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Expressing the equation of hydrostatic equilibrium

  1. Mar 17, 2015 #1
    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. jcsd
  3. Mar 17, 2015 #2
    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
     
  4. Mar 17, 2015 #3
    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.
     
  5. Mar 17, 2015 #4
    Using either the product or the quotient rule, what is the derivative with respect to r of m2/r4?

    Chet
     
  6. Mar 17, 2015 #5
    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
     
  7. Mar 17, 2015 #6
    m2(r) is a function of r, so it can be differentiated. Have you heard of the chain rule for differentiation?

    Chet
     
  8. Mar 17, 2015 #7
    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.
     
  9. Mar 17, 2015 #8
    No. The derivative of m2 with respect to r is 2m(dm/dr). You are in serious need of reviewing calculus.

    Chet
     
  10. Mar 17, 2015 #9
    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
     
  11. Mar 17, 2015 #10
    Recheck your algebra, and where did d/dr (dP/dr) come from?

    Chet
     
  12. Mar 17, 2015 #11
    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
     
  13. Mar 17, 2015 #12
    I think that rechecking your work would be a good idea. Now your issue is algebra.

    Chet
     
  14. Mar 17, 2015 #13
    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?
     
  15. Mar 17, 2015 #14
    Your error is right here. Read your problem statement.

    Chet
     
  16. Mar 17, 2015 #15
    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)
     
  17. Mar 17, 2015 #16
    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
     
  18. Mar 17, 2015 #17
    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
     
  19. Mar 17, 2015 #18
    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.
     
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