Mean radius of the Earth = 6371 km
Using this to calculate the Earth's surface area yields A = 4*pi*r^2
= 5.10*10^14 m^2
The annual carbon emissions in 2002 due to fossil fuel burning = 6.73
GT Carbon/year (1 GT = 10^15 grams). If cement production is
included, then the annual anthropogenic emissions = 6.975 GT C/year.
Neither of these figures include carbon emissions (or removals) due to
land use changes (e.g., forest clearing). Data from
Density of carbon (graphite) = 2.267 gm/cm^3
If all the carbon emitted by fossil fuel burning in 2002 were
converted to a single block of graphite, that block would have a
volume of (6.73*10^15 gm)/(2.267 gm/cm^3) = 2.97*10^15 cm^3. That's
the volume of a cube that's about 1.44 km (about 0.9 miles) on a side.
If that volume were spread out over the *entire* surface of the Earth,
it would form a layer (2.97*10^15 cm^3)/(5.10*10^14 m^2) = 5.82*10^-6
meters (i.e., 5.82 micrometers) thick. For comparison, a human hair
is 100-200 micrometers in diameter, so the layer would be 20 to 40
times thinner than a hair.
Another interesting number is how many grams of carbon (graphite) per
unit area this thickness corresponds to. That's simply given by
(6.73*10^15 gm)/(5.1*10^14 km^2) = 13.2 gm/m^2, or 1.32*10^5
gm/hectare (1 hectare = 10^4 m^2).
The mean carbon content of the vegetation on a temperate grassland =
7*10^6 gm carbon/hectare
If the entire Earth were covered with grasslands (including the
oceans, polar regions, etc.), there would be (7*10^6 gm
carbon/hectare)*( 5.1*10^14 km^2) = 3.57*10^17 grams of carbon tied up
in the vegetation. The ratio of the carbon emitted by fossil fuel
burning in 2002 to the total amount of carbon in this hypothetical
"grassy Earth" would be (6.73*10^15 gm)/(3.57*10^17 gm) = 0.019, or
1.9% percent of the total. This implies that for the (admittedly
unrealistic) case you've posed, the vegetation would have to get about
2% longer to soak up all the carbon emitted in a single year.