What is the Correct Diameter of the Critical Object?

[mystery_witch]

Diameter of "critical" object

Homework Statement

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 $\rho = 1 \textrm{g}/ \textrm{cm} ^{3}$.

Homework Equations

I was told to use the equation for hydrostatic equilibrium

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

where

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

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.

 

[Jhero]

Hello,

To estimate the diameter of the "critical" object, we need to use the equation for hydrostatic equilibrium as you have correctly done. However, there are a few errors in your calculations. Here is the correct solution using both densities:

For \rho = 3 \textrm{g}/ \textrm{cm} ^{3} :

M_{r} = \frac{4\pi {r}^{3}\rho}{3} = \frac{4\pi {(0.5r)}^{3}(3)}{3} = 2\pi r^{3}

Substituting this into the equation for hydrostatic equilibrium, we get:

\frac{dp}{dr} = -\frac{GM_{r}\rho}{{r}^{2}} = -\frac{2\pi G r^{3} (3)}{r^{2}} = -6\pi G r

Integrating this, we get:

p = \int_{2r}^{r} -6\pi G r dr = -3\pi G r^{2}

Solving for r, we get:

r = \sqrt{\frac{p}{-3\pi G}} = \sqrt{\frac{(\textrm{2} \cdot \textrm{10}^{9} Pa)(\textrm{10}^{9} yr)}{-3\pi (\textrm{6.67} \cdot \textrm{10}^{-11} Nm^{2}/kg^{2})}} = 1.0 \cdot \textrm{10}^{10} m

Therefore, the diameter of the "critical" object is approximately 20,000 km (4 times the radius). This is significantly larger than the 500-600 km range, which may indicate that the density used is not appropriate for this type of object.

For \rho = 1 \textrm{g}/ \textrm{cm} ^{3} :

Following the same steps as above, we get:

r = \sqrt{\frac{p}{-3\pi G}} = \sqrt{\frac{(\textrm{2} \cdot \textrm{10}^{9} Pa)(\textrm{10}^{9} yr)}{-3\pi (\textrm{