Posts Tagged ‘Newton’s gravitational theory’

What is Earth’s Mass?

Friday, November 7th, 2014

      Last time we learned how Henry Cavendish used Christiaan Huygens’ work with pendulums to determine the value of g, the acceleration of gravity factor for Earth, to be 32.3 ft/sec2, or 9.8 m/sec2.    From there Cavendish was able to go on and arrive at values for other factors in Isaac Newton’s gravity formula, namely G, the universal gravitational constant, and M, Earth’s mass.   Today, we’ll discuss how Cavendish was able to calculate the Earth’s mass.

      Newton’s formula for gravity, once again, is:

M = (g × R2) ÷ G

where M stands for the mass of the heavenly body being quantified.   For our case today M will represent the mass of Earth, which was originally quantified in slugs, a British unit of measurement.   Today the measurement unit of choice in most parts of the world is the kilogram, which is the metric equivalent of a slug.

      With regard to the other variables in Newton’s gravity formula, namely, R and G, their values had previously been determined.   Eratosthenes’ measurement of shadows cast by the sun on Earth’s surface had revealed Earth’s radius, R, to be 6,371 kilometers, or 6,371,000 meters.   And Cavendish’s experiments led him to conclude that the universal gravitational constant, G, was 6.67 × 10-11 cubic meters per kilogram-second squared.   Plugging these values into Newton’s equation, we calculate Earth’s mass to be:

M = ((9.8 m/sec2) × (6,371,000 m)2) ÷ (6.67 × 10-11 m3/kg-sec2)

M = 5.96 × 1024 kilograms

      Incidentally, 5.96 × 1024 is scientific notation, or mathematical shorthand, for the number 5,960,000,000,000,000,000,000,000.   That’s a whole lot of zeros!

Engineering expert witness blog

      Calculating the mass of Earth was an impressive accomplishment.   Now that its value was known, scientists would be able to calculate the mass and acceleration of gravity for any heavenly body in the universe.   We’ll see how that’s done next time.