Impossible hydrostatic scenario

In summary: The function p(r) is asked, not the pressure at r_E.dp/dr =constant/r2. What is the antiderivative of the function constant/r2?Or: add integral symbols to both sides. Integrate the left side from r to infinity, the right side from p to zero:##\int_r^∞\ {G*\frac{M*(\rho)}{r^2}dr}=\int_p^0 dp ##Note that there is a sign...
  • #1
RiotRick
42
0

Homework Statement


TankPhysik.JPG

There is an infnite high hydrostatic head in an infite high tank on the surface of earth. How big is the pressure p(r) in tank at a distance r. Ignore the rotation of the Earth and assume the water stays liquid. ( So basically ignore it's an impossible scenario )
Instruction: Look at a mass element with thickness dr and establish the condition of equilibrium between gravitational force and pressure force. You'll get a differential equation. Then use the fact that ##p(r \rightarrow \infty )## is 0.
Try to give an answer without G and M. G is the gravitational constant and M the mass of the earth

Homework Equations



Given:
##g,r,r_E,\rho,M## M is the mass of the earth

##G*\frac{M*m}{r^2}=F_g##

The Attempt at a Solution


[/B]
I'don't fully understand how set up the equillibrium condition.
So the Earth is pulling the water:
##G*\frac{M*m}{(r)^2}=m*g##
but the water dr away pulls the water in the opposite direction but is also pulled towards the earth. I don't see it.
 

Attachments

  • TankPhysik.JPG
    TankPhysik.JPG
    9.9 KB · Views: 516
Physics news on Phys.org
  • #2
RiotRick said:

Homework Statement


View attachment 236766
There is an infnite high hydrostatic head in an infite high tank on the surface of earth. How big is the pressure p(r) in tank at a distance r. Ignore the rotation of the Earth and assume the water stays liquid. ( So basically ignore it's an impossible scenario )
Instruction: Look at a mass element with thickness dr and establish the condition of equilibrium between gravitational force and pressure force. You'll get a differential equation. Then use the fact that ##p(r \rightarrow \infty )## is 0.
Try to give an answer without G and M. G is the gravitational constant and M the mass of the earth

Homework Equations



Given:
##g,r,r_E,\rho,M## M is the mass of the earth

##G*\frac{M*m}{r^2}=F_g##

The Attempt at a Solution


[/B]
I'don't fully understand how set up the equillibrium condition.
So the Earth is pulling the water:
##G*\frac{M*m}{(r)^2}=m*g##
but the water dr away pulls the water in the opposite direction but is also pulled towards the earth. I don't see it.
The liquid is in equilibrium. Write the force balance equation for a thin layer of thickness dr.

upload_2019-1-4_8-50-23.png
 

Attachments

  • upload_2019-1-4_8-50-23.png
    upload_2019-1-4_8-50-23.png
    9.4 KB · Views: 288
  • #3
##-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)##
##-G*\frac{M*(\rho*A)}{r^2}=\frac{A*p(r)-A*p(r+dr)}{dr}##
The right side would be a perfect differential equation. Does this work?
 
  • #4
RiotRick said:
##-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)##
##-G*\frac{M*(\rho*A)}{r^2}=\frac{A*p(r)-A*p(r+dr)}{dr}##
The right side would be a perfect differential equation. Does this work?
The equation is almost correct, (check the signs), but one side of an equation is not an equation. Have you meant perfect differential? (Better, it is negative of the derivative of p with respect r) If so, what is the function p(r)?
 
Last edited:
  • #5
I thought p(r) is a function which gives me the pressure with respect to the distance r
 
  • #6
RiotRick said:
I thought p(r) is a function which gives me the pressure with respect to the distance r
Yes, and what is its form? How does it depend on r?
 
  • #7
ehild said:
Yes, and what is its form? How does it depend on r?
##p(r)=\rho*g(r-r_E)##
 
  • #8
RiotRick said:
##p(r)=\rho*g(r-r_E)##
No.
Divide the whole equation by A.Then the numerator on the right side is -(p(r+dr)-p(r))= -dp, so the right side is -dp/dr. Look at the left side. dp/dr=?
 
