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"Progress, therfore, is not and accident,¡K" ..."Herbert Spencer(1820-1903, British philosopher)"
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Author Topic: Space shuttle in a circular orbit with a small satellite held above  (Read 5765 times)
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Fu-Kwun Hwang
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on: March 02, 2011, 09:47:16 pm » posted from:Taipei,T\'ai-pei,Taiwan

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.

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