At what height from the ground can one see exactly 1/4 th of the Earth's surface ? Take earth to be sphere of radius r.
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@nbubis. The question comes from NASA. – Claude Leibovici Mar 02 '14 at 08:01
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Actually for me, this question came in an Indian Scholarship Examination named KVPY. – Mar 02 '14 at 08:19
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I didn't even know how to proceed with the question :( so haven't tried anything – Mar 02 '14 at 08:20
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@Adeetya. I was just joking ! – Claude Leibovici Mar 02 '14 at 08:54
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ {1 \over 4}\,\pars{4\pi r^{2}} = 2\pi r^{2}\int_{0}^{\alpha}\sin\pars{\theta}\,\dd\theta = 4\pi r^{2}\sin^{2}\pars{\alpha \over 2}\quad\imp\quad \sin^{2}\pars{\alpha \over 2} = {1 \over 4} $$
$$ \cos\pars{\alpha} = 1 - 2\sin^{2}\pars{\alpha \over 2} = \half ={r \over r + h}\quad\imp\quad \color{#00f}{\large h = r} $$
Felix Marin
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@Adeetya Yes. You are right. I wrote $\pi r^{2}$ as the sphere surface instead of the correct value $4\pi r^{2}$. I already rewrote the answer. Thanks a lot. – Felix Marin Mar 02 '14 at 10:44
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Let Earth's radius be $R$, and the height above it be $r$. Moreover, denote the radius of the spherical cap visible as $a$.
Form the right triangle with side $R$ and hypotenuse $R+r$, whose second side has length $\sqrt{(R+r)^2-R^2}$. By triangle similarity you have:
$$\frac{a}{\sqrt{(R+r)^2-R^2}}=\frac{R}{R+r}$$
Solve for $a$ and use the fact that the area of a spherical cap is:
$$A=\pi(a^2+(R-\sqrt{R^2-a^2})^2)=\pi (a^2+(R-\sqrt{R^2-a^2})^2)$$
And find for what $r$, $A= \pi R^2$ as requested.
Nathaniel Bubis
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