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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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