Expressing the equation of hydrostatic equilibrium

[Jdraper]

Homework Statement

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

Homework Equations

n/a

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.

 

[Chestermiller]

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

 

[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.

 

[Chestermiller]

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

Chet

 

[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

 

[Chestermiller]

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

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

Chet

 

[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.

 

[Chestermiller]

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.

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

Chet

 

[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

 

[Chestermiller]

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

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

Chet

 

[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

 

[Chestermiller]

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

Chet

 

[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?

 

[Chestermiller]

ok starting from the beginning

dP/dr =- Gm/4πr⁴ dm/dr

=-G/4π d/dr (m²/r⁴)

Your error is right here. Read your problem statement.

Chet

 

[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)

 

[Chestermiller]

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)

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

 

[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

 

[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.