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Illusion [34]
3 years ago
8

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?
Physics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Answer:

a = 9.94 m/s²

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 = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})

M = \pi R^3(\dfrac{\rho_0}{3}+\rho)

we know,

a = \dfrac{GM}{R^2}

a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}

a =\pi RG(\dfrac{\rho_0}{3}+\rho)

a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)

a = 9.94 m/s²

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