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 atleast one satellite is visible from any point on the equator.

`[M = 6 xx 10^(24) kg, R = 6400 km, T = 24 h, G = 667 x 10-^(11)` SI unit]

Explanation

`implies` Consider the adjacent diagram,
Given mass of the earth, `M = 6 xx 10^(24)` kg
Radius of the earth, `R = 6400 km = 6.4 xx 10^(6) m`
Time period, T = 24 h
`= 24 xx 60 xx 60 = 86400 s`
`G = 6.67 xx 10^(-11) Nm^2//kg^2`
`v_0 = (2pir)/T`
`:. T = (2pir)/v_0`
`:. T = (2pi(R+h))/(sqrt((GM)/((R+h))))`
(a) `T = 2pisqrt((R+h)^3/(GM))`
`:. T^2=4pi^2((R+h)^3)/(GM)`
`:. (R+h)^3=(T^2GM)/(4pi^2)`
`:. R+h=((T^2GM)/(4pi^2))^(1/3)`
`:. h =((T^2GM)/(4pi^2))-R`
`:. h =[((24xx 60 xx 60)^2 xx 6.67 xx 10^(-11) xx 6 xx 10^(24))/(4xx(3.14)^2)]`
`- 6.4 xx 10^6 `
` = 4.23 xx 10^7 - 6.4xx 10^6`
`=(42.3 -6.4 )xx10^6`
` = 35.9 xx10^6 m`
`=3.59 xx 10^7m`
(b)
`cos theta = R /(R+h) =1/((1+h/R))= 1/ ((1+(3.59 xx 10^7)/(6.4 xx 10^6)))`
`= 1/ (1+5.61 ) = 1/(6.61) = 0.1513 = cos 81^@18.`
`theta = 81^@ 18.`
`:.2theta = 2xx (81^@ 18.) =162^@ 36.`
If n is the number of satellites needed to cover the entire earth , then
`n = (360^@)/(2theta) = (360^@)/(162^@ 36.) =2.31`
`:.` Minimum 3 satellites are required to cover the entire earth.