Planet Formation

A planet forms by gravitational condensation of an initially cold, spherical dust cloud.  Find an expression for the radius of the planet if the resulting temperature rise causes the planetary material to reach its melting point.  Estimate this radius using reasonable values of the quantities involved and comment on the result.  Neglect radiation energy losses.

Solution:

Assumptions and symbols: The 3 K background of space implies that space is very cold so for simplicity we take “cold” to mean essentially 0 K in this context.  Let the spherical dust cloud have an initial radius R and let TM be the melting temperature of the dust material.  We assume spherical symmetry is retained throughout planet formation and we ignore rotational energy.  Angular momentum of the cloud would be conserved but the same is not true for rotational energy.  We also assume the density of the dust cloud is always uniform.

The Physics:  As the cloud falls in on itself, gravitational potential energy is turned into kinetic energy which in turn creates heat once the material has coalesced.  So, we equate potential energy loss to heat gained. 

     The form (equation) for potential energy is tricky.  Imagine the spherical dust cloud is composed of many thin, concentric shells.  The gravitational force at each shell is then due only to the sphere of dust inside it.  For a shell of radius r, the mass inside it is

So the gravitational force acting on it is

Thus, the force is linear in distance from the center, a Hooke’s Law force and the associated potential (kx2/2) is

This formula applies to successive spherical shells of dust.

The heat energy goes into a temperature change

M is the total mass of the dust cloud.

The Math:

The initial potential energy is therefore the sum (integral) over all the shells making up the spherical dust cloud.   The mass of a shell of thickness dr and radius r is

So the total initial potential energy is

Note that r here is the initial density of the dust cloud.  We set this equal to the heat and get the result

More Physics:

Since the initial temperature is 0 K, DT = TM.  The dust reaches the melting point but does not actually melt.  The radius of the cloud is then

To get some numbers, r of a dust cloud is about 10-18 kg/m3 and cTM of quartz is about 1.5 x 106 J/kg.  This gives a value of R = 1.2 x 1017 m.  This is a hundred million times greater than the radius of the Solar system.  It seems that planets will not melt as they form.  Such a cloud would contain 7 x 1033 kg, several thousand times the mass of the Sun.

Eddy Current Brakes

Problem:

One can imagine using a very strong magnetic field to stop a fighter jet as it lands on an aircraft carrier.  The jet would land and then be stopped by eddy current damping as it passes between the poles of a huge magnet.

a) If the plane is traveling 55 m/s as it enters the magnetic field and it is stopped in 30 m, what g force will the plane and pilot experience?

b) What would the kinetic energy loss be if the jet mass is 7500 kg?

c) If the eddy currents are active in 1% of the mass (the skin is most strongly affected and "screens" the interior), what temperature rise should that part of the jet experience (assume a specific heat of 0.4 J/g)?

d) Can you estimate the requisite magnetic field?

Solution:

• Δv2 = 2as so 552 = 2 x 3 0 x acceleration and acceleration = 50.42 m/s2 or 5.14 g’s.
• ½mΔv2 = 3750 x 552 = 11,343,750 Joules.  That’s a lot!
• mcΔT = 11,343,750 J  leads to ΔT = 378oC.  1% is actually a high estimate so this greatly underestimates actual temperature rises.
• The time required to stop the plane is 55/50.42 = 1.1 s so the power loss is 11,343,750 J/1.1 s ≈ 107 watts.  Because the voltage and current decline approximately linearly, P ≈ ½V2/R  ≈  (Blv)2/2R.  In most metals R/l ≈ 1 Ωm so P ≈ B2v2 implies B ≈ 58 Tesla, extremely high even for superconducting magnets.

Atomic Stability on a Neutron Star

Problem:

Surface magnetic fields of a neutron star easily range as high as 108 T.

a) What is the maximum radius circle an electron can orbit in such a field if it has a speed comparable to that of an electron in an atomic orbit (2 x 106 m/s)?  How does this compare with the 10-10 m radius of an atomic orbit?

b) What is the ratio of this magnetic force to the electric force holding an electron in an atomic orbit?

c) What is the difference in energy between electrons with magnetic moment aligned and anti-aligned with the magnetic field (the magnetic moment of an electron is the Bohr magneton, 9.27 x 10-24 J/T)? How does it compare with the energy of an electron in orbit in an atom (the total energy of an orbiting electron is half the kinetic energy because the potential energy is negative)?

d) If the neutron star has a radius of 5 km and mass equal to that of the sun, what is the gravitational potential energy of an electron at the surface of the star?

e) What do you think the survival prospects are for an atom at the surface of a neutron star?  Why is it called a neutron star?

Solution:

• r = mv/qB = 9.11 x 10-31 x 2 x 106/1.6 x 10-19 x 108 = 1.139 x 10-13 m.   This is much smaller than 10-10 m and brings the electron perilously close to the nucleus and to K capture by the nucleus.
• Fmag = qvB = 1.6 x 10-19 x 2 x 106 x 108 = 3.2 x 10-5 N while Felec =  e2 x 9 x109/10-20  =  2.3 x 10-8 N and the ratio is 1390 and the magnetic force is much greater.
• The energy difference is 2μBH = 2 x 9.27 x 10-24 x 108 = 1.854 x 10-15 J.  This compares with energies of around 13.6 eV = 2.2 x 10-18 J, about 1000 times more energy.
• PEg = GmM/R = 6.67 x 10-11 x 9.11 x 10-31 x 2 x 1030/5000 = 2.43 x 10-13 J.  This is hundreds and thousands of times the energies above so gravity will dominate and overcome electric and magnetic forces.
• The strong gravitation should crush the electron into the nucleus, creating neutrons.  Hence the name “neutron star.”