- #1
compwiz3000
- 17
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If we launch a satellite to a circular orbit around the Earth at height 357.1 km, to find the velocity needed at launch, do we just set the energies equal?:
[tex]
- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}
[/tex]
and then solve for [tex]v_L[/tex]?
[tex]\mu = GM[/tex], where [tex]M[/tex] is the mass of the Earth.
Then, in space, the mechanical energy is
[tex]
\frac {v^2}{2} - \frac {\mu}{r}.
[/tex]
Using centripetal force, we have
[tex]
\frac {m v^2}{r} = \frac {\mu m}{r^2} \implies v^2 = \frac {\mu}{r},
[/tex]
so the mechanical energy is
[tex]
- \frac {\mu}{2r}.
[/tex]
Since the mechanical energy is conserved at the surface, we set it equal to that mechanical energy to find the launch velocity.
[tex]
- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}
[/tex]
and then solve for [tex]v_L[/tex].
Is my reasoning for the launch velocity correct?
[tex]
- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}
[/tex]
and then solve for [tex]v_L[/tex]?
[tex]\mu = GM[/tex], where [tex]M[/tex] is the mass of the Earth.
Then, in space, the mechanical energy is
[tex]
\frac {v^2}{2} - \frac {\mu}{r}.
[/tex]
Using centripetal force, we have
[tex]
\frac {m v^2}{r} = \frac {\mu m}{r^2} \implies v^2 = \frac {\mu}{r},
[/tex]
so the mechanical energy is
[tex]
- \frac {\mu}{2r}.
[/tex]
Since the mechanical energy is conserved at the surface, we set it equal to that mechanical energy to find the launch velocity.
[tex]
- \frac {\mu}{2\left(r_E + h\right)} = \frac {v_L^2}{2} - \frac {\mu}{r_E}
[/tex]
and then solve for [tex]v_L[/tex].
Is my reasoning for the launch velocity correct?
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