Question #cb61c

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Question #cb61c

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Answer 1 · 2016-11-22T00:01:01

Start with $r = \sqrt[3]{G M \frac{T^{2}}{4 π^{2}}}$ $r = root(3)(GMT^2/(4pi^2))$

Explanation:

Reference Orbital period

$r = \sqrt[3]{G M \frac{T^{2}}{4 π^{2}}}$ $r = root(3)(GMT^2/(4pi^2))$

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

The distance, d, above the Earth is $r - R_{e a r t h}$ $r - R_(earth)$

$d = \sqrt[3]{G M \frac{T^{2}}{4 π^{2}}} - R_{e a r t h}$ $d = root(3)(GMT^2/(4pi^2)) - R_(earth)$

where $R_{e a r t h}$ $R_(earth)$ is the radius of the Earth.

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

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

Earth's Mass
$M = 5.972 \times 10^{24} k g$ $M = 5.972xx10^24kg$

Earth's Radius
$R_{e a r t h} = 6371000 m$ $R_(earth) = 6371000 m$

$d = \sqrt[3]{(6.67 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}) (5.972 \times 10^{24} k g) \frac{{(21600 s)}^{2}}{4 π^{2}}} - 6371000 m$ $d = root(3)((6.67 xx 10^-11 m^3kg^-1s^-2)(5.972xx10^24kg)(21600 s)^2/(4pi^2)) - 6371000 m$

$d = 10388631 m$ $d = 10388631 m$

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Question #cb61c

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