Accretion Energy

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/r² while the element is still at distance r. The volume of the pile is 4/3πr₀³ = m/ρ where r₀ is the pile’s radius.
It will reach the surface at radius r₀ = (4/3πρ/m)^–1/3 having released energy = de =
∫_r0^∞f dr = G m dm/r₀ = (4/3πρ/m)^1/3 G m dm = G dm (4/3πρ)^1/3 m^2/3.
As the mass increases the total energy is
∫₀^Mde = ∫₀^MG dm (4/3πρ)^1/3 m^2/3 = G (4/3πρ)^1/3∫₀^M m^2/3 dm
= G (4/3πρ)^1/3 (3/5)M^5/3 = 2^2/33^2/35^–1 (πρ)^1/3G M^5/3 = 6^2/3/5 (πρ)^1/3G 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)r³ρ
ρ = M(3/(4π))r^–3
E = 6^2/3/5 (πρ)^1/3G M^5/3
= 6^2/3/5 (πM(3/(4π))r^–3)^1/3G M^5/3
= 2^2/3–2/33^2/3+1/3/5π⁰M^6/3G r^–1
= 3/5M²G/r
= 1.6667 M²G/r
= 1.1124 10^–10 M²/r m³ kg^–1 s^–2

Accretion Energy of Earth:
Mass = 5.98 × 10²⁴ kg.
Radius of Earth = 6.37× 10⁶ m.
E = 6.2449 × 10³² Joules