# Expressing the equation of hydrostatic equilibrium

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1. Mar 17, 2015

### Jdraper

1. The problem statement, all variables and given/known data

This problem has been stumping me for days now, I'm sure i'm missing something simple as it's only worth a small number of marks on the coursework. Any help would be appreciated.

I've been asked to re-express the equation of hydrostatic equilibrium:

dP/dr = - Gm/4πr4 dm/dr

as, -Gm/4πr4 dm/dr = -d/dr (Gm2/8πr4) - Gm2/2πr5

2. Relevant equations
n/a

3. The attempt at a solution
The most obvious thing seemed to try to differentiate with respect to r using product rule. This however leaves me with the wrong answers and an imbalance in my equation.

I've also tried integrating by parts but i don't believe this is the correct way to tackle the problem as you can't get the r5 by integration, only by differentiation.

Any help would be appreciated, I'm sure i'm missing something really simple.

Thanks, John.

2. Mar 17, 2015

### Staff: Mentor

It looks to me like differentiation by the product rule works just fine. Show us what you did. Also, it might have helped if you had written your original equation in a more readable form:
$$\frac{dP}{dr}=-\frac{Gm}{4πr^4}\frac{dm}{dr}$$

Chet

3. Mar 17, 2015

### Jdraper

ok so differentiating the RHS, tried writing this using LaTex but really struggling to get it to work. Sorry, hope this format is ok.

d/dr ( -Gm/4πr4 dm/dr ) =

Completing this would give me a triple product differentiation rule, I'm happy to write it out but it seems to me that it would give me three terms which i would be unable to simplify down to two terms which is what i need.

The only other way i can see seems to be mathematically unsound.

dP/dr = -Gm/4πr4 dm/dr manipulating the differential in the LHS we get

dP/dr = -G/4π d/dr (m2/r4) , as mass is a function of radius we get

dP/dr =-G/4πr4 d/dr(m^2) + Gm2/πr5

Which is also not the required answer, Are either of these on the right track to the solution?

Thanks, John.

4. Mar 17, 2015

### Staff: Mentor

Using either the product or the quotient rule, what is the derivative with respect to r of m2/r4?

Chet

5. Mar 17, 2015

### Jdraper

Using the product rule i get:

=m2 d/dr (1/r4) + 1/r4 d/dr (m2)
=-4m2/r5 + 1/r4 d/dr(m2)

as the m2 cannot be differentiated with respect to r as we are not given the mass profile.

John

6. Mar 17, 2015

### Staff: Mentor

m2(r) is a function of r, so it can be differentiated. Have you heard of the chain rule for differentiation?

Chet

7. Mar 17, 2015

### Jdraper

Ahh ok, that didn't seem apparent at first, so now i get:

=-4m2/r5 + 1/r4 d/dr(m2)
using chain rule
=-4m2/r5 + 1/r4 (2m(r)*1)

=-4m2/r5 + 2m/r4

Is this correct?

Thanks, John.

8. Mar 17, 2015

### Staff: Mentor

No. The derivative of m2 with respect to r is 2m(dm/dr). You are in serious need of reviewing calculus.

Chet

9. Mar 17, 2015

### Jdraper

Yeah, i agree. Most have my modules have strayed away from differential calculus is recent years so a review is needed.

So that gives me:

d/dr (dP/dr) =-Gm2 / πr5 +Gm/2πr4 dm/dr

10. Mar 17, 2015

### Staff: Mentor

Recheck your algebra, and where did d/dr (dP/dr) come from?

Chet

11. Mar 17, 2015

### Jdraper

Oh sorry, Forgot we manipulated the differential in the RHS rather than differentiating both sides.

So we have

dP/dr =-Gm2 / πr5 +Gm/2πr4 dm/dr

Manipulating the differentials again I get

dP/dr = Gm/πr5 - d/dr (Gm2/2πr4)

It seems that some factors are missing, i'll recheck my workings

12. Mar 17, 2015

### Staff: Mentor

I think that rechecking your work would be a good idea. Now your issue is algebra.

Chet

13. Mar 17, 2015

### Jdraper

ok starting from the beginning

dP/dr =- Gm/4πr4 dm/dr

=-G/4π d/dr (m2/r4) = -G/4π (m2 d/dr (r-4) + 1/r4 d/dr (m2)) using product rule

then we get

=-G/4π(-4m2/r5 + 2m/r4 dm/dr )

so this give us

=Gm2/πr5 -d/dr (Gm2/2πr4)

Sorry but i don't understand where my missing factors have gone?

14. Mar 17, 2015

### Staff: Mentor

Chet

15. Mar 17, 2015

### Jdraper

I know i'm probably being really dense but i dob't see it. It can't be from the original statement as that's fine so there must be something wrong with the

=-G/4π d/dr (m2/r4)

16. Mar 17, 2015

### Staff: Mentor

Here are the first two equations of your problem statement combined into one:

Show that

dP/dr = - Gm/4πr4 dm/dr = -d/dr (Gm2/8πr4) - Gm2/2πr5

If you can't get the answer from this, I'm at a loss for how I can help you further.

Chet

17. Mar 17, 2015

### Jdraper

I understand you are doing your best to help but it's very unclear to me what the source of my error is.

At the moment i have an answer in which one term is a factor of -½ out and another which is a factor of ¼ out. I see nothing wrong with my algebraic working so could you please tell me whether this step in my workings is wrong?

dP/dr =- Gm/4πr4 dm/dr =-G/4π d/dr (m2/r4)

Thanks, John

18. Mar 17, 2015

### Jdraper

Finally shown it, didn't realize you were working from the RHS to the LHS. I was trying to do the opposite hence the confusion.

Thanks anyway.