Solution to II.2.8-II.2.10: Textbook Equations

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Homework Help Overview

The discussion revolves around equations II.2.8, II.2.9, and II.2.10 from a textbook, focusing on the transition between these equations and the implications of certain parameters within them. The subject area appears to involve mathematical physics, particularly in the context of differential equations and their applications.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants express confusion regarding the transition from II.2.8 to II.2.9, particularly questioning the introduction of an extra 'r' in the denominator. There are discussions about the correctness of the equations, with some participants pointing out potential errors in notation and signs. The introduction of a new parameter (r0) in II.2.10 raises further questions about its relevance and correctness.

Discussion Status

The discussion is ongoing, with participants actively questioning assumptions and interpretations of the equations. Some guidance has been offered regarding the differentiation process and the need for careful attention to signs and parameters. There is no explicit consensus yet, as participants continue to explore the implications of their findings.

Contextual Notes

Participants note potential issues with the clarity of the equations and the notation used, particularly regarding the differentiation process and the introduction of new parameters. There is also mention of the need to correct previous work, indicating a collaborative effort to clarify the problem.

Mechdude
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Homework Statement


please look at equations II.2.8 , II.2.9, II.2.10 in the attached scan,

Homework Equations


none

The Attempt at a Solution


im unable to get from II.2.8 to II.2.9, i wonder where they got the extra r in the denominator of the stuff in the brackets,
here is what i get for II.2.10:
\frac{1}{18} \left( \frac{ (n+2) n C} { r^{n+3}} - \frac{3 M e^2}{r^4} \right)
what I am i doing wrong?
 

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Looks like sloppy work to me:

1. in going from II.2.8 to II.2.9 somehow E changed to e.

2. There should be a minus sign in front of II.2.9, otherwise it's correct.

remember: d/dr(rn) = nrn-1 etc.

No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.
 
rude man said:
Looks like sloppy work to me:

1. in going from II.2.8 to II.2.9 somehow E changed to e.

2. There should be a minus sign in front of II.2.9, otherwise it's correct.

remember: d/dr(rn) = nrn-1 etc.

No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.

how did u do that? in II.2.8 there is (for stuff after partial differentiation***{EDIT})
\frac{1}{6r^2} \left( ME^2 \frac{1}{r} - \frac{nC}{r^n} \right)
how does this end up being
\frac{1}{6} \left( ME^2 \frac{1}{r^4} - \frac{nC}{r^{n+3}} \right)
when the stuff just outside the bracket gets taken in? r is only to the inverse second power, why does r in II.2.9 end up to the inverse fourth power? 2+1 =4 ? remember the partial diff hasn't yet been done.

***EDIT
i mean partial differentiation sign not after performing the operation.
 
Last edited:
though i think the r_o is the r corresponding to minimum P.E., looking at II.2.7 and the paragraph preceeding it.
 
Oops - I hadn't noticed - in II.2.9 there should not be a partial differential anymore. The derivative is taken already in going from II.2.8 to II.2.9.

Like I said - sloppy work! And there still needs to be a minus sign in front of the corrected II.2.9.
 
rude man said:
Oops - I hadn't noticed - in II.2.9 there should not be a partial differential anymore. The derivative is taken already in going from II.2.8 to II.2.9.

Like I said - sloppy work! And there still needs to be a minus sign in front of the corrected II.2.9.

Ok, i see that now, i had not noticed that the differentiation had already been done, so that's why you kept on insisting on the negative sign, ( i admit i was clueless why until now).
thanks. Let me work on it and see whether i can better this guy's work.
cheers.
 
one quick last one, has anyone seen this question before? is n=8.1 really? is that the correct answer that i need to find when i correct this guy's unholy mess?
 

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