A geostationary satellite is at a height h ab | vedclass

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(D) From the geometry of the problem,consider a right-angled triangle formed by the center of the Earth,the satellite,and the point on the horizon vis

Vedclass · Accepted Answer

$(A)$ and $(C)$ both.

Vedclass · Answer

(D) From the geometry of the problem,consider a right-angled triangle formed by the center of the Earth,the satellite,and the point on the horizon visible from the satellite. The hypotenuse is $(R+h)$ and the side opposite to the angle $	heta$ is $R$. Thus,$\sin 	heta = \frac{R}{R+h}$.The colatitude is the angle measured from the pole. The satellite can communicate up to a certain latitude,which corresponds to a minimum colatitude of $	heta = \sin^{-1} \left( \frac{R}{R+h} ight)$.The area of the spherical cap visible from the satellite is $A_{visible} = 2\pi R^2 (1 - \cos \alpha)$,where $\alpha$ is the angular radius of the visible region. In this geometry,$\cos \alpha = \frac{R}{R+h} = \sin 	heta$. Thus,the visible area is $2\pi R^2 (1 - \sin 	heta)$.The area on Earth that is $NOT$ visible (escaped) from the satellite is the total surface area minus the visible area:$A_{escaped} = 4\pi R^2 - 2\pi R^2 (1 - \sin 	heta) = 2\pi R^2 (1 + \sin 	heta).$Therefore,both statements $(A)$ and $(C)$ are correct.

$A$ geostationary satellite is at a height $h$ above the surface of the Earth. If the Earth's radius is $R$,which of the following statements is correct?

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