A satellite is to be placed in equatorial geostationary orbit around 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.
`[M = 6 xx 10^(24) kg, R = 6400 km, T = 24 h, G = 6.67 xx 10^(-11) SI units]`

Options

A. `3.57xx10^(5)m`
B. `3.57xx10^(6)m`
C. `3.57xx10^(7)m`
D. `3.57xx10^(8)m`

Answer

C

Explanation

( c) Time period of satellite,
`T=(2pi(R_(E)+h))/(sqrt((GM_(E))/((R_(E)+h))))=(2pi(R_(E)+h)^(3//2))/(sqrt(GM_(E)))`
Squaring both sides, we get
`T^(2)=(4pi^(2)(R_(E)+h)^(3))/(GM_(E))`
`(R_(E)+h)^(3)=(GM_(E)T^(2))/(4 pi^(2))`
`(R_(E)+h)=((GM_(E)T^(2))/(4 pi^(2)))^(1//3) or h=((GM_(E)T^(2))/(4 pi^(2)))^(1//3)-R_(E)`
Here, `M_(E)=6xx10^(24) kg`
`R_(E)=6400 km=6400xx10^(3) m=6.4xx10^(6) m`
`T=24 h =24xx60xx60 s=86400 s`
`G=6.67xx10^(-11) N m^(2) kg^(-2)`
On substituting the given values, we get
`h=((6.67xx10^(-11)xx6xx10^(24)xx(86400)^(2))/(4xx(3.14)^(2)))^(1//3)-6.4xx10^(6)=4.21xx10^(7)-6.4xx10^(6)=3.57xx10^(7) m`