Question

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.60 x 10^4 kg/m^3 at the center and 2100 kg/m^3 at the surface.

What is the acceleration due to gravity at the surface of this planet?

Answer 1

a = 9.94 m/s²

Step-by-step explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

`\rho(r) = ar^2 - br^3`

r is the radius of the spherical shell

dr is the thickness

volume of shell

`dV = 4 \pi r^2 dr`

mass of shell

`dM = \rho(r)dV`

`\rho = \rho_0 - br`

now,

`dM = (\rho_0 - br)(4 \pi r^2)dr`

integrating both side

`M = \int_0^(R) (\rho_0 - br)(4 \pi r^2)dr`

`M = (4\pi)/(3)R^3\rho_0 - \pi R^4((\rho_0-\rho)/(R))`

`M = \pi R^3((\rho_0)/(3)+\rho)`

we know,

`a = (GM)/(R^2)`

`a = (G( \pi R^3((\rho_0)/(3)+\rho)))/(R^2)`

`a =\pi RG((\rho_0)/(3)+\rho)`

`a =\pi (6.674* 10^(-11)* 6.38 * 10^6)((1.60* 10^4)/(3)+2.1* 10^3)`

a = 9.94 m/s²