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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: $\Delta v = v_e - v_o = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2}-1)$
Let $S=\left\{t\in\mathbb{R}: f(x)=|x-\pi|\cdot(e^{|x|}-1)\sin|x|\right.$ is not differentiable at $t\}$. Find $S$.
Check $x=0$: $f(x)=|x-\pi|(e^{|x|}-1)\sin|x|$. Near $x=0$, both $(e^{|x|}-1)$ and $\sin|x|$ have factors of $|x|$, so $f(x)\approx |x-\pi|\cdot|x|^2$ near 0, which is differentiable at 0.
Check $x=\pi$: near $\pi$, $(e^{|x|}-1)\sin|x|$ is smooth and non-zero ($e^\pi\sin\pi=0$!). Since $\sin\pi=0$, the product $(e^{|x|}-1)\sin|x|$ vanishes at $\pi$, making $f(x)=0$ near $\pi$ — wait, $\sin\pi=0$, so $f(\pi)=0$. Checking differentiability more carefully shows $f$ is differentiable at $\pi$ too.
Therefore $S=\boldsymbol{\emptyset}$.
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