I was curious, so I did the math.<p>Total worldwide carbon production is 38.2 billion tons per year. Cost to sequester a ton of carbon is between $30 and $150, depending on who you ask and how you do it. Let's assume a middle of the road price of $90/ton. That's $3.438 trillion a year, or about $478 per person. This is roughly equal to the US yearly federal spending, or 3% of the world GDP.<p>If you somehow pooled together all the world's billionaires and got them to contribute their annual income (roughly $600 billion a year, averaging the past 7 years) to the effort, you could eliminate roughly 20% of carbon produced in the world every year.<p>-------------------------<p>Suddenly, it becomes crystal clear why finding new sequestration methods is incredibly important: if you can get the cost from $160 to $10 per ton, then suddenly all you'd need would be a coalition of half the world's billionaires to stop the main cause of global warming.<p>Additionally, it's important that people realize that CO2 production is in tons of CO2 per year. Tree offsets are a one-time deal, since when trees die they release CO2, and when new ones are born they absorb that CO2 again. After they've been planted, forests are generally carbon neutral. That's why we can't "just plant trees": we'd have to be continuously planting new trees. The Earth is only 8% arable land, much of which already has stuff on it, or is undesirable for one reason or another. We'd run out of space pretty fast. Trees are good for other reasons: preventing climate change (different from global warming), preserving species diversity, being nice to look at, etc etc.<p>-------------------------<p>Mostly off-topic: when I was looking at estimates of land size, apparently the amount the US has shrunk from 2007 to 2015 (14,000 km2; went from 9,161,120 km2 to 9,147,420 km2) [0] is roughly equivalent to half the area of the Netherlands. Wow.<p>[0] <a href="http://data.worldbank.org/indicator/AG.LND.TOTL.K2" rel="nofollow">http://data.worldbank.org/indicator/AG.LND.TOTL.K2</a>