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 Author Topic: Space shuttle in a circular orbit with a small satellite held above  (Read 4404 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
ahmedelshfie
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 « Embed this message on: March 03, 2011, 05:42:20 pm » posted from:Uberaba,Minas Gerais,Brazil

Applet design by prof Hwang, Modified layout by Ahmed.
Original Applet Space shuttle in a circular orbit with a small satellite held above

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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