# Question #f0549

Jan 13, 2018

$7427.18 m {s}^{- 1}$

#### Explanation:

The centripetal force is given by:

$F = \frac{{m}_{2} {v}^{2}}{r}$

and the force of gravity by:

$F = \frac{G {m}_{1} {m}_{2}}{r} ^ 2$

As the centripetal force is provided solely by the force of gravity in orbital mechanics then set these equal to each other:

$\frac{G {m}_{1} {m}_{2}}{r} ^ 2 = \frac{{m}_{2} {v}^{2}}{r}$

Rearrange for $v$:

$\to \frac{G {m}_{1}}{r} = {v}^{2} \to v = \sqrt{\frac{G {m}_{1}}{r}}$

The values are:

$G = 6.67 \times {10}^{- 11} N k {g}^{-} 2 {m}^{2}$ (The universal gravitational constant)
$r = \left(850 + 6371\right) k m = 7221 \times {10}^{3} m$ (Radius of the earth plus the height of the satellite)
${m}_{1} = 5.972 \times {10}^{24} k g$

Sub into formula for $v$.

$v = \sqrt{\frac{\left(6.67 \times {10}^{- 11}\right) \left(5.972 \times {10}^{24}\right)}{7221 \times {10}^{3}}} = 7427.18 m {s}^{- 1}$