## Related questions with answers

Certain cold stars (called white dwarfs) are stabilized against gravitational collapse by the degeneracy pressure of their electrons. Assuming constant density, the radius R of such an object can be calculated as follows: (a) Write the total electron energy in terms of the radius, the number of nucleons (protons and newtons) N, the number of electrons per nucleon q, and the mass of the electron m. (b) Look up, or calculate, the gravitational energy of a uniformly dense sphere. Express your answer in terms of G (the constant of universal gravitation), R, N, and M (the mass of a nucleon). Note that the gravitational energy is negative. (c) Find the radius for which the total energy, (a) plus (b), is a minimum. (d) Determine the radius, in kilometers, of a white dwarf with the mass of the sun. (e) Determine the Fermi energy, in electron volts, for the white dwarf in (d), and compare it With the rest energy of an electron.(Note that this system is getting dangerously relativistic.

Solution

Verifieda) From equation 5.45 we have (with $V=\dfrac{4\pi R^3}{3}$):

$\begin{align*} E &= \frac{\hbar^2 (3\pi^2Nq)^{5/3}}{10\pi^2m}\left( \frac{3}{4\pi R^3} \right)^{2/3}\\ &=\frac{A}{R^2}. \end{align*}$

b) Gravitational energy needed to build layer of material mass $dm$:

$\begin{align*} E_{\text{grav}} &= - \int \frac{Gm}{r}dm\\ m=\rho \cdot V,& \quad V = \frac{4\pi r^3}{3} \\ m&= \frac{4 \pi \rho}{3} r^3 \\ dm &= 4\pi \rho r^2 dr\\ \Rightarrow E_{\text{grav}}&= - \int \frac{G}{r} \frac{4 \pi \rho}{3}r^3 \cdot 4 \pi \rho r^2 dr\\ &= -\frac{16\pi^2G\rho^2}{3} \int_0^R r^4 dr\\ &=-\frac{16\pi^2G\rho^2R^5}{15}\\ \rho &= \frac{NM}{V} = \frac{3NM}{4\pi R^3}\\ \Rightarrow E_{\text{grav}}&= - \frac{3G}{5}\frac{N^2M^2}{R}=-\frac{B}{R}. \end{align*}$

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