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Easy Java Simulations (2001- ) => dynamics => Topic started by: ahmedelshfie on March 03, 2011, 05:42:20 pm

 Title: Space shuttle in a circular orbit with a small satellite held above Post by: ahmedelshfie on March 03, 2011, 05:42:20 pm Applet design by prof Hwang, Modified layout by Ahmed.Original Applet Space shuttle in a circular orbit with a small satellite held above (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2129)a space shuttle is in a circular orbit at a height H above the Earth.A small satellite is held above the shuttle(i.e directly away from the Earth) by means of a rod of length h.i.e with a total height H+h above the Earth. The satellite is then released . The heights of the new orbit of the space shuttle and the satellite are R and r respectively.Compare R and r with H and H+h.Assume mass of shutter is $M_s$, mass of satellite is m_s, Mass of earth is $M$.The tension along the rode is $T$. The radius of earth is $R_e$.Let $r=R_e+H$, the angular velocity is $\omega$Before the satellite is released.For space shutter: $M_s r \omega^2+T = \frac{G M M_s}{r^2}$For satellite: $m_s (r+h) \omega^2 -T =\frac{G M m_s}{(r+h)^2}$From the above two equations:$\omega =\sqrt{\frac{G M(M_s/r^2+m_s/(r+h)^2)}{M_s r+m_s(r+h)}}$When the satellite is released, tension T become zero.For space shutter: the gravitation force $\frac{G M M_s}{r^2}> M_s r \omega^2$ , so space shutter move inward.For satellite: the gravitation force$\frac{G M m_s}{(r+h)^2}< m_s (r+h) \omega^2$, so satellite move outward.Both of them become an ellipse orbit.