Gravitational Pressure

Since the Jupiter is modeled as a sphere with constant density $\rho$ , we can think that it consists of many concentric shells. Outside the sphere, we assume that there is no pressure, because Jupiter was in vacuum. The gravitation pressure is built up shell by shell from the outermost one to the center.

The shell theorem says that if the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell. This means that if a point mass is placed inside the sphere at a distance $r$ from the center, where $r < R$ and $R$ is the radius of the sphere, the net gravitational force exerted on the point mass can be attributed only to the sphere of radius $r$ and all the spherical shells of radius greater than $r$ will not influence the gravitational force on the point mass.

Hence, the gravitational field at $r$ from the center is $g=-\frac{G\rho (\frac{4}{3}\pi r^3)}{r^2}=-\frac{4}{3}G\rho\pi r$ , where $\rho$ is the constant density of Jupiter.

Force exerted on each thin shell is: $F= \rho (4\pi r^2 \Delta r)g$

The additional pressure exerted on each thin shell is: $\Delta p=\frac{F}{A}=\frac{\rho (4\pi r^2 \Delta r)g}{4\pi r^2}=\rho g \Delta r=-\frac{4}{3}G\pi \rho^2 r\Delta r$

Take the limit and integrate both sides, we get: $\int_{p(R)}^{p(0)} \mathrm {d}p=\int_R^0 -\frac{4}{3}G\pi\rho^2 r\mathrm{d} r$

$\implies p(0)-p(R)=\frac{2}{3}\pi G \rho^2 R^2$

Since we assume $p(R)=0$ , $p(0)=\frac{2}{3}\pi G\rho^2 R^2$ , where $\rho=\frac{M}{\frac{4}{3}\pi R^3}=1322.4 kgm^{-3}$

The final answer is $p(0)=1.1970 \times 10^{12} Pa$

Let's consider our problem on a general basis for some planet with mass $M$ and radius $R$ . Let the planet be made of infinite layers with thickness $\mathrm d r$ . At the top of the layer the weight per unit area which is compressing the material be $p_r$ at a distance $r$ from the centre. Let the difference in weight at the bottom and top of the layer be $\mathrm d w$ . If $\rho$ is density of the material of planet (assumed to be uniform),area of layer one unit, and change in pressure between top and bottom layer be $\mathrm d p$ , we have

$\mathrm d w= \mathrm d p= \rho \cdot g_r \cdot \mathrm d r$

where $g_r$ is planet's internal gravity. On integrating the above equations from $r$ $\text{to} R$ we have

$\int \mathrm d w = \int \limits_r^R \rho \cdot g_r \cdot \mathrm d r$

But $\rho= \frac{M}{\frac{4}{3} \pi \cdot R^3} = \frac{3M}{4 \pi \cdot R^3}$

$g_r=g_R \cdot \frac{r}{R} = \frac{GMr}{R^3}$ .

So our integral reduces to, after some rearrangements,

$p_r = \frac{3G}{4 \pi} \cdot (\frac{M}{R^3})^2 \int \limits_r^R r \mathrm d r$

After some integration and applying limits, regrouping like terms, we have

$p_r=\frac{3G}{8 \pi} \cdot (\frac{M}{R^2})^2 \cdot (1-\frac{r^2}{R^2})$ which is our general formula.

At the centre $r=0$ . And plugging in the given values, we have gravitational compression pressure at centre of Jupiter to be $1.2E+12$

(Don't forget to convert radius from kilometres to metres)

Gravitational field at a distance 'r' from center E= (4/3) pi G D r. D=density of planet. So force on a differential shell of mass dF = E 4 pi (r^2) D dr. Pressure at a distance r from center dP = F/A = dF/4 pi*r^2. Integrating we get pressure at center = (3G(M^2))/(8pi(R^4))

We have that the gravitational field inside a planet is given by

g(r) = GMr/R^3

r is the distance from the center, M is the mass of the planet, R is the radius of the planet and G is the gravitational constant. This formula can be proven using the Gauss law.

