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

density = 5520 kg/m^3

Step-by-step 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 = (GM)/(r^2)`

mass of earth

`M = (gr^2)/(G)`

`M = (9.8 * (6378 * 10^3)^2)/(6.67 * 10^(-11))`

M = 5.972 x 10²⁴ kg

density =
`(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 =
`(5.972* 10^(24))/(1.08* 10^(21))`

density = 5.52 x 10³ kg/m^3

density = 5520 kg/m^3