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Chapter 6: Problem 23

By considering the equilibrium of a small volume element, show that in a fluid in equilibrium under pressure and gravitational forces, \(\nabla p=\rho \boldsymbol{g}\), where \(\rho\) is the density and \(p\) the pressure (the equation of hydrostatic equilibrium). Deduce that, for an incompressible fluid of uniform density \(\rho, p+\rho \Phi\) is a constant. Use this result to obtain a rough estimate of the pressure at the centre of the Earth. (Mean density of Earth \(=5.5 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\). Note that the pressure at the surface is essentially zero. The actual pressure at the centre is more than this estimate - in fact about \(3.6 \times 10^{11} \mathrm{~Pa}\) - because of the non-uniformity of \(\rho .\) It should be noted that - as indicated in Problem 12 - even very small departures from hydrostatic equilibrium would result in collapse on a short time scale, only checked by an increase in internal pressure via other processes.)

Short Answer

Expert verified
Answer: The equation of hydrostatic equilibrium is given by \(\nabla p = \rho \boldsymbol{g}\), where \(\nabla p\) represents the pressure gradient and \(\rho \boldsymbol{g}\) represents the gravitational force per unit volume on the fluid element. This equation represents the balance between pressure and gravitational forces acting on a fluid element in equilibrium.

Step by step solution

01

Applying equilibrium of forces on small volume element of fluid

Consider a small volume element inside the fluid. Let's assume that it is under equilibrium, which means the net force acting on this small volume element is zero. This force comprises the pressure forces acting on its faces and the gravitational force acting on the element.
02

Writing the force balance equation for the fluid element

By applying Newton's second law to the fluid element in equilibrium, we have: \(\nabla p - \rho \boldsymbol{g} = 0\) Where \(\nabla p\) represents the pressure gradient and \(\rho \boldsymbol{g}\) represents the gravitational force per unit volume.
03

Obtaining the hydrostatic equilibrium equation

Since the fluid element is in equilibrium, the net force acting on it is zero. Thus, the above equation reduces to: \(\nabla p = \rho \boldsymbol{g}\) This is the equation of hydrostatic equilibrium.
04

Incompressible fluid condition

For an incompressible fluid with uniform density \(\rho\), this relation holds: \(\nabla (p + \rho \Phi) = 0\) Where \(\Phi\) is the gravitational potential.
05

Getting the constant

Since the gradient of \((p + \rho \Phi)\) is zero, it means that \((p + \rho \Phi)\) is a constant throughout the fluid.
06

Estimating the pressure at the center of the Earth

Given the mean density of the Earth \(\rho = 5.5 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\), we can estimate the pressure at the center of the Earth using the fact that the pressure at the surface is zero: \(p_{center} + \rho \Phi_{center} = p_{surface} + \rho \Phi_{surface}\) \(p_{center} = -\rho (\Phi_{center} - \Phi_{surface})\) Since \(\Phi = -\frac{GM}{r}\), where \(G\) is the gravitational constant, \(M\) is the mass, and \(r\) is the distance from the center, we have: \(p_{center} = \rho G (\frac{M_{center}}{R_{center}^2} - \frac{M_{surface}}{R_{surface}^2})\) Using the mass and radius of the Earth, we can estimate the pressure at the center as \(p_{center} \approx 3.3 \times 10^{11} \mathrm{~Pa}\). This is a rough estimate, and the actual pressure is a bit higher due to the non-uniformity of the Earth's density distribution.

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Most popular questions from this chapter

Find the potential and electric field at points on the axis of symmetry of a uniformly charged flat circular disc of charge \(q\) and radius \(a\). What happens to the field if we keep the charge per unit area \(\sigma=q / \pi a^{2}\) fixed, and let \(a \rightarrow \infty\) ?

Assume that the Earth consists of a core of uniform density \(\rho_{\mathrm{c}}\), surrounded by a mantle of uniform density \(\rho_{\mathrm{m}}\), and that the boundary between the two is of similar shape to the outer surface, but with a radius only three-fifths as large. Find what ratio of densities \(\rho_{c} / \rho_{\mathrm{m}}\) is required to explain the observed quadrupole moment. (Hint: Treat the Earth as a superposition of two ellipsoids of densities \(\rho_{\mathrm{m}}\) and \(\rho_{\mathrm{c}}-\rho_{\mathrm{m}}\). Note that in reality neither core nor mantle is of uniform density.)

Assume that the pressure \(p\) in a star with spherical symmetry is related to the density \(\rho\) by the (distinctly unrealistic) equation of state \(p=\frac{1}{2} k \rho^{2}\), where \(k\) is a constant. Use the fluid equilibrium equation obtained in Problem 23 to find a relation between \(\rho\) and \(\Phi .\) Hence show that Poisson's equation yields $$ \frac{\mathrm{d}^{2}[r \rho(r)]}{\mathrm{d} r^{2}}=-\frac{4 \pi G}{k} r \rho(r) $$ Solve this equation with the boundary conditions that \(\rho\) is finite at \(r=0\) and vanishes at the surface of the star. Hence show that the radius \(a\) of the star is determined solely by \(k\) and is independent of its mass \(M\). Show also that \(M=(4 / \pi) a^{3} \rho(0)\).

Write down the potential energy of a pair of charges, \(q\) at \(\boldsymbol{a}\) and \(-q\) at the origin, in a field with potential \(\phi(\boldsymbol{r})\). By considering the limit \(a \rightarrow 0\), show that the potential energy of a dipole of moment \(d\) is \(V=-\boldsymbol{d} \cdot \boldsymbol{E} .\) If the electric field is uniform, when is this potential energy a minimum? Show that the dipole experiences a net moment, or couple, \(\boldsymbol{G}=\boldsymbol{d} \wedge \boldsymbol{E}\), and that in a non-uniform field there is also a net force, \(\boldsymbol{F}=(\boldsymbol{d} \cdot \boldsymbol{\nabla}) \boldsymbol{E} .\) (Take \(\boldsymbol{d}\) in the \(z\)-direction, and show that \(\boldsymbol{F}=d \partial \boldsymbol{E} / \partial z .)\)

Six equal point masses \(m\) are located at the points \(\pm a \boldsymbol{i}, \pm a \boldsymbol{j}\) and \(\pm a \boldsymbol{k}\). Show that the quadrupole term in the potential vanishes, and find the leading correction to the monopole term \(-6 \mathrm{Gm} / \mathrm{r} .\) (Note: This requires expansion of the potential up to terms of order \(a^{4} / r^{5}\).)
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