\\ &= \left( {6.38 \times {{10}^3} + 20372} \right)\,\,{\rm{km}} \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right)\\ &= 2.68 \times {10^4}\;{\rm{m}}\end{aligned}\)

The speed of the satellite can be calculated as,

\(\begin{aligned}v &= \sqrt {\frac{{G{M_{{\rm{Earth}}}}}}{r}} \\ &= \sqrt {\frac{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}/{\rm{k}}{{\rm{g}}^2}} \right)\left( {5.97 \times {{10}^{24}}\;{\rm{kg}}} \right)}}{{2.68 \times {{10}^7}\;{\rm{m}}}}} \\ &= 3.86 \times {10^3}\;{\rm{m/s}}\\ \approx 3.9 \times {10^3}\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of satellite is \(3.9 \times {10^3}\;{\rm{m/s}}\).

(b)

The period of satellite can be calculated as,

\(\begin{aligned}v &= \frac{{2\pi r}}{T}\\T &= \frac{{2\pi r}}{v}\\ &= \frac{{2\pi \left( {2.68 \times {{10}^7}\;{\rm{m}}} \right)}}{{3.9 \times {{10}^3}\;{\rm{m/s}}}}\\ &= 4.4 \times {10^4}\;{\rm{s}}\left( {\frac{{1\;{\rm{hr}}}}{{3600\;{\rm{s}}}}} \right)\\ \approx 12\;{\rm{h}}\end{aligned}\)

Substitute the values in the above equation,

\(\begin{aligned}T &= \frac{{2\pi \left( {2.68 \times {{10}^7}\;{\rm{m}}} \right)}}{{3.9 \times {{10}^3}\;{\rm{m/s}}}}\\ &= 4.4 \times {10^4}\;{\rm{s}}\left( {\frac{{1\;{\rm{hr}}}}{{3600\;{\rm{s}}}}} \right)\\ \approx 12\;{\rm{h}}\end{aligned}\)

Thus, the period of satellite is \(12\;{\rm{h}}\).