Question #03b84

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Question #03b84

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Answer 1 · 2015-08-05T09:38:22

$ρ_{e a r t h} = \frac{3 g}{4 G \cdot π \cdot R}$ $rho_(earth)=(3g)/(4G*pi*R)$

Just don't forget that $d_{e a r t h} = \frac{ρ_{e a r t h}}{ρ_{w a t e r}}$ $d_(earth)=(rho_(earth))/(rho_(water))$ and $ρ_{w a t e r} = 1000 k g$ $rho_(water)=1000kg$ / $m^{3}$ $m^3$

Explanation:

Knowing that a body's density is calculated as:

$\frac{body's volumic mass}{water's volumic mass}$ $"body's volumic mass"/"water's volumic mass"$

Knowing that water's volumic mass expressed in $k g$ $kg$ / $m^{3}$ $m^3$ is $1000$ $1000$ .

In order to find earh's density, you need to calculate
$ρ_{e a r t h} = \frac{M_{e a r t h}}{V_{e a r t h}}$ $rho_(earth)=M_(earth)/V_(earth)$

Knowing that $g = \frac{G \cdot M_{e a r t h}}{{(R_{e a r t h})}_{e a r t h}^{2}} \to \frac{g}{G} = \frac{M_{e a r t h}}{{(R_{e a r t h})}_{e a r t h}^{2}}$ $g=(G*M_(earth))/((R_(earth))^2) rarr g/G=(M_(earth))/((R_(earth))^2)$

A sphere's volume is calculated as:
$V = \frac{4}{3} \cdot π \cdot R^{3} = \frac{4}{3} \cdot π \cdot R \cdot (R^{2})$ $V=4/3*pi*R^3=4/3*pi*R*(R^2)$

Therefore:

$ρ_{e a r t h} = \frac{g}{G \cdot \frac{4}{3} \cdot π \cdot R} = \frac{3 g}{4 G \cdot π \cdot R}$ $rho_(earth)=g/(G*4/3*pi*R)=(3g)/(4G*pi*R)$

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Question #03b84

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