Respuesta :
Answer:
[tex]g=13.42\frac{m}{s^2}[/tex]
Explanation:
1) Notation and info given
[tex]\rho_{center}=13000 \frac{kg}{m^3}[/tex] represent the density at the center of the planet
[tex]\rho_{surface}=2100 \frac{kg}{m^3}[/tex] represent the densisty at the surface of the planet
r represent the radius
[tex]r_{earth}=6.371x10^{6}m[/tex] represent the radius of the Earth
2) Solution to the problem
So we can use a model to describe the density as function of the radius
[tex]r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}[/tex]
[tex]r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}[/tex]
So we can create a linear model in the for y=b+mx, where the intercept b=[tex]\rho_{center}=13000 \frac{kg}{m^3}[/tex] and the slope would be given by [tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}[/tex]
So then our linear model would be
[tex]\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r [/tex]
Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be [tex]dV=r^2 sin\theta d\phi d\theta dr[/tex].
And the total mass would be given by the following integral
[tex]M=\int \rho (r) dV[/tex]
Replacing dV we have the following result:
[tex]M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)[/tex]
We can solve the integrals one by one and the final result would be the following
[tex]M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})[/tex]
Simplyfind this last expression we have:
[tex]M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})[/tex]
[tex]M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})[/tex]
[tex]M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}][/tex]
And replacing the values we got:
[tex]M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg[/tex]
And now that for any shape the gravitational acceleration is given by:
[tex]g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}[/tex]
