# Question #cb61c

Nov 22, 2016

Start with $r = \sqrt[3]{G M {T}^{2} / \left(4 {\pi}^{2}\right)}$

#### Explanation:

Reference Orbital period

$r = \sqrt[3]{G M {T}^{2} / \left(4 {\pi}^{2}\right)}$

Where r is the radius of the orbit, measured from the center of the Earth.

The distance, d, above the Earth is $r - {R}_{e a r t h}$

$d = \sqrt[3]{G M {T}^{2} / \left(4 {\pi}^{2}\right)} - {R}_{e a r t h}$

where ${R}_{e a r t h}$ is the radius of the Earth.

Six hours converted to seconds:
$T = 21600 s$

Gravitational Constant
$G = 6.67 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$

Earth's Mass
$M = 5.972 \times {10}^{24} k g$

${R}_{e a r t h} = 6371000 m$
$d = \sqrt[3]{\left(6.67 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2\right) \left(5.972 \times {10}^{24} k g\right) {\left(21600 s\right)}^{2} / \left(4 {\pi}^{2}\right)} - 6371000 m$
$d = 10388631 m$