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A satellite is to be placed in equatorial geostationary orbit around the earth for communication

a) calculate height of such a satellite

b) find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator

Answers (1)

$M=6\times10^{24}\; kg$

$R=6400\; km=6.4\times10^{6}m$

$Time \; Period \; (T)=24h=24\times3600s=24\times36\times10^{2}s$

$G=6.67\times10^{-11}Nm^{2}kg^{-2}$

Orbital Radius = R+h

Orbital velocity is $v_{0}=\sqrt{\frac{GM}{R+h}}\Rightarrow v_{0}^{2}=\frac{GM}{R+h}$

$T= \frac{2\pi (R+h)}{v_{0}}$

$T^{2}= \frac{4\pi^{2} (R+h)^{2}}{v_{0}^{2}}= \frac{4\pi^{2} (R+h)^{3}}{GM}$

$R+h=\left [ \frac{GT^{2}M}{4\pi^{2}} \right ]^{\frac{1}{3}}$

$h=\left [ \frac{GT^{2}M}{4\pi^{2}} \right ]^{\frac{1}{3}}-R$

$h=\left [ \frac{6.67\times10^{-11}\times(24\times36)^{2}\times(10^{2})^{2}\times6\times10^{24}}{4\times3.14\times3.14} \right ]^{\frac{1}{3}}-6.4\times10^{6}$

$h=35940\; km$

Let a satellite S is at h m above the earth surface. Let the angle subtended by it at the centre of the earth be 2 $\theta$

$\cos\; \theta =\frac{R}{R+h}=\frac{1}{[1+\frac{h}{R}]}$

$h=3.59\times10^{7}m$ (height of geostationary satellite)

$\cos\; \theta =\frac{1}{[1+\frac{3.59\times10^{7}}{6.4\times10^{6}}]}=0.1515$

$\theta =81.28$

$2\; \theta =2\times 81.28$

Number of satellites required to cover $360^{0}=\frac{360}{2\times 81.28}=2.21$

Number of satellites required=3

Posted by

infoexpert23

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