Pressure Halfway through a planet

[dlc_iii]

Homework Statement

"In a galaxy far far away, a planet composed of an incompressible liquid of uniform mass density ρ has mass m_planet and radius R. Determine the pressure midway between the surface and the center of the planet."
I used M=mass of planet, m=mass of shell, R=radius of planet, r=radius of shell, V=volume of planet, dr=thickness of shell, ρ=density of planet

Homework Equations

P=ρgh
g=Gm/r^2
Area of sphere = 4/3πr^3

The Attempt at a Solution

I solved this problem by adding up all the pressures of thin shells from R/2 to R.
P=ρgh=∫(3GM²r)/(4πR⁶)dr= (3GM²)/(4πR⁶)∫rdr from R/2 to R
final answer= (9GM²)/(32πR⁴)
This is the answer that I got and I found a few other places for a normal planet, but my textbook says it should be 45/64 instead of 9/32 in front of the (GM²)/(πR⁴). Is it different for fluids or am I or my textbook wrong?

 

[haruspex]

P=ρgh=∫(3GM2r)/(4πR6)dr

Please explain in detail how you arrive at that integral.

 

[dlc_iii]

Please explain in detail how you arrive at that integral.

How I arrived at this integral:
We needed dP, the pressure under an infinitesimally thin shell of the liquid at any point, r, away from the center of the planet. dP=ρgh so we need to have ρ, g, and h in terms of r and various constants.
h=dr, since the height of the shell, in this case would be the thickness and therefore dr
ρ=M/((4/3)πR³)
g(r)=Gm/r²), we need it in terms of M, r, and R since m is variable so through proportions we get m=M(r/R)³ and plug that into the equations for g, getting GMr/R³
Plugging this all into dP and simplifying we get dP=(GMr/R³)(M/(4/3)πR³)(dr)= (3GM²r)/(4πR⁶)dr
and P=∫dP

 

[haruspex]

How I arrived at this integral:
We needed dP, the pressure under an infinitesimally thin shell of the liquid at any point, r, away from the center of the planet. dP=ρgh so we need to have ρ, g, and h in terms of r and various constants.
h=dr, since the height of the shell, in this case would be the thickness and therefore dr
ρ=M/((4/3)πR³)
g(r)=Gm/r²), we need it in terms of M, r, and R since m is variable so through proportions we get m=M(r/R)³ and plug that into the equations for g, getting GMr/R³
Plugging this all into dP and simplifying we get dP=(GMr/R³)(M/(4/3)πR³)(dr)= (3GM²r)/(4πR⁶)dr
and P=∫dP

Yes, that all looks right to me.

 

[siyujiang81]

I agree fully with your answer. Just to show that it is correct, I will solve the same problem by using the definition of pressure as the ratio of the perpendicular force applied to a surface to the area of the surface.

Consider a thin cylinder sliced from the surface ($r = R$) to an arbitrary location within the planet ($r = r_0$). We will apply $r_0 = (1 / 2) R$ at the end of the problem. Let the radius of the cylinder be $x << R - r_0$. Because of this restriction, variation of the gravitational acceleration $g$ is negligible across individual surfaces of circular cross sections taken from the cylinder, and the spherical curvature of the planet at $r = R$ can be safely ignored. The area of the surface at $r = r_0$ is the same for all cross sections of the cylinder; specifically, it is $\pi x^2$.

The perpendicular force at $r = r_0$ involves a slightly more complicated calculation. Consider the force exerted by an infinitesimally thin disk on the bottom of the cylinder. This is exactly equal to the weight of the disk, which is given by $g \cdot dm$. The differential mass is equal to the density times the differential volume, which in turn is equal to the density times the cross sectional area times a differential $dr$, or $\rho \pi x^2 dr$. Further, $g$ is not constant; it is a function of $r$ given specifically by Gauss' Law applied to gravitation: $g(r) = \frac{\mu r}{R^3}$, where $\mu$ is the gravitational parameter $G M$.

Now we integrate. The total downward force experienced at the bottom of the cylinder is equal to the contributions from each of the disks:
$$F = \int _{r_0} ^{R} \rho \pi x^2 \frac{\mu r}{R^3} dr$$
$$= \frac{\rho \pi x^2 \mu}{R^3} \int _{r_0} ^{R} r dr$$
$$= \frac{\rho \pi x^2 \mu}{R^3} \left[ \frac{1}{2} r^2 \right] _{r_0} ^{R}$$
$$= \frac{\rho \pi x^2 \mu}{2 R^3} \left[ R^2 - r_0^2 \right]$$.

It immediately follows that the desired pressure is
$$P = \frac{F}{A}$$
$$= \frac{\rho \mu}{2 R^3} \left[ R^2 - r_0^2 \right]$$.

If we wanted this exclusively in terms of the parameters $M$, $R$, and $G$, we can simply substitute:
$$P = \frac{\frac{M}{\frac{4}{3} \pi R^3} G M}{2 R^3} \left[ R^2 - r_0^2 \right]$$
$$= \frac{3 G M^2}{8 R^6} \left[ R^2 - r_0^2 \right]$$.

To complete the problem, we evaluate using $r_0 = \frac{1}{2} R$.
$$P = \frac{3 G M^2}{8 R^6} \left( R^2 - \frac{1}{4} R^2 \right)$$
$$= \frac{3 G M^2}{8 R^6} \left( \frac{3}{4} R^2 \right)$$
$$= \frac{9 G M^2}{32 R^4}$$.

This is exactly the same result obtained by adding partial pressures.