Question

7 A satelite is launched to orbit the Earth at an altitude of 2.35 x 10 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 1024 kg and its radius 6.38 x 106 m. (a) What is the orbital period, in hours, of this GPS satellite? (b) With what speed, in m/s, does it orbit the Earth? m/s

Answer 1

For the GPS satellite, the gravitational force will be equal to centripetal force

$F_{g}=F_{c}$

$F_{g}=GM_{e}m_{s}/R^{2}$ and $F_{c}=m_{s}v^{2}/R$

where $R=R_{e}+Altitude=6.38*10^{6}m+2.35*10^{7}m=2.988*10^{7}m$

$F_{g}=F_{c}$

$GM_{e}m_{s}/R^{2}=m_{s}v^{2}/R$

$GM_{e}/R=v^{2}$

$v=\sqrt{GM_{e}/R}$

$v=\sqrt{6.674*10^{-11}*5.97*10^{24}kg/2.988*10^{7}m}$

(b)ANSWER: ${\color{Red} v=3651.657m/s}$

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Time Period

Speed =Distance/Time

Here Distance is circumference of orbit =$2\pi R$

$v=2\pi R/T$

$T=2\pi R/v$

$T=2\pi (2.988*10^{7}m)/3651.657m/s$

$T=51412.71s$

(a)ANSWER: ${\color{Red} T=14.28h}$

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