Nonuniform density of a planet

• WraithM
In summary: Good luck!In summary, we discussed a non-uniform density planet with an object falling through it. Using the standard trick of writing dr/dt = v, we were able to simplify the differential equation to \frac{d^2 r}{dt^2} = v \frac{dv}{dr}. However, when trying to solve for r as a function of t, we encountered a difficult integral that may not have a closed form solution. It is possible to solve it numerically or with the use of elliptical functions.

WraithM

I was thinking about a situation like this: You have a planet that has a nonuniform density (mass) like this $$\rho = a - r$$ where a is just some number and r is the radius away from the center of the planet. Then you have an object falling through the planet. I had no idea how to solve this differential equation. I don't even know if there is a solution. You'd have an F=ma like this: $$-\frac{G(a - r)(\frac{4}{3}\pi r^3)m}{r^2} = ma$$. That becomes through simplification: $$\frac{d^2 r}{dt^2}=\frac{4}{3}\pi a G(r^2 - r)$$. I don't really know much beyond solving simple differential equations. Is this solvable? If there is a solution, how do you get it?

WraithM said:
$$\frac{d^2 r}{dt^2}=\frac{4}{3}\pi a G(r^2 - r)$$. I don't really know much beyond solving simple differential equations. Is this solvable? If there is a solution, how do you get it?

Hi WraithM!

Use the standard trick of writing dr/dt = v, and then

d2r/dt2

= dv/dt

= dv/dr dr/dt [chain rule]

= v dv/dr

Does that actually work? Do you just separate variables from there? I see the logic of making $$\frac{d^2 r}{dt^2} = v \frac{dv}{dr}$$, but I don't want to believe that it's that easy.

For example, take $$\frac{d^2 r}{dt^2} = g$$ (constant acceleration). Separating variables using v dv/dr: v dv = g dr, that's: v^2/2 = gr
$$\frac{dr}{dt} = \sqrt{2gr}$$
$$\frac{dr}{\sqrt{2gr}} = dt$$
$$1/g \sqrt{2gr}=t$$
$$r = \frac{gt^2}{2}$$
Wow, so that actually works?

:/ I'm feeling stupid for not knowing this.

EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?

Last edited:
WraithM said:
Wow, so that actually works?

Woohoo!

(btw, that's how 1/2 mv2 gets into energy equations )
EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?

hmm … let's see …

you didn't get vdv = 4/3 πaG (r2 - ar) dr ?

Well, yes. I did get: $$v \, dv = \frac{4}{3}Ga(r^2 - r)dr$$
You integrate from ($$r_{0}$$ to $$r$$) that to get $$v = \sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}$$. That's all good, but then if you try to go another step further, it blows up. Now I'm solving for r(t). You get: $$\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt$$ I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?

WraithM said:
… You get: $$\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt$$ I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?

ah, I didn't realize you wanted to go as far as solving for r as a function of t

Yes, that looks correct … unfortunately, not all integrals are "doable"

If the density is not uniform, the force acting on the mass when is at distance r from the center will be
G*m*M(r)/r^2
where M(r) is the mass contained in the sphere of radius r.
This is not 4/3pi*r^3*density . Density is not constant. You need to integrate from 0 to r in order to find the mass.
M(r) will be 4*pi(a/3*r^3-r^4/4) - you should do it again do double check.
But before you do this, check again your density. Of course, if you are interested in a math problem only, it does not matter.
But for a physics problem...
density=a-r ?
So density is zero at r=a and then it becomes negative for r<a.
Not mentioning the units. You will have density equal to distance. (a and r are disatnces, I suppose) Not very physical...

:( You're absolutely right... I completely forgot about that... I was more interested in the math at this point. It's sort of analogous to non-uniform charge distributions and using Gauss's law. You need to integrate, but thank you! I was being stupid :/

You still end up with an integral in the same sort of form as above.
$$\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{2}{3} \pi G(r^3 - 2ar^2-r_{0}^3+2ar_{0}^2)}}=\int_{0}^{t}dt$$

If you expect a nice formula as a result then you'll be disappointed. Few problems will allow for such a thing. I suppose this is not one of them.
The integral that you show in the last post can be solved numerically or maybe can be expressed in term of elliptical functions, I am not sure. Try Mathematica or Maple.

1. What is nonuniform density of a planet?

The nonuniform density of a planet refers to the variation in the distribution of mass throughout the planet. This means that different parts of the planet have different densities, which can affect its overall gravitational pull and structure.

2. What causes a planet to have nonuniform density?

There are a few factors that can contribute to nonuniform density in a planet. These include the planet's formation process, geological activity, and the presence of different materials such as dense iron in the core and lighter rock in the crust.

3. How is nonuniform density measured on a planet?

Scientists can use various methods to measure the density of a planet, including gravitational measurements, seismic waves, and remote sensing techniques. These methods allow scientists to map out the variation in density throughout the planet.

4. What are the implications of nonuniform density for a planet?

The nonuniform density of a planet can have significant implications for its overall structure and composition. It can affect the planet's rotation, magnetic field, and even the possibility of supporting life. Understanding a planet's density is essential for studying its evolution and potential habitability.

5. Can nonuniform density change over time on a planet?

Yes, the density of a planet can change over time due to various factors such as tectonic activity, volcanic eruptions, and impacts from other celestial bodies. These changes can also have a ripple effect on other aspects of the planet, such as its climate and surface features.