Astronomy - interstellar dust grains

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Homework Help Overview

The problem involves calculating the number of dust particles in a molecular cloud and determining the total surface area of these dust grains. The context is within the field of astronomy, specifically related to interstellar dust and molecular clouds.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the interpretation of the mass of the molecular cloud and the percentage of that mass attributed to dust grains. There are attempts to calculate the number of dust particles based on given mass values and to find the surface area of the dust grains.

Discussion Status

Some participants have identified potential errors in calculations and interpretations. There is an ongoing exploration of the definitions and assumptions regarding the mass of the molecular cloud and the nature of the dusty gas. Guidance has been provided regarding the methods used, but no consensus has been reached on the correct interpretation of the problem.

Contextual Notes

There is a discussion about the definition of "dusty gas" and its implications for the total mass of the molecular cloud. Participants are questioning the relationship between the mass of the cloud and the mass of the dust grains, indicating a need for clarification on these terms.

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Homework Statement


A typical dust grain has a radius of 10-7 m and a mass of 10-14 g. How many dust particles are there in a molecular cloud containing 1000*Masssun of dusty gas if 1% of the cloud's mass is in the form of dust grains?


Homework Equations





The Attempt at a Solution


1000*Msun = Mcloud * 0.01
Masscloud = 100,000*Msun = 2 * 1038 g.

Then do I just do:
(2 * 1038 g) / (10-14 g) = 2 * 1051 dust particles.

Is that correct?

Then, it asks to find the total surface area these (spherical) grains would take up if side by side. Do I just find S.A. = 4*pi*r2 = 4*pi*(10-7 m)2 = 1.2566*10-11 m2. So:
1.2566*10-11 m2 * (2 * 1051 particles) = 2.5 * 1040 m2 for the surface area. Is that how I'm supposed to do it?
 
Last edited:
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Hi,

I have spotted a few errors, but for the most part your methods are correct.

mbradar2 said:

Homework Statement


A typical dust grain has a radius of 10-7 m and a mass of 10-14 g. How many dust particles are there in a molecular cloud containing 1000*Masssun of dusty gas if 1% of the cloud's mass is in the form of dust grains?


Homework Equations





The Attempt at a Solution


1000*Msun = Mcloud * 0.01
I interpret the question statement differently:
Mcloud = 1000*Msun
and
Mdust = 1% of Mcloud

Masscloud = 100,000*Msun = 2 * 1038 g.

Then do I just do:
(2 * 1038 g) / (10-14 g) = 2 * 1051 dust particles.

Is that correct?
No, but once you get a value for Mdust, your method should work.

Then, it asks to find the total surface area these (spherical) grains would take up if side by side. Do I just find S.A. = 4*pi*r2 = 4*pi*(10-7 m)2 = 1.2566*10-11 m2.
Correct method, but you should redo the calculation.
So:
1.2566*10-11 m2 * (2 * 1051 particles) = 2.5 * 1040 m2 for the surface area. Is that how I'm supposed to do it?
Yes, if you had the correct surface area for one dust grain, and the correct number of dust grains, that is how you would do it.
 
Well, it says that the molecular cloud contains 1000*M_sun of *dusty gas* - so dust particles. So that's only the dust particles, there's still all the other stuff that makes up a molecular cloud that isn't included in that, so the total M_cloud mass can't just be 1000*M_sun. It says 1% of the molecular cloud's total mass is dust grains... so 1% of the molecular cloud is the 1000*M_sun of *dusty gas* - dust grains. I guess I don't understand why you get 1000*M_sun = M_cloud when you look at the question, so I'm not sure which to use. Are you sure my way is not correct?
 
Oh, interstellar dust is defined in our textbook as "tiny, SOLID particles" so then 'dusty GAS' means the whole cloud, not just the solid particles.

Okay, thank you!
 
You're welcome :smile:
 

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