# Diameter of critical object

1. Oct 13, 2007

### mystery_witch

Diameter of "critical" object

1. The problem statement, all variables and given/known data

A pressure of $\textrm{2} \cdot \textrm{10}^{9}$ Pa on a time scale of $\textrm{10}^{9}$ yr, holds half way to the center of the "critical" objects.
Estimate the diameter using two densities: $\rho = 3 \textrm{g}/ \textrm{cm} ^{3}$ and $\rho = 1 \textrm{g}/ \textrm{cm} ^{3}$.

2. Relevant equations

I was told to use the equation for hydrostatic equilibrium

$$\frac{dp_}{dr} = -\frac{GM_{r}\rho}{{r}^{2}}$$

where

$$M_{r} = \frac{4\pi {r}^{3}\rho}{3}$$

3. The attempt at a solution

I chose 2r as the radius for the entire object, then r is the radius half way.
I then inserted $M_{r}$ into the equation for hydrostatic equilibrium and integrated the new equation

$$p = \int_{2r}^r -\frac{{4\pi G} \rho^{2}}{3}r^\prime\, dr^\prime = \frac{{4\pi G} \rho^{2}}{3} ((2r)^{2}-r^{2}) = 2 \pi G \rho^{2} r^{2}$$

Then I just inserted all the constants and solved the equation for r.

$$\rho = 3 \textrm{g}/ \textrm{cm} ^{3} \Rightarrow r \approx 728,183 km \Rightarrow d=4r \approx 3000km$$

and

$$\rho = 1 \textrm{g}/ \textrm{cm} ^{3} \Rightarrow r \approx 2184,550 km \Rightarrow d=4r \approx 8740 km$$

I was told that I should get a diameter between 500-600 km. Obviously I'm no where near that, so I really need help with this because I have to present a solution on October 18th.