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# A satellite is in an elliptic orbit around the Earth (Fig. 8-51). Its speed at the perigee A is $8650\mathrm{~m/s}$. $(a)$ Use conservation of energy to determine its speed at B.

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$\mathrm{r_B=\sqrt{(13900\;km)^2+(8230\;km)^2}=16150\;km}$

Applying energy conservation at points A and B:

$\mathrm{E_A=E_B}$

$\mathrm{\dfrac{1}{2}mv_A^2+\left(-\dfrac{GM_Em}{r_A}\right)=\dfrac{1}{2}mv_B^2+\left(-\dfrac{GM_Em}{r_B}\right)}$

$\mathrm{v_B=\sqrt{v_A^2+2GM_E\left(\dfrac{1}{r_B}-\dfrac{1}{r_A}\right)}}$

$\mathrm{=\sqrt{(8650\;m/s)^2+2(6.67\times10^{-11}\;Nm^2/kg^2)(5.98\times10^{24}\;kg)\left(\dfrac{1}{1.615\times10^7\;m}-\dfrac{1}{8.23\times10^6\;m}\right)}}$

$\mathrm{=5220\;m/s}$

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