Ncert Exercises & Solutions

Q. 36 A satellite is to be placed in an equatorial geostationary orbit around the earth for communication.

(a) Calculate the height of such a satellite.

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

[

M = 6 \times 10^{24} k g, R = 6400 k m, T = 24 h, G = 6.67 \times 10^{- 11}

SI unit]

Hint: A geostationary satellite is a satellite that revolves with the rotation of the earth.

Consider the adjacent diagram.

Given, the mass of the earth M =

6 \times 10^{24} kg

The radius of the earth, R = 6400 km =

64 \times 10^{6} m

Time period, T = 24 h = 24x60x60 = 86400 sec

G = 6.67 \times 10^{- 11} N - m^{2} / {kg}^{2}

Step 1: Find the height of the satellite.

(a) Time period,

T = 2 π \sqrt{\frac{{(R + h)}^{3}}{GM}} [∵ v_{o} = \sqrt{\frac{GM}{R + h}} and T = \frac{2 π (R + h)}{v_{0}}]

\Rightarrow T^{2} = 4 π^{2} \frac{{(R + h)}^{3}}{GM} \Rightarrow {(R + h)}^{3} = \frac{T^{2} GM}{4 π^{2}}

\Rightarrow R + h = {(\frac{T^{2} GM}{4 π^{2}})}^{1 / 3} \Rightarrow h = {(\frac{T^{2} GM}{4 π^{2}})}^{1 / 3} - R \Rightarrow h = {[\frac{{(24 \times 60 \times 60)}^{2} \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{4 \times {(3.14)}^{2}}]}^{1 / 3} - 6.4 \times 10^{6} = 4.23 \times 10^{7} - 6.4 \times 10^{6} = (42.3 - 64) \times 10^{6} = 35.9 \times 10^{6} m = 3.59 \times 10^{7} m

Step 2: Find the no. of satellites.

(b) If the satellite is at height h from the earth’s surface, then according to the diagram:

cosθ = \frac{R}{R + h} = \frac{1}{(1 + \frac{h}{R})} = \frac{1}{(1 + \frac{3.59 \times 10^{7}}{6.4 \times 10^{6}})} = \frac{1}{1 + 5.61} = \frac{1}{6.61} = 0.1513 = \cos 81 ° 18' θ = 81 ° 18' ∴ 2 θ = 2 \times (81 ° 18') = 162 ° 36'

If n is the number of satellites needed to cover the entire earth, then:

n = \frac{360 °}{2 θ} = \frac{360 °}{162 ° 36'} = 2.31

∴

Minimum 3 satellites are required to cover the entire earth.

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