Gravitation of non-uniform density

In summary, we are given a gaseous planet with a radially dependent density function and asked to find the magnitude of force for a mass both inside and outside of the planet. Using the equations for force and mass, we can determine that the force inside the planet is proportional to the distance from the center, while the force outside is inversely proportional to the distance. This solution is similar to a Gauss' law problem for a ball of non-uniform charge. Overall, the solutions seem reasonable.
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Homework Statement


A gaseous planet, with radius R, has a radially dependent density function
[tex]\rho (r') = \rho_0 [\frac{r'}{R}]^2[/tex]
where r' is the distance from the center planet. Find the magnitude of force for a mass m inside and outside of the planet.


Homework Equations


[tex] F = -\frac{GmM}{r^2} [/tex]
[tex] F = -G m \int_V \frac{\rho(r') \hat{e_r}}{r^2}dv'[/tex]


The Attempt at a Solution


I'm pretty sure I did it right, but would like some confirmation. The mass inside the planet, taking a sample shell, would be

[tex] M = \int_0^r 4\pi r'^2 dr' \rho(r')[/tex]

which would give

[tex] M = \int_0^r 4\pi r'^2 dr' \rho_0 (\frac{r'}{R})^2[/tex]

and evaluates to

[tex] M = \frac{4}{5} \pi r^5 \frac{\rho_0}{R^2}[/tex]

such that the force inside will be

[tex] F = \frac{-Gm r^3}{r^2}\rho_0 \hat{e_r}[/tex]

Outside

the same idea applies

[tex] M = \int_0^R 4 \pi r'^2 dr' \rho(r')[/tex]

and gives

[tex] M = \frac{4}{5} \pi R^5 \frac{\rho_0}{R^2}[/tex]

so that the force is

[tex] F = \frac{-GmR^2}{r^2}\rho_0 \hat{e_r}[/tex]

Both answers, more or less, make sense. Inside, as you move further away that the force will increase. Outside, the force will decrease as you move further away.
 
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  • #2
It's the same problem as a Gauss' law problem for a ball of non-uniform charge, and that's how the solution goes, so that all looks right.
 

FAQ: Gravitation of non-uniform density

What is the concept of "gravitation of non-uniform density"?

The concept of "gravitation of non-uniform density" refers to the gravitational pull exerted by an object with a non-uniform distribution of mass. This means that the density of the object is not the same throughout, resulting in varying levels of gravitational force.

How does the density of an object affect its gravitational pull?

The density of an object directly affects its gravitational pull. Objects with a higher density will have a stronger gravitational pull compared to objects with a lower density. This is because the higher the density, the more mass is concentrated in a smaller space, resulting in a stronger gravitational force.

Can the gravitational force of an object with non-uniform density be calculated?

Yes, the gravitational force of an object with non-uniform density can be calculated using the same formula as for objects with uniform density. However, the calculation becomes more complex due to the varying density throughout the object. It requires integration and other mathematical techniques to accurately calculate the gravitational force.

What are some real-life examples of objects with non-uniform density?

Some real-life examples of objects with non-uniform density include the Earth's atmosphere, oceans, and other planets in our solar system. Each of these objects has a varying density due to differences in composition and structure, resulting in different levels of gravitational force.

How does the concept of "gravitation of non-uniform density" relate to the theory of general relativity?

The theory of general relativity, proposed by Albert Einstein, explains gravity as the curvature of space-time caused by the presence of mass and energy. This means that objects with non-uniform density will have a different curvature of space-time, resulting in variations in gravitational force. Therefore, the concept of "gravitation of non-uniform density" is closely related to the theory of general relativity.

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