  • #9
ehild said:
No.
Divide the whole equation by A.Then the numerator on the right side is -(p(r+dr)-p(r))= -dp, so the right side is -dp/dr. Look at the left side. dp/dr=?
I don't see how your approach is different than mine?
##-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)##
= ##-G*\frac{M*(\rho*dr)}{r^2}=p(r)-p(r+dr)##
= ##-G*\frac{M*(\rho)}{r^2}=\frac{p(r)-p(r+dr)}{dr}##
= ##G*\frac{M*(\rho)}{r^2}=\frac{dp(r)}{dr}##
= ##G*\frac{M*(\rho)}{r^2}*dr=dp(r)##
integrate from r_E to infinity, with infinity = 0
##-G*\frac{M*(\rho)}{r^2}*r_E=p(r)##
 
  • #10
RiotRick said:
I don't see how your approach is different than mine?
##-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)##
= ##-G*\frac{M*(\rho*dr)}{r^2}=p(r)-p(r+dr)##
= ##-G*\frac{M*(\rho)}{r^2}=\frac{p(r)-p(r+dr)}{dr}##
= ##G*\frac{M*(\rho)}{r^2}=\frac{dp(r)}{dr}##
= ##G*\frac{M*(\rho)}{r^2}*dr=dp(r)##
integrate from r_E to infinity, with infinity = 0
##-G*\frac{M*(\rho)}{r^2}*r_E=p(r)##
The function p(r) is asked, not the pressure at r_E.
dp/dr =constant/r2. What is the antiderivative of the function constant/r2?
Or: add integral symbols to both sides. Integrate the left side from r to infinity, the right side from p to zero:
##\int_r^∞\ {G*\frac{M*(\rho)}{r^2}dr}=\int_p^0 dp ##
Note that there is a sign error.
 
Last edited:
  • #11
I see

##-G*\frac{M*(\rho)}{r}+C(maybe?)=p(r)##
 
  • #12
RiotRick said:
I see

##-G*\frac{M*(\rho)}{r}+C(maybe?)=p(r)##
This is OK, but the minus sign. See the original equation, with the forces. The water column above the layer pushes it downward, just like gravity, that pulls it downward. The hydrostatic pressure below the layer pushes it upward. Choose C so as p(∞)=0.
 
  • #13
##G*\frac{M*(\rho)}{r}+C=p(r)##
##p(r=\infty)=0 =C##
how do I make G and M disappear now?
 
  • #14
RiotRick said:
##G*\frac{M*(\rho)}{r}+C=p(r)##
##p(r=\infty)=0 =C##
how do I make G and M disappear now?
You know that the gravitational acceleration is g at the surface of the Earth, when r=RE. What is the gravitational acceleration at RE, expressed with G, M and RE?
 
  • Like
Likes RiotRick
  • #15
##G=g*r_E^2/M##
## p(r)=\frac{\rho*g*r_E^2}{r} ##

Thank you very much
 

Related to Impossible hydrostatic scenario

1. What is an impossible hydrostatic scenario?

An impossible hydrostatic scenario is a situation where the principles of hydrostatics, which govern the behavior of fluids at rest, are violated. This means that the pressure, density, and volume of the fluid do not follow the expected relationship, leading to an impossible or physically unrealistic result.

2. What are some examples of impossible hydrostatic scenarios?

Some examples of impossible hydrostatic scenarios include a liquid flowing uphill, a fluid with a negative pressure, or a fluid with a higher pressure at the top than at the bottom of a container. These situations go against the fundamental laws of hydrostatics and are physically impossible.

3. How do impossible hydrostatic scenarios occur?

Impossible hydrostatic scenarios can occur due to a variety of factors, such as errors in measurement or calculation, incorrect assumptions about the fluid properties, or the presence of external forces that disrupt the equilibrium of the system. They can also occur in hypothetical or theoretical scenarios that go against the laws of physics.

4. What are the implications of impossible hydrostatic scenarios?

Impossible hydrostatic scenarios have significant implications in the fields of engineering, physics, and fluid mechanics. They can lead to incorrect design and construction of structures, inaccurate predictions in fluid behavior, and a lack of understanding of the fundamental principles of hydrostatics.

5. How can impossible hydrostatic scenarios be avoided?

To avoid impossible hydrostatic scenarios, it is crucial to have a thorough understanding of the principles of hydrostatics and to use accurate and precise measurements and calculations. It is also essential to consider all external forces and factors that may affect the fluid system. In cases where impossible hydrostatic scenarios are encountered, it is important to re-evaluate the assumptions and calculations to identify and correct any errors.

Similar threads

  • Introductory Physics Homework Help
2
Replies
60
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
882
  • Introductory Physics Homework Help
Replies
5
Views
9K
  • Introductory Physics Homework Help
Replies
19
Views
849
  • Introductory Physics Homework Help
Replies
19
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
986
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top