Question

Co-latitude of any position on earth is the angle the line from center of earth to the position forms with the axis of rotation of earth.

A geostationary satellite is at a height h above the surface of earth. If earth radius is R, choose the correct statement(s).

• A. the minimum colatitude on earth upto which the satellite can be used for communication is sin${}^{-1}$ (R/R + h)
• B. the maximum colatitudes on earth upto which the satellite can be used for communication is sin${}^{-1}$ (R/R + h)
• C. the area on earth escaped from this satellite is given as 2$\pi$R${}^2$ (1 + sin$\theta$)
• D. the area on earth escaped from this satellite is given as 2$\pi$R${}^2$ (1 + cos$\theta$)

Answer 1

Answer: (A), (C)

the minimum colatitude on earth upto which the satellite can be used for communication is sin${}^{-1}$ (R/R + h)
the area on earth escaped from this satellite is given as 2$\pi$R${}^2$ (1 + sin$\theta$)

Co-latitude of any position on earth is the angle the line from center of earth to the position forms with the axis of rotation of earth.

As shown in the figure, the minimum co-latitude that the satellite reaches is r.

Hence $\cos (\frac{\pi }{2}-r)=\frac{R}{R+h}$

Hence, $r=si{n}^{-1}\left(\frac{R}{R+h}\right)$

Infinitesimal area on sphere is given by

$dA=R^2\sin \theta d\theta d\varphi$($\theta$ is latitude of place)

Therefore the area of the reach of the satellite is

${\int }_0^{A}dA={\int }_0^{2\pi }{\int }_0^{\theta }{R}^2\sin \theta d\theta d\varphi$

Hence $A=2\pi R^2(1-\cos \theta )$

Hence area of earth not in the reach=$4\pi R^2-A=2\pi R^2(1+cos\theta )$

$=2\pi R^2(1+\sin r)$

But in the question, colatitude has been termed $\theta$