# Lower limit to the diameter

1. ### duder1234

4
I have a homework question that I am having troubles with.

Q: By equating the pressure at the centre of an icy planetesimal to the maximum pressure that cold ice can sustain without deforming, about 40 MPa, find a lower limit to the diameter of an icy minor planet.

The part I dont understand is the "lower limit to the diameter"
Do I use: Pcentral >$\frac{GM^{2}}{8πr^{4}}dm$

I just dont know how to get the diameter....

### Staff: Mentor

I don't know where your formula comes from, but you need some relation between size (like the radius r in your formula?) and pressure in the center, and then let the pressure in the center be 40MPa.
This looks more like an upper limit, however.

3. ### duder1234

4
I did:
$\frac{dP}{dr}$=$ρg$
and $g$=$\frac{GM}{r^2}$
so $\frac{dP}{dr}=\frac{-GMρ}{r^2}$ (Hydrostatic equilibrium equation)
and $\frac{dM}{dr}$=$4πr^{2}ρ$ (equation of mass conservation)

by dividing the two equations: $\frac{dP/dr}{dM/dr}$=$\frac{dP}{dM}$=$\frac{-GM}{4πr^4}$

integration: $P_{c}-P_{s}$=-$\int^{M_{c}}_{M_{s}}$($\frac{GM}{4πr^4}$)$dM$
$P_{c}$ and $P_{s}$ are pressure at centre and surface of the planet
and by setting $M_{c}=0$ and by switching the intergral:

$P_{c}-P_{s}$=$\int^{M_{s}}_{0}$($\frac{GM}{4πr^4}$)$dM$

and

$\int^{M_{s}}_{0}$($\frac{GM}{4πr^4}$)$dM$ > $\int^{M_{s}}_{0}$($\frac{GM}{4πr^{4}_{s}}$)$dM$ = $\frac{GM^{2}_{s}}{8πr^{4}_{s}}$

hence

$P_{c}-P_{s}$>$\frac{GM^{2}_{s}}{8πr^{4}_{s}}$

and by approximating that the pressure at the surface to be zero ($P_{s}=0$)
we get:

$P_{c}$>$\frac{GM^{2}_{s}}{8πr^{4}_{s}}$

So are you saying I should do:

$40MPa$>$\frac{GM^{2}_{s}}{8πr^{4}_{s}}$?

I figure I have to solve for r and then obtain the diameter from there but Im stuck because I am not given the mass... (or do I use a mass of an icy minor planet like Ceres?)

Last edited: Nov 3, 2013

### Staff: Mentor

Your formula differs from the one in the first post now.

You know the density of ice, this gives the relation radius<->mass,

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