Question #03b84

Question

966 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T09:38:22

$rho_(earth)=(3g)/(4G*pi*R)$

Just don't forget that $d_(earth)=(rho_(earth))/(rho_(water))$ and $rho_(water)=1000kg$ / $m^3$

Knowing that a body's density is calculated as:

$"body's volumic mass"/"water's volumic mass"$

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

In order to find earh's density, you need to calculate
$rho_(earth)=M_(earth)/V_(earth)$

Knowing that $g=(G*M_(earth))/((R_(earth))^2) rarr g/G=(M_(earth))/((R_(earth))^2)$

A sphere's volume is calculated as:
$V=4/3*pi*R^3=4/3*pi*R*(R^2)$

Therefore:

$rho_(earth)=g/(G*4/3*pi*R)=(3g)/(4G*pi*R)$