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For $x\in\mathbb{R}$, let $f(x)=|\log 2-\sin x|$ and $g(x)=f(f(x))$. Which of the following is correct?
Since $\log 2 \approx 0.693$ and $\sin 0=0$, we have $f(0)=\log 2 > 0$.
Near $x=0$: $f(x)=\log 2-\sin x$ (since $\log 2>\sin x$ locally). So $f$ is smooth near $x=0$.
$g(x) = f(f(x)) = |\log 2 - \sin(f(x))|$. At $x=0$: $f(0)=\log 2$, so $\sin(f(0))=\sin(\log 2)$. Since $\log 2 < 1 < \pi$, $\sin(\log 2)>0$.
$g(0) = \log 2 - \sin(\log 2) > 0$, so near $x=0$, $g(x)=\log 2 - \sin(f(x))$.
$g'(0)=-\cos(f(0))\cdot f'(0)=-\cos(\log 2)\cdot(-\cos 0)=\cos(\log 2)$
A satellite orbits at height $h$ from Earth's surface where $h \ll R$ ($R$ = Earth's radius). The minimum increase in speed for the satellite to escape Earth's gravitational field is:
For $h \ll R$, orbital radius $\approx R$:
Orbital speed: $v_o = \sqrt{gR}$
Escape speed from surface (approximately same radius): $v_e = \sqrt{2gR}$
Minimum increase: $\Delta v = v_e - v_o = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2}-1)$
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