Question

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of the earth, G=6.67×10−11m3/(kg⋅s2), and a value of g at the surface of 9.80m/s2.Express your answer to three significant figures.

Answer 1

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

$g = \dfrac{GM}{r^2}$

mass of earth

$M = \dfrac{gr^2}{G}$

$M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}$

M = 5.972 x 10²⁴ kg

density =$\dfrac{mass}{volume}$

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378  x 10³)³

V = 1.08 x 10²¹ m³

density = $\dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}$

density = 5.52 x 10³  kg/m^3

density = 5520 kg/m^3