How to Calculate the Lower Limit of the Diameter of an Icy Minor Planet?

In summary, the conversation discusses finding the lower limit for the diameter of an icy minor planet by equating the pressure at its center to the maximum pressure that cold ice can sustain without deforming, which is about 40 MPa. The formula for this is derived using the hydrostatic equilibrium equation and the equation of mass conservation. The conversation also mentions the need for a relation between size and pressure, and using the density of ice to find the mass for a given pressure.
  • #1
duder1234
4
0
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 don't understand is the "lower limit to the diameter"
Do I use: Pcentral >[itex]\frac{GM^{2}}{8πr^{4}}dm[/itex]

I just don't know how to get the diameter...
 
Astronomy news on Phys.org
  • #2
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
mfb said:
I don't know where your formula comes from
I did:
[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]


mfb said:
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.

So are you saying I should do:

[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 I am stuck because I am not given the mass... (or do I use a mass of an icy minor planet like Ceres?)
 
Last edited:
  • #4
Your formula differs from the one in the first post now.

You know the density of ice, this gives the relation radius<->mass,
 
  • Like
Likes 1 person
  • #5


A: The "lower limit to the diameter" refers to the smallest possible diameter that an icy minor planet can have while still maintaining its shape and not deforming under the maximum pressure of 40 MPa. In order to find this lower limit, you will need to use the equation you mentioned, Pcentral >\frac{GM^{2}}{8πr^{4}}dm, which relates the central pressure (Pcentral) to the mass (M) and radius (r) of the planet. By equating this pressure to the maximum pressure of 40 MPa, you can solve for the minimum radius (and therefore diameter) that the planet can have. This will give you the lower limit to the diameter of an icy minor planet.
 

FAQ: How to Calculate the Lower Limit of the Diameter of an Icy Minor Planet?

What does "lower limit to the diameter" mean?

The lower limit to the diameter refers to the smallest possible diameter that a particular object or material can have.

Why is it important to know the lower limit to the diameter?

Knowing the lower limit to the diameter is important in many scientific and engineering applications, as it can determine the feasibility and functionality of a design or product.

How is the lower limit to the diameter calculated?

The lower limit to the diameter is typically calculated by considering the physical properties and limitations of the material or object in question, such as its strength, flexibility, and manufacturing capabilities.

Can the lower limit to the diameter be exceeded?

In most cases, the lower limit to the diameter is a strict boundary that cannot be exceeded without compromising the integrity and functionality of the object or material. However, with advancements in technology and materials, this limit may be pushed in certain circumstances.

How does the lower limit to the diameter affect engineering and design processes?

The lower limit to the diameter plays a crucial role in engineering and design processes, as it helps determine the optimal size and dimensions for a product or structure to ensure its functionality and safety.

Back
Top