Note that a infinitesimal layer of mass at a distance r from the center increases the pressure at the surface right below it by the value of

dP = P(r) - P(r + dr) = -dm.g(r)/A

Since g(r) depends only on r, we can assure that we can apply this for this entire spherical layer

Note that

dm = 4πρ(r^2)dr Where ρ is the planet density

A = 4πr^2

dm/A = ρ.dr

dP = - GMρr.dr/R^3

P(R) - P(0) = - GMρ/2R

Assuming P(R) = 0

P(0) = GMρ/2R

But

ρ = 3M(4πR^3)

Then

P(0) = 3G(M^2)/(8π(R^4))

P(0) = 1.197E+12

The gravitational field $\vec {g }\,$ is related locally to the mass density $\rho\,$ by the Gauss' law:

$\nabla\cdot\vec {g } =-4\pi G\rho\,$ and for a closed surface $\Sigma\,$ : $\int_{\Sigma}\vec {g }\cdot{d\vec s}=-4\pi GM_{\Sigma}\,$ where $d\vec s\,$ is the vectorial element of surface perpendicular to the surface and outward-oriented, $M_{\Sigma}\,$ is the mass internal to the surface;

if we take a system of spherical coordinates with center $O\,$ in the center of Jupiter, and spherical surfaces $\Sigma\,$ with center in $O\,$ if $\Sigma\,$ contains the entire planet $r>R\,$ :

$\vec g = -\,G\frac M {r^2} \hat {e_r}$

if the surface is internal to the planet, so that $r\le R\,:$

$M(r)={\frac 4 3}\pi r^3;\,g(r)4\pi r^2=4\pi GM(r);\,g(r)={\frac 4 3}\pi G\rho r$

the pressure $p(r)\,$ is related to $r\,$ by Stevin's law:

$dp=-\rho g(r)dr,\,$ substituting $g(r)\,\Rightarrow\,dp=-{\frac 4 3}\pi G{\rho}^2rdr\,$ integrating between $R\,$ and $r\,$ and taking in account that $p(R)=0\,$ :

$p(r)={\frac 2 3}\pi G {\rho}^2(R^2-r^2)\,\Rightarrow\,p(0)={\frac 2 3}\pi G {\rho}^2R^2$

if $V=\frac 4 3 \pi R^3\,$ is the volume of the planet $\rho = \frac M V = \frac 3 {4\pi}\frac M {R^3}\,$ hence:

$p(r)=\frac {3G}{8\pi}{(\frac {M^2}{R^2})}^2(1-\frac {r^2}{R^2})\,\Rightarrow$

$p(0)=\frac {3G}{8\pi}{(\frac {M^2}{R^2})}^2\approx {1.197\times 10^{12}{Pa}}$

Let us consider a mass element $\Delta m$ at a distance $r$ from the center of the planet. The gravitational force acting on this element equals $\Delta F(r)= \frac{G {\Delta m} M(r)}{r^{2}}$ where $M(r)=M_{J}\frac{ r^{3}}{R_{J}^{3}}$ is the the enclosed mass (this is the mass contained in the sphere of radius $r$ ). It attracts the element $\Delta {m}$ towards the center of the planet. The mass elements outside this sphere do not contribute; their contributions cancel out. This follows from Gauss Law for gravitation. In order to compute the pressure at a distance $r$ we need to find the "weight" of a column from $r$ to $R_{J}$ and base area $\Delta {A}$ . That is, $p(r)= \frac{1}{\Delta{A}} \sum{\Delta{F}}= \frac{G M_{J}\rho_{J}}{R_{J}^{3}} \int_{r}^{R_{J}} r' dr'= \frac{3 G M_{J}^{2}}{8 \pi R_{J}^{6}} (R_{J}^{2}-r^{2}).$ Thus, the pressure at the center of the planet is $p(0)= \frac{3 G M_{J}^{2}}{8 \pi R_{J}^{4}}=1.2 \times 10^{12}~\mbox{Pa}.$ Note that even though Jupiter mass is, roughly, 300 times that of the Earth, the pressure at the center of our planet is $p_{Earth}(0)= 1.72\times 10^{11}~\mbox{Pa}$ . That is to say, Jupiter's gravitational pressure is only 7 times that of the Earth. Why?