Force on Particle in Dust Cloud

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SUMMARY

The discussion focuses on calculating the force on a particle of mass m located within a dust cloud surrounding a planet, as described in Thorton & Marion's Classical Dynamics, Ch. 5 Problem 5-13. The correct approach involves determining the mass of the dust cloud that is within the radius of the particle, leading to the formula F = (4/3) * G * m * π * [(R1^3) + (r - R1)^3] / r^2. Participants clarified that only the mass of the dust cloud between the particle and the planet should be considered, correcting initial miscalculations regarding the mass contributions from both the planet and the dust cloud.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F = -GmM/r²
  • Familiarity with volume calculations for spheres, V = (4/3)πr³
  • Knowledge of density calculations, ρ = m/V
  • Basic calculus concepts for handling variable radius in force calculations
NEXT STEPS
  • Review gravitational force derivations in classical mechanics
  • Study the implications of density variations in astrophysical contexts
  • Learn about the integration of mass distributions in spherical coordinates
  • Explore advanced topics in gravitational fields and potential theory
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Students and educators in physics, particularly those studying classical mechanics and gravitational dynamics, as well as researchers in astrophysics dealing with planetary formation and dust cloud interactions.

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

Homework Equations


<br /> V_{sphere}=\frac{4}{3}\pi \ r^{3}<br />
<br /> F = \frac{-GmM}{r^{2}}<br />
<br /> \rho = \frac{m}{v}<br />

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.
<br /> M_{1} = \frac{4}{3}\pi\rho_{1} \ {R_{1}}^{3} <br />
for the mass of the planet, and:
<br /> M_{2} = \frac{4}{3} \pi\rho_{2} {R_{2}}^{3}<br />

substituting M with M_{1} + M_{2} and a bit of factoring, I get:
<br /> F = \frac{4}{3} \frac{Gm \pi ({R_{1}}^{3}\rho_{1} + {R_{2}}^{3}\rho_{2})}{r^{2}}<br />
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:
<br /> M_{2} = \frac{4}{3} \pi\rho_{2} (r-R_{1})^{3}<br /><br /> F = \frac{4}{3} \frac{Gm \pi ({R_{1}}^{3}\rho_{1} + (r - R_{1})^{3}\rho_{2})}{r^{2}}<br />

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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