Question

1) A satellite is to be placed in equatorial geostationary orbit around earth for communication. Calculate height of such a satellite.
$M=6\times {10}^{24}kgR=6400kmT=24hoursG=6.67\times {10}^{-11}N{m}^{2}K{g}^{-2}$

2) A satellite is to be placed in equatorial geostationary orbit around earth for communication. 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.
$M=6\times {10}^{24}kgR=6400kmT=24hoursG=6.67\times {10}^{-11}N{m}^{2}K{g}^{-2}$

Answer 1

1) Step 1: Draw the diagram.


Step 2: Calculate the height of the satellite.
As we know,
Time period of satellite is given by,

$T=\frac{2\pi R}{v}$ ….(i)

According to the given diagram, velocity of satellite is given by

$v=\sqrt{\frac{G{M}_E}{R_E+h}}$

Therefore, by putting the values of v in equation (i)

Time period of satellite,

$T=\frac{2\pi (R_E+h)}{\sqrt{\frac{G{M}_E}{(R_E+h)}}}=\frac{2\pi (R_E+h{)}^{\frac{3}{2}}}{\sqrt{G{M}_E}}$

Squaring both sides, we get

$T^2=\frac{4{\pi }^2(R_E+h{)}^3}{G{M}_E}$

$(R_e+h{)}^3=\frac{G{M}_E{T}^2}{4{\pi }^2}$
$(R_E+h)={\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}$

$h={\left(\frac{G{M}_E{T}^2}{4{\pi }^2}\right)}^{\frac{1}{3}}-R_E$ .....(ii)

Step 3 : Putting the values in the equation of the height.

Here, $M_E=6\times {10}^{24}kg$

$R_E=6400km=6400\times {10}^{3}m$

$=6.4\times {10}^{6}m$

$T=24h=24\times 60\times 60s=86400s$

$G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

On substituting the given values, we get $h$

$={\left(\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times (86400{)}^2}{4\times (3.14{)}^2}\right)}^{\frac{1}{2}}$$-6.4\times {10}^6$

$h=4.21\times {10}^7-6.4\times {10}^6$
$h=3.59\times {10}^{7}m$

Hence final answer: $h=3.59\times {10}^{7}m$


2) Step 1: Draw the diagram.


Step 2: Calculate the number of the satellites.

Let a satellite $S$ is at a height h above the earth surface, as given in the diagram.

Let the angle subtended by it at the centre of the earth be $2\theta$

$\cos \theta =\frac{R}{R+h}=\frac{1}{[1+\frac{h}{R}]}$
$h=3.59\times {10}^{7}m$

(Height of geostationary satellite)

$\cos \theta =\frac{1}{[1+\frac{3.59\times {10}^7}{6.4\times {10}^6}]}=0.1515$

$\theta =co{s}^{-1}0.1515$


$\theta =81.28$

$2\theta =2\times 81.28$

Therefore,
Number of satellites required to cover $360°$
$\frac{360}{2\times 81.28}=2.21$

Approximately,
Number of satellites required = 3