Force on Particle in Dust Cloud

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Homework Help Overview

The problem involves calculating the force on a particle of mass m located within a dust cloud surrounding a planet. The planet has a defined density and radius, while the dust cloud has its own density and radius. The challenge lies in accurately determining the mass contributing to the gravitational force acting on the particle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to calculate the mass of the dust cloud and the planet separately, considering the correct radius for the calculations. There is uncertainty about whether to include the entire mass of the dust cloud or only the mass within a certain radius.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the correct mass to consider for the dust cloud. Some participants suggest adjustments to the radius used in calculations, while others express uncertainty about the correctness of their approaches. There is an ongoing dialogue about potential mistakes and the need for careful verification.

Contextual Notes

Participants note the complexity of the problem and the potential for errors in reasoning, emphasizing the importance of not rushing to conclusions. The original poster expresses a desire for further clarification and guidance.

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[SOLVED] Force on Particle in Dust Cloud

The following problem is from Thorton & Marion's Classical Dynamics, Ch. 5 Problem 5-13 (p. 205 in the 5th edition of the text)

Homework Statement


A planet of density [tex]\rho_{1}[/tex] (spherical core, radius [tex]R_{1}[/tex]) with a thick spherical cloud of dust (density [tex]\rho_{2}[/tex], radius [tex]R_{2}[/tex]) is discovered. What is the force on a particle of mass [tex]m[/tex] placed within the dust cloud?

Homework Equations


[tex] V_{sphere}=\frac{4}{3}\pi \ r^{3}[/tex]
[tex] F = \frac{-GmM}{r^{2}}[/tex]
[tex] \rho = \frac{m}{v}[/tex]

The Attempt at a Solution


So my intuition for this one is to solve for big M and add the mass of the cloud with the mass of the planet.
[tex] M_{1} = \frac{4}{3}\pi\rho_{1} \ {R_{1}}^{3} [/tex]
for the mass of the planet, and:
[tex] M_{2} = \frac{4}{3} \pi\rho_{2} {R_{2}}^{3}[/tex]

substituting [tex]M[/tex] with [tex]M_{1} + M_{2}[/tex] and a bit of factoring, I get:
[tex] F = \frac{4}{3} \frac{Gm \pi ({R_{1}}^{3}\rho_{1} + {R_{2}}^{3}\rho_{2})}{r^{2}}[/tex]
Yet somehow this doesn't feel right...

The text provides answers for the even numbers only, so I don't know how to verify this. I feel like there's something else I should be doing and it might involve calculus...

Any suggestions? Thanks in advance for any help.
 
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Well, the dust cloud isn't in the planet and you should only be using the mass of the dust cloud inside of the radius of the particle. So maybe it would be better to use (r-R1) instead of R2?
 
Ah, alright!

So the mass of the dust cloud (as far as we are concerned, which is the radius of the particle) is now:
[tex] M_{2} = \frac{4}{3} \pi\rho_{2} (r-R_{1})^{3}[/tex][tex] F = \frac{4}{3} \frac{Gm \pi ({R_{1}}^{3}\rho_{1} + (r - R_{1})^{3}\rho_{2})}{r^{2}}[/tex]

and I'm assuming there should be a few things that end up cancelling when all is said and done.

Is there anything else, or is thing solved?
 
You know what? That's not right either. You only want the mass of the dust cloud between you and the planet. Make that (r^3-R1^3)*rho2. Sorry. Don't close the thread and call it solved until you are happy. Obviously, I make mistakes. Don't agree with me too fast.
 
Last edited:
If the particle is at a distance r such that R1<r<R2, the froce on the particle due to the planet and dust cloud can be calculated by finding the mass of the sphere of radius r.
F = 4/3*G*m*pi*[R1*3 + (r-R1)^3]/r^2
 
rl.bhat said:
If the particle is at a distance r such that R1<r<R2, the froce on the particle due to the planet and dust cloud can be calculated by finding the mass of the sphere of radius r.
F = 4/3*G*m*pi*[R1*3 + (r-R1)^3]/r^2

Great. You made the same mistake I did.
 
Yes Dick, I realized it later on.
 
Yes I think that makes more sense. Thanks for your help!
 

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