Lower limit to the diameterby duder1234 Tags: central pressure, diameter, limit, lower limit, minor, planet 

#1
Nov313, 07:51 AM

P: 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: P_{central} >[itex]\frac{GM^{2}}{8πr^{4}}dm[/itex] I just dont know how to get the diameter.... 



#2
Nov313, 08:19 AM

Mentor
P: 10,813

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
Nov313, 03:57 PM

P: 4

[itex]\frac{dP}{dr}[/itex]=[itex]ρg[/itex] and [itex]g[/itex]=[itex]\frac{GM}{r^2}[/itex] so [itex]\frac{dP}{dr}=\frac{GMρ}{r^2}[/itex] (Hydrostatic equilibrium equation) and [itex]\frac{dM}{dr}[/itex]=[itex]4πr^{2}ρ[/itex] (equation of mass conservation) by dividing the two equations: [itex]\frac{dP/dr}{dM/dr}[/itex]=[itex]\frac{dP}{dM}[/itex]=[itex]\frac{GM}{4πr^4}[/itex] integration: [itex]P_{c}P_{s}[/itex]=[itex]\int^{M_{c}}_{M_{s}}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex] [itex]P_{c}[/itex] and [itex]P_{s}[/itex] are pressure at centre and surface of the planet and by setting [itex]M_{c}=0[/itex] and by switching the intergral: [itex]P_{c}P_{s}[/itex]=[itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex] and [itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex] > [itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^{4}_{s}}[/itex])[itex]dM[/itex] = [itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex] hence [itex]P_{c}P_{s}[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex] and by approximating that the pressure at the surface to be zero ([itex]P_{s}=0[/itex]) we get: [itex]P_{c}[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex] [itex]40MPa[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex]? 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?) 



#4
Nov413, 09:03 AM

Mentor
P: 10,813

Lower limit to the diameter
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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