Impossible hydrostatic scenario

[RiotRick]

Homework Statement

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.

 

[ehild]

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.

 

[RiotRick]

$-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)$
$-G*\frac{M*(\rho*A)}{r^2}=\frac{A*p(r)-A*p(r+dr)}{dr}$
The right side would be a perfect differential equation. Does this work?

 

[ehild]

$-G*\frac{M*(\rho*A*dr)}{r^2}=A*p(r)-A*p(r+dr)$
$-G*\frac{M*(\rho*A)}{r^2}=\frac{A*p(r)-A*p(r+dr)}{dr}$
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)?

 

[RiotRick]

I thought p(r) is a function which gives me the pressure with respect to the distance r

 

[ehild]

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?

 

[RiotRick]

Yes, and what is its form? How does it depend on r?

$p(r)=\rho*g(r-r_E)$

 

[ehild]

$p(r)=\rho*g(r-r_E)$

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=?

 

[RiotRick]

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
$-G*\frac{M*(\rho)}{r^2}*r_E=p(r)$

 

[ehild]

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.

 

[RiotRick]

I see

$-G*\frac{M*(\rho)}{r}+C(maybe?)=p(r)$

 

[ehild]

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.

 

[RiotRick]

$G*\frac{M*(\rho)}{r}+C=p(r)$
$p(r=\infty)=0 =C$
how do I make G and M disappear now?

 

[ehild]

$G*\frac{M*(\rho)}{r}+C=p(r)$
$p(r=\infty)=0 =C$
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=R_E. What is the gravitational acceleration at R_E, expressed with G, M and R_E?

 

[RiotRick]

$G=g*r_E^2/M$
$p(r)=\frac{\rho*g*r_E^2}{r}$

Thank you very much