Engineering Expert Witness Blog

      Last time we calculated the sun’s force of gravity acting upon Earth.   It was the final unknown quantity within Newton’s equation to determine the mass of the sun, an equation we’ve been working with for some time now.   Today we’re set to discover just how big the sun is.

      Newton’s formula, introduced in a past blog in this series entitled, Gravity and the Mass of the Sun is again,

M = (F_g × r²) ÷  (m × G)

where G is the universal gravitational constant as determined by Henry Cavendish and discussed in our blog, How Big is the Earth? and is equal to,

G = 6.67 × 10^-11 meters per kilogram • second²

      As discussed in last week’s blog, The Sun’s Gravitational Force, Earth’s mass, m, its distance from the sun, r, and the force of the sun’s gravity acting upon Earth, F_g , are respectively,

m = 5.96 × 10²⁴ kilograms

r = 149,000,000,000 meters

F_g = 3.52 × 10²² Newtons

      Inserting these values into Newton’s equation to determine the mass, M, of the sun we get:

M = [(3.52 × 10²²) × (149,000,000,000)²] ÷ [(5.96 × 10²⁴) × (6.67 × 10^-11)]

M = 1.96 × 10³⁰ kilograms

      So how big is 1.96 × 10³⁰ kilograms?   To get a better idea, let’s divide the sun’s mass, M, by the Earth’s mass, m,

(1.96 × 10³⁰ kilograms) ÷ (5.96 × 10²⁴ kilograms) = 328,859.06

      That’s a big number, and it translates to the sun being over 300,000 times more massive than Earth.   The picture below displays this comparison in stunning visual terms.

      Once 19th Century scientists had calculated the mass of the sun, they went on to calculate the masses of other heavenly bodies in our solar system and the gravitational forces at play on each of them.   Armed with this information mankind was able to subsequently build exploratory probes capable of extending their reach into the far unknowns of our solar system and beyond.

      This ends our discussion on gravity within our solar system.   Next time we’ll return to Earth and begin exploring the physics behind falling objects.

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