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

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

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

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

$g = 9.8m/s^2\\G = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m$

Replacing at this equation we have that

$M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg$

The Volume of a Sphere is equal to

$V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3$

Therefore using the relation between mass, volume and density we have that

$\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3$