What is the rate of mass increase for the last 5 M?

[Florian]

To estimate the increase in mass ∆m over the last 5 Ma, we can assume that the density of Earth (ρ=5.5 g/cm³) did not change significantly during that time period. It follows that ∆m=ρ*∆V. The increase in volume ∆V can be calculated from the increase in surface ∆S as follow:

S₀, V₀ and R₀ are the current surface, volume and radius of Earth.
S₅, V₅ and R₅ are the surface, volume and radius of Earth 5 Ma ago.

∆S=S₀-S₅
S₀=4π*R₀²
S₅=4π*R₅²=S₀-∆S
S₅/S₀=R₅²/R₀²
R₅=R₀(S₅/S₀)^1/2

V₀=4π/3*R₀³
V₅=4π/3*R₅³
∆V=V₀-V₅=4π/3*(R₀³-R₅³)=4π/3*(R₀³-R₀³(S₅/S₀)^3/2)=4π/3*R₀³(1-(S₅/S₀)^3/2)=4π/3*R₀³(1-((S₀-∆S)/S₀)^3/2)


Acknowledging that active margins and orogens are the place of material outflows, the surface increase of Earth corresponds approximatively to the surface of new seafloor accreted at Mid Ocean Ridges.

Isochron data freely available from Earthbyte.org allow to calculate the additional surface of seafloor for 5-Ma time slices:

surface-5My-period.jpg (41.52 KiB) Viewed 10103 times

The increase in surface ∆S is about 14 Millions km² for the last 5 Ma.
Knowing that the current surface and radius of Earth are S₀=510 million km² and R₀=6371 km, the increase in volume ∆V is:
∆V=4π/3*6371³(1-((510-14)/510)^3/2)=4.43 10¹⁰ km³=4.43 10²² dm³

Finally, ∆m=ρ*∆V=5.5*4.43 10²²=2.44 10²³ kg for 5 Ma.
It follows that 4% of the mass of Earth (5.97 10²⁴ kg) was gained during the last 5 Ma.

Note that 2.44 10²³ kg for 5 Ma is an average rate over geological time. Extrapolation to an instantaneous rate at our timescale (e.g. kg/year) assumes a steady input of mass at our timescale, which is not established.