Force on Particle in Dust Cloud

AI Thread Summary
The discussion revolves around calculating the gravitational force on a particle within a dust cloud surrounding a planet. Initially, the participant attempted to combine the masses of the planet and the dust cloud but realized this approach was incorrect. The correct method involves considering only the mass of the dust cloud that is within the radius of the particle, leading to a revised formula for force. The final expression for the force incorporates both the planet's mass and the mass of the dust cloud up to the particle's distance from the planet. The conversation highlights the importance of accurately defining the relevant mass in gravitational calculations.
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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.
 
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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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