A 1.0 kg mass weighs 9.8 N on Earth's surface, and the radius of the Earth is roughly 6.4 x 10^6 m. Calculate the average density of Earth. (Show how you would calculate for density; don't just give the answer because L wan't to understand how you arrived at your answer).

Answer 1

F = 9.8 N = 1*Me * G /(6.4*10^6)^2
so
Me = 9.8 *41*10^12 / 6.67*10^-11
= 60.2 * 10^23 kg

Ve = (4/3) pi Re^3 = (4/3)(pi)(6.4*10^6)^3
= 1098*10^18

Rho = mass/volume = (6.02/1.098)10^(24-21)
= 6020/1.098 = 5483 kg/m^3

about 5.5 times density of water, lots of molten metal and heavy rock in there.

Answer 2

density ρ =M/V

mg=W=GmM/R² =>
M= WR²/Gm.

V=4πR³/3.

ρ =3 WR²/4πR³Gm =
=3W/4πRGm =
=3•9.8/4•3.14•6.4•10⁶•6.67•10⁻¹¹•1=
=5.48•10³ kg/m³