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← Revision 3 as of 2024-08-09 05:38:18 ⇥
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| How much volume of Earth material would we need to remove to make the sea level circumference PRECISELY 40,000 km? :-) | How much seawater would we need to remove to make the sea level circumference PRECISELY 40,000 km? :-) |
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| Lets assume the Earth's mass is proportional to the cube of that radius, and that the shape and density are not changed by reducing the mass. Making the radius 7.863 km smaller ( a factor of 0.999803463 ) would make the volume smaller by a factor of 0.999803463³ or 0.999410506, a volume reduction of 5.895e-4. Assuming an oblate spheroid with a polar radius of 6356.7523 km and an equatorial radius of 6379.137 km, the volume is (4π/3) * Re² Rp or 1.0835e12 km³, so the volume reduction is ( 5.895e-4 * 1.0835e12 km³ ) or 6.387e8 km³. For comparison, the volume of the world's oceans oceans is 1.37e9 km³ ... we can remove half the world's oceans (send the water to settlements on asteroids, perhaps ) and achieve our calibrate-the-planet-to-the meter radius reduction. |
7.863 km / 2π ; 1250 meters. During the last ice age, the sea dropped 120 meters as water became glaciers. So, we either make 10 ice ages worth of ice, or we launch the water to colonies in space. |
Correcting The Earth To Match The Meter
The kilometer was intended to be 1/40,000 of the polar circumference of the Earth. Due to small surveying error, the polar circumference is actually 40,007.863 km.
How much seawater would we need to remove to make the sea level circumference PRECISELY 40,000 km? 
7.863 km / 2π ; 1250 meters. During the last ice age, the sea dropped 120 meters as water became glaciers. So, we either make 10 ice ages worth of ice, or we launch the water to colonies in space.