Astronomy - interstellar dust grains

AI Thread Summary
A typical dust grain in a molecular cloud has a radius of 10^-7 m and a mass of 10^-14 g. To determine the number of dust particles in a cloud containing 1000 solar masses of dusty gas, where 1% of the mass is dust, the total mass of the cloud is calculated as 100,000 solar masses, equating to 2 x 10^38 g. The correct calculation yields approximately 2 x 10^51 dust particles. For the total surface area of these grains, the method is correct, but the surface area of one grain must be recalculated accurately. The discussion highlights confusion regarding the interpretation of the molecular cloud's mass and the distinction between dusty gas and solid dust particles.
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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?
 
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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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