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Two satellites are launched at a distance $R$ from a planet of negligible radius. Both satellites are launched in the tangential direction. The first satellite launches correctly at a speed $v_0$ and enters a circular orbit. The second satellite, however, is launched at a speed $\frac {1}{2}v_0$ . What is the minimum distance between the second satellite and the planet over the course of its orbit?
$\frac {1}{2}R$
$\frac {1}{3}R$
$\frac {1}{4}R$
$\frac {1}{7}R$
Solution
$v_{0}=\sqrt{\frac{G M}{R}}$
For satellite launches with velocity $\frac{V_{0}}{2}$
$\mathrm{T} \mathrm{E}=\mathrm{KE}+\mathrm{PE}$
$=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{m}\left(\frac{\mathrm{v}_{0}}{2}\right)^{2}$
$=-m v_{0}^{2}+\frac{1}{8} m v_{0}^{2}=-\frac{7}{8} m v_{0}^{2}$
By energy conservation
$(\mathrm{TE})_{\mathrm{i}}=(\mathrm{TE})_{\mathrm{f}}$
$-\frac{7}{8} \mathrm{mv}_{0}^{2}=-\frac{\mathrm{GMm}}{\mathrm{R}_{1}}+\frac{1}{2} \mathrm{mv}_{1}^{2}$
Since at lowest distance velocity is along tangential direction again.
$\therefore \mathrm{L}_{i}=\mathrm{L}_{\mathrm{f}}$
$\frac{\mathrm{mv}_{0} \mathrm{R}}{2}=\mathrm{mv}_{1} \mathrm{R}_{1}$
$\mathrm{v}_{1}=\frac{\mathrm{R}_{\mathrm{V}_{0}}}{2 \mathrm{R}_{1}}$
$-\frac{7}{8} v_{0}^{2}=-\frac{R v_{0}^{2}}{R_{1}}+\frac{1}{2}\left(\frac{R V_{0}}{2 R_{1}}\right)^{2}$
$-8 \mathrm{RR}_{1}+\mathrm{R}^{2}=-7 \mathrm{R}_{1}^{2}$
$\left(7 \mathrm{R}_{1}-\mathrm{R}\right)\left(\mathrm{R}_{1}-\mathrm{R}\right)=0$
$\mathrm{R}_{1}=\mathrm{R}_{1} \mathrm{R} / 7$
Lowest point is $\mathrm{R}_{1}=\mathrm{R} / 7$
