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Marianna [84]
3 years ago
15

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 1.30×104 kg/m3 at the center and 2100 kg/m3 at the surface. Part A What is the acceleration due to gravity at the surface of this planet?
Physics
1 answer:
elena-s [515]3 years ago
4 0

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m 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

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

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 dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

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)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

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}

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