Q. 36

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 × 1024 kg, R = 6400 km, T = 24h, G = 6.67 × 10 - 11 SI units]

Answer :

(a) For a geostationary satellite, the centripetal force must equal the gravitational force. Let the mass of earth be M and the satellite be m, then



Where r is the radius of the orbit of the satellite and ω is its angular velocity



where T is the time period of revolution, then



Now the height h of the satellite from the surface of the earth,



where R is the radius of earth.


For a geostationary satellite, the time period is 24 hours or 86400 seconds,




(b) A geostationary satellite is visible form the equator in the region shown in the figure having an angle 2θ,



Now from the figure we have





Now 2θ = 162.58, therefore to cover the earth with such satellites so that from any point on the equator a satellite can be seen,



Therefore, we need a minimum of 3 satellites.


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