Question

Suppose that the earth is a sphere of radius $6400$ kilometers. The height from the earth's surface from where exactly a fourth of the earth's surface is visible, is:

• A. $3200km$
• B. $3200\sqrt{2}km$
• C. $3200\sqrt{3}km$
• D. $6400km$

Answer 1

The correct option is D $6400km$

Let the height above the Earth be $h$ as shown in the figure.

Only the spherical cap bounded by $A$ and $A^{\prime }$ will be visible from that height. The total surface area of Earth is $4\pi R^2$.

If $\frac{1}{4}$ area of Earth is visible from $B$, then, area of spherical cap $A{A}^{\prime }$ is $\pi R^2$

We know that, area of spherical cap, that subtends an semi-apex angle $\theta$ is $2\pi R^2(1-\cos \theta )$

Here, $\theta$ is $\angle AOB$

From the Right triangle $△OAB$, $\cos \theta =\frac{OA}{OB}=\frac{R}{R+h}$

And, $2\pi R^2(1-\cos \theta )=\pi R^2\Rightarrow \cos \theta =\frac{1}{2}$

Therefore, $\frac{R}{R+h}=\frac{1}{2}$

$h=R$