Question

The space shuttle releases a satellite into a circular orbit 740 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?

Answer 1

Let us consider v speed of shuttle on orbit and v_0 launched speed. Note that the centripetal force is gravitational force, Thus
\[F=m\frac{v^2}{r}=G\frac{mM}{r^2}\]\[\Rightarrow v=\sqrt{GMr}=\sqrt{GM(R+h)}\]
where m and M is mass of the shuttle and Earth respectively. radius of Earth R and h=740km
Applying energy conversation's law we have
\[\frac{1}{2}m{v_0}^2+G\frac{mM}{R}=\frac{1}{2}mv^2+G\frac{mM}{R+h}\]
since, \[{v_0}^2=2GM(\frac{2}{R+h}-\frac{1}{R})=\frac{2GM(R-h)}{R(R+h)}\]

Answer 2

total energy of the space shuttle is conserved...

Answer 3

\[\frac{ 1 }{ 2 }mv ^{2}_{0}-\frac{ GmM }{ R }=\frac{ 1 }{ 2 }mv ^{2}-\frac{ GmM }{ R+h }\]