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

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

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 3564 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.
If 50 million believe in a fallacy, it is still a fallacy. Sam W Carey
Florian