deconnected. stuff i like. or don't.

dphiffer:

Left: All the water in the world (1.4087 billion cubic kilometres of it) including sea water, ice, lakes, rivers, ground water, clouds, etc. Right: All the air in the atmosphere (5140 trillion tonnes of it) gathered into a ball at sea-level density. Shown on the same scale as the Earth.

Update: this image has received a lot of attention and several people have emailed me asking whether I created it. Both the text and image were found on an internet message board, so I’m not sure who made it or whether it’s accurate. A commenter at Boing Boing credits this to Adam Nieman and the Science Photo Library.

Andrew Nowicki writes:

The mass of the oceans is approximately 1.35 × 10^18 tonnes, or about 1/4400 of the total mass of the Earth (ranges reported: 1.347 × 10^21 to 1.4 × 10^21 kg.) source

The average mass of the atmosphere is about 5 quadrillion metric tons or 1/1,200,000 the mass of Earth. According to the National Center for Atmospheric Research, “The total mean mass of the atmosphere is 5.1480 × 10^18 kg with an annual range due to water vapor of 1.2 or 1.5 × 10^15 kg depending on whether surface pressure or water vapor data are used; somewhat smaller than the previous estimate. The mean mass of water vapor is estimated as 1.27 × 10^16 kg and the dry air mass as 5.13 × 10^18 kg.” source

Andrew’s numbers are a good starting point, but I’m not sure they refute the image’s depiction or its caption. I’m currently traveling so I can’t do the math right now, but I will make an attempt to verify this once I have a chance. If anyone else wants to take a shot, I’d be happy to post your results.

Update #2 Pat Stanton writes:

My son sent me a link to the image of spheres representing all the water and air on Earth. He was skeptical and suspected it was an example of “tree-hugger shock media.”

I decided to do the math, starting with the data provided by Andrew Nowicki. The math appears to verify the posted image. Although I applaud my son’s skepticism, tree-hugger shock media sometimes brings us an important and informative message.

General approach:

1. Measure the spheres representing Earth, water and air in the image. Obtain the diameter in pixels of each sphere. Also, identify relevant physical constants.

2. Starting with data independently provided by Andrew Nowicki, calculate the diameters of spheres that would contain Earth’s water and air.

3. Normalize the results from Step 2 into pixels and compare with measurements from Step 1.

Detailed results:

1. Image Measurements

The image was imported into Adobe Photoshop Elements, enhanced to better reveal spherical outlines, and measured using the Info tool. Measurements from several trials were averaged.

Earth diameter = 185 pixels Water diameter = 22 pixels Air diameter = 30 pixels

Relevant physical constants and relationships:

Spherical volume = (1/6) x Pi x diameter^3 Earth diameter = 12,732 km Mean molar mass of air = 28.94 g / mol Atmospheric pressure (sea level) = 101,325 Pa Temperature (sea level) = 20 C = 293 K Universal gas law constant = R = 8.3145 m^3 Pa / (mol K)

2. Water Calculations

Ocean mass = 1.35 x 10^18 tonnes (Andrew Nowicki) = 1.35 x 10^21 kg

Ocean volume = 1.35 x 10^21 L = 1.35 x 10^24 cm^3 = 1.35 x 10^9 km^3

Solving for diameter, Sphere diameter = 1.371 x 10^3 km

Normalizing to pixels, Sphere diameter = 1.371 x 10^3 km x (185 pixels / 12,732 km) = 20 pixels

This value compare favorably with the measured diameter of 22 pixels. Note that the preceding calculations are based on ocean mass, while the image also included ice, water vapor, fresh water, etc.

3. Air Calculations

Air mass = 5.1480 x 10^18 kg (Andrew Nowicki) = 5.1480 x 10^21 g

Air molecules = 5.1480 x 10^21 g / (28.94 g / mol) = 1.777 x 10^20 mol

Using the Ideal Gas Law, PV = nRT (101,325 Pa) x (Air volume) = (1.777 x 10^20 mol) x (8.3145 m^3 Pa / (mol K)) x 293 K

Air volume = 4.273 x 10^18 m^3 = 4.273 x 10^9 km^3

Solving for diameter, Sphere diameter = 2.013 x 10^3 km

Normalizing to pixels, Sphere diameter = 2.013 x 10^3 km x (185 pixels / 12,732 km) = 29 pixels

This value compare very favorably with the measured diameter of 30 pixels.

Conclusion: The posted image is correct.

Well done, thanks Pat!