# Text book solution

1. Apr 18, 2012

### Mechdude

1. The problem statement, all variables and given/known data
please look at equations II.2.8 , II.2.9, II.2.10 in the attached scan,

2. Relevant equations
none

3. 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 im i doing wrong?

#### Attached Files:

• ###### MihalyAndMartin.pdf
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2. Apr 18, 2012

### rude man

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.

3. Apr 19, 2012

### Mechdude

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: Apr 19, 2012
4. Apr 19, 2012

### Mechdude

though i think the $r_o$ is the r corresponding to minimum P.E., looking at II.2.7 and the paragraph preceeding it.

5. Apr 19, 2012

### rude man

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.

6. Apr 19, 2012

### Mechdude

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.

7. Apr 19, 2012

### Mechdude

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?