Potential Energy to bring in sphere from infinity

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Homework Help Overview

The problem involves calculating the work done to bring a uniform sphere of mass M and radius R from infinity, using integral calculus. The original poster attempts to derive the self-potential energy of the sphere and expresses uncertainty about their approach, particularly regarding the treatment of the differential mass and the force involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of integral calculus to find the work done, with some questioning the interpretation of the problem's wording. There is a focus on the uniform mass distribution and its implications for the calculations. The original poster expresses confusion about the role of the unit vector and the work-energy theorem.

Discussion Status

Some participants have provided clarifications and suggestions for approaching the problem, particularly regarding the integration process and the interpretation of the mass distribution. The discussion is ongoing, with multiple interpretations being explored, and the original poster indicates they have solved the problem, suggesting some productive direction has been achieved.

Contextual Notes

There are indications that the problem's wording may lead to confusion, and participants are considering how to properly set up the integral based on the changing radius of the sphere as mass is added. The original poster's uncertainty about the treatment of the differential mass and the force suggests a need for further clarification on these concepts.

Sentin3l
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Homework Statement



Given a uniform sphere of mass M and radius R. Use integral calculus and start with a mass dm in the sphere. Calculate the work done to bring in the remainder of the mass from infinity. By this technique show that the self-potential energy of the mass is:

P = -\frac{3}{5} \frac{GM^{2}}{R}​



Homework Equations



W = \int\vec{F} \bullet d\vec{r}
F = \frac{GMm}{r^{2}}​


The Attempt at a Solution



First let me say that this is a cosmology question. I began by considering a differential mass near or at the center of the sphere. Using the above equations for force and work, I derived:

W = - \int \frac{GM(dm)}{r^{2}} \hat{r}​

Since the sphere is uniform, it has a constant mass to radius ratio λ = \frac{M}{R} = \frac{dm}{dr}. So using this I found:


W = -λ\int \frac{GM}{r^{2}}dr = -3λ \frac{GM}{r^{3}}​

If we substitute λ = \frac {M}{R} and r=R, we get the result:

W = -3 \frac{GM^{2}}{R^{3}}​

Here is where i think I went wrong, I don't know if I need to deal with \hat{r} and if so, I'm not sure how to approach that.

I think that once I get the work, you use the work-energy theorem, and intial/final KE is 0 so the potential energy equals the work, please correct me if I'm wrong in that.
 
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Sentin3l said:
Since the sphere is uniform, it has a constant mass to radius ratio λ = \frac{M}{R} = \frac{dm}{dr}.

Uniform mass distribution means that the mass to volume ratio is constant.
 
The problem is worded kind of strangely. Consider when the mass that's already been brought in occupies a sphere of radius ##r##. How much work is required to bring in the mass to increase the radius by dr?
 
TSny said:
Uniform mass distribution means that the mass to volume ratio is constant.
Should I use my previous integral, taking note that dV = 4 \pi r^{2} dr ?
vela said:
The problem is worded kind of strangely. Consider when the mass that's already been brought in occupies a sphere of radius ##r##. How much work is required to bring in the mass to increase the radius by dr?

I agree that the problem is worded strangely, it's a recurring problem with this textbook. Also I'm not sure what your mean in your suggestion, did you mean "How much work is required to bring in the mass to decrease the radius by dr"?
 
Last edited:
No. Why would you want to decrease the amount of mass there when you're building it up?
 
I have solved the problem, thanks to you both for the assistance.
 

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