# Planet made entirely of liquid

1. Jun 17, 2007

### ApeXaviour

Okay obviously a hypothetical situation, this planet has a radius $$R$$, uniform density, and it doesn't rotate.

Pressure in a liquid is given by
$$p=\rho g d$$ where $$d$$ is the depth.

So the liquid pressure a distance $$r$$ from the centre of the planet is.
$$p=\rho g (R-r)$$ (where r<R)
But $$g$$ is also a function of $$r$$. A little bit of fiddling with newton's gravitation law gives:
$$g(r)=GM\frac{r}{R^3}$$
So...
$$p=\rho GM\frac{r}{R^3}(R-r)$$

This would mean, at $$r=0$$ the core of the planet is under zero pressure...! Em, I don't believe this to be correct, so what am I missing? am I slipping up somewhere? Am I putting too much stock in the formula: $$p=\rho g d$$?

2. Jun 17, 2007

### billiards

your first equation is actually the result of a more general equation after some integration, it refers specifically to a column of liquid in which g remains constant. You need to go right back to the roots.

The reason you found pressure to be zero is because g is zero at the centre.

Last edited: Jun 17, 2007
3. Jun 18, 2007

### ApeXaviour

Hmm.. not sure why this was moved, it's not homework, I've actually just graduated. It's something I was debating with a friend.

Okay I see how it doesn't work as it assumes g is constant.

Okay back to basics, let me try again.
Pressure is force over area, in this case "weight" over area.

Taking a column of water between r and R, of height h and base area A. Taking an infinitesimal segment of the liquid at the top, it's weight is given by:
$$dF=\rho .A .g(r). dh$$
where $$g(r)=GM\frac{r}{R^3}$$ and $$h = R-r$$
so $$dh=-dr$$
so now the total force/weight is $$F=\rho A GM \int_{R}^{r} -g(r) dr$$

So now pressure is:
$$p=\frac{F}{A}=\rho GM \frac{1}{2}(\frac{1}{R}-\frac{r^2}{R^3})$$

How does that look??? I'm happier with it, as it appears to be max when r is zero and min when r is max.

Last edited: Jun 18, 2007
4. Jun 18, 2007

### billiards

It looks wrong to me, p should equal zero for r=R.

5. Jun 19, 2007

### ApeXaviour

Good point, that didn't occur to me. But look at it again, it actually does equal zero when r=R. The bracketed part goes to (1/R - 1/R).

Hmm I'm pretty confident in it now. What do people think?

6. Jun 19, 2007

### Office_Shredder

Staff Emeritus
Why did you take a column of water? Especially if you're near the center, I assume you'd need to look at a sphere, as the force wouldn't all be pointing in the same direction as you move to points near where you're calculating

7. Jun 19, 2007

### ApeXaviour

I don't quite see what you're getting at. It's static so the only force is gravitational, which is radial. Imagine the column as being very thin and pointing in a radial direction.

8. Jun 19, 2007

### billiards

Oh yes so it is! silly me. Note that because M=rho*V you should be able to find a more elegant form of this equation.

9. Jun 19, 2007

### gabee

It looks like he took infinitesimal concentric spheres of liquid and totaled up the force from each of these thin shells, which looks to me like it should work since the force is uniform and always pointing toward the center. Interesting problem!

Last edited: Jun 19, 2007
10. Sep 25, 2007

### Agua88

So how much?

Sooo . . . assuming an earth mass world, how high is the pressure at the center? What is the density of the water there? Would it still be water or some sort of metal-like substance or what?

11. Sep 25, 2007

### genneth

That would also depend on the temperature. We could put it in by hand, which would mean we could choose what phase we wanted. Or we could try coming up with a full model of the planet, but then we'd be constrained by reality.

12. Sep 26, 2007

### Agua88

Temperature

Well with NASA about to publish a list of world types that might support life, and one of the types being a model made purely of water, I would think we would want to arrange it so that the surface temperature, under an earthlike atmosphere, would be around 80 degrees farenheit or thereabouts. No idea what temperature that would make the interior.

We'd want to assume a standard rotation, not too different from that of Earth, to avoid wierd freezing problems on the backside and such.