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Question

A satellite is in an elliptic orbit around the Earth (Fig. 8-51). Its speed at the perigee A is 8650 m/s8650\mathrm{~m/s}. (a)(a) Use conservation of energy to determine its speed at B.

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Answered 2 years ago
Answered 2 years ago
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rB=(13900  km)2+(8230  km)2=16150  km\mathrm{r_B=\sqrt{(13900\;km)^2+(8230\;km)^2}=16150\;km}

Applying energy conservation at points A and B:

EA=EB\mathrm{E_A=E_B}

12mvA2+(GMEmrA)=12mvB2+(GMEmrB)\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)}

vB=vA2+2GME(1rB1rA)\mathrm{v_B=\sqrt{v_A^2+2GM_E\left(\dfrac{1}{r_B}-\dfrac{1}{r_A}\right)}}

=(8650  m/s)2+2(6.67×1011  Nm2/kg2)(5.98×1024  kg)(11.615×107  m18.23×106  m)\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)}}

=5220  m/s\mathrm{=5220\;m/s}

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