### Accretion Energy

I compute here the gravitational energy released in the assembly of material from infinity to form a spherical ruble pile.
This is also the energy required to disperse it again.
Let ρ be the pile’s constant density.
Let m be the mass of the pile which will increase as the pile accretes.
When a new mass element, dm, falls in from infinity the attraction is f = G m dm/r^{2} while the element is still at distance r.
The volume of the pile is (4/3)πr_{0}^{3} = m/ρ where r_{0} is the pile’s radius.

It will reach the surface at radius r_{0}
= ((4/3)πρ/m)^{−1/3}
having released energy = de =

∫_{r0}^{∞}f dr
= ∫_{r0}^{∞}G m dm r^{−2} dr
= G m dm/r_{0}
= ((4/3)πρ/m)^{1/3} G m dm
= G dm ((4/3)πρ)^{1/3} m^{2/3}.

As the mass increases the total energy is

∫_{0}^{M}de
= ∫_{0}^{M}G dm ((4/3)πρ)^{1/3} m^{2/3}
= G ((4/3)πρ)^{1/3}∫_{0}^{M} m^{2/3} dm

= G ((4/3)πρ)^{1/3} (3/5)M^{5/3}
= 2^{2/3}3^{2/3}5^{−1} (πρ)^{1/3}G M^{5/3}
= 6^{2/3}/5 (πρ)^{1/3}G M^{5/3}

where M is the total mass of the pile.

This agrees almost with this.

Expressing the answer in M and the radius r using M = (4π/3)r^{3}ρ

ρ = M(3/(4π))r^{−3}

E = 6^{2/3}/5 (πρ)^{1/3}G M^{5/3}

= 6^{2/3}/5 (πM(3/(4π))r^{−3})^{1/3}G M^{5/3}

= 2^{2/3−2/3}3^{2/3+1/3}/5π^{0}M^{6/3}G r^{−1}

= 3/5M^{2}G/r

= 0.6 M^{2}G/r

= (4.0044 10^{−11} M^{2}/r) m^{3} kg^{−1} s^{−2}

Holding the density constant, the energy varies as the 5th power of the radius.

Accretion Energy of Earth:

Mass = 5.98 × 10^{24} kg.

Radius of Earth = 6.37 × 10^{6} m.

G = 6.6734∙10^{−10} m^{3}kg^{−1}sec^{−2}

E = 2.248 × 10^{32} Joules

P.S., Today, 2007 July 24, I fixed a bug that crept in as I expressed E in terms of M and r.
These calculations are trivial yet very error prone for me.
My old equation had the wrong units.
I need a calculator that does units!!
(So does NASA!)

(If you plan to disassemble the Earth you should first check my calculations.)
And on 2008 Nov 10 I found and fixed another bug.
Very embarrassing!

A neutron star of radius 5 km and density 10^{15} g cm^{−3} would have a mass of (4/3) π (5 km)^{3} 10^{15} g cm^{−3} (10^{5} cm/km)^{3}

= (500/3) π 10^{30} g

= 5.24 10^{32} g.
By the above formula the accretion energy is
(4.004 10^{−11} 27.5 10^{64} g^{2} /
(5 10^{3} m)) m^{3} kg^{−1} s^{−2}
(10^{3} g/kg)^{−2}

= 22 10^{51} kg (m/s)^{2}
= 2.2 10^{52} Joules.

This does not include the substantial energy that is required to overcome degeneracy pressure.