A rod, a ball, garvitational Potential Energy (U), and the power series expns.

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SUMMARY

The discussion focuses on calculating the gravitational potential energy (GPE) of a rod and ball system using the power series expansion for ln(1+x). The relevant equation for GPE is U = -GMm/r, where M is the mass of the rod, m is the mass of the ball, and r is the distance between them. Participants express confusion about integrating the GPE and applying the power series expansion effectively, particularly in relation to the limits of integration and the approximation of ln(1+x) for small x. The integration approach discussed involves using the integral of 1/r and setting limits from infinity to x.

PREREQUISITES
  • Understanding of gravitational potential energy (GPE) and its formula U = -GMm/r
  • Familiarity with power series expansions, specifically ln(1+x)
  • Basic calculus concepts, including integration techniques
  • Knowledge of limits and their application in calculus
NEXT STEPS
  • Study the derivation and applications of the power series expansion for ln(1+x)
  • Learn about integrating functions involving square roots, particularly in the context of gravitational problems
  • Explore the concept of limits in calculus, especially in relation to infinity
  • Practice problems involving gravitational potential energy calculations for various systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and gravitational systems, as well as educators looking for examples of applying calculus to physical problems.

TFM
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Homework Statement



Mass of rod: M
mass of ball: m
Length of Rod: L
distance between rod and ball: x
GPE is zero at infinty

The questiopn asks to take the GPE of the rod/ball system, using the Power Series Expansion for ln(1+x) .

Homework Equations



U = -GMm/r

The Attempt at a Solution



I'm not quite sure where to start - the Power Series expansion has confused me slightly, as otherwise I wouold hqave just put the variables in the above equation?

TFM :confused:
 
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It just means you can approximate ln(1+x) with x, for x << 1.
 
Where does the Ln(x+1)come from?

TFM
 
It'll probably appear as you solve the problem. They say to use the power series to make it easier to solve.
 
Do you still use the U=-GMm/r, with U = 0, r = infinty, giving:

0=-GMm/infinity?

TFM
 
I guess you want to integrate over the length of the rod and end up with the integral of
(1/l+1) dl from 0 to L... that should give you ln(x+1)

U= -GmM/L [integral 0 to L (dl /sqrt. of x^2+l^2)] and I guess you can say that the square root of x^2+l^2 is x+C where C is some constant, though this makes no sense I can't think of any other way

btw, the way I got that is by saying dU= -Gm(dM)/r.. then setting dM=dl(M/L) and r=sqrt. (x^2+l^2)
 
Last edited:
take the intergral of 1/r, which is ln(r), setting your limits from infinity to x. Do you know the expansion to ln(r)?
 
from infinity to x? how in the world do you integrate from infinity to x?
 
t-money said:
take the intergral of 1/r, which is ln(r), setting your limits from infinity to x. Do you know the expansion to ln(r)?

Why do I need to take the integral of 1/r?

TFM
 
  • #10
I'm Still rather cionfused about what I should be doing:frown:

Any Help?

TFM
 

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