Question #280700

A satellite was launched in 1983. It followed an elliptical orbit that varied from about 2.00 x 10^3 km to 5.00 x 10^3 km above earth. What was its orbital speed at its closest approach to earth?


1
Expert's answer
2021-12-17T11:50:02-0500

Explanations & Calculations


  • If the satellite was launched at the highest margin: 5000km\small 5000\,km in 1983, then it was meant to stay in that orbit (ideally) moving at the corresponding speed of that orbit.
  • That was,

GMEmr12=mv12r1v1=GMEr1\qquad\qquad \begin{aligned} \small G\frac{M_Em}{r_1^2}&=\small m\frac{v_1^2}{r_1}\\ \small v_1&=\small \sqrt{\frac{GM_E}{r_1}} \end{aligned}


  • When it approaches the lowest orbit, its speed is meant to increase according to the law of conservation of angular momentum (I=mvr\small I=mvr ).
  • Then, it can be calculated as follows.

mv1r1=mv2r2v2=v1.r1r2=GMEr1.(r1r2)\qquad\qquad \begin{aligned} \small mv_1r_1&=\small mv_2r_2\\ \small v_2&=\small v_1.\frac{r_1}{r_2}\\ &=\small \sqrt{\frac{GM_E}{r_1}}.\Big(\frac{r_1}{r_2}\Big) \end{aligned}

  • Here G and ME\small M_E are the Universal gravitational constant and the mass of the earth respectively.

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