In this blog series we’ve been working our way towards determining mass values for heavenly bodies in our universe. Today we’ll explore the inverse relationship phenomenon that’s present when examining the acceleration of gravity between two heavenly bodies, for our purposes Earth and its moon. To demonstrate the inverse relationship phenomenon, we’ll use Newton’s Law of Gravitation formula: g = (G × M) ÷ r^{2} (1) In this formula g represents Earth’s acceleration of gravity that’s acting upon the moon. The center of the moon is located at a distance r from Earth’s center. Earth’s mass is represented by M, and G is the universal gravitational constant, which never varies from its value of 3.49 × 10^{8} cubic feet per slug per second squared. We’ll solve for Earth’s g factor relative to the moon, which we’ll position at varying distances from Earth’s core, thereby demonstrating how the attracting force of Earth’s gravity becomes weaker as the moon’s distance from Earth’s core increases — hence the inverse relationship. To begin, we know that the present distance from Earth’s center to the center of the moon is about 238,900 miles, or 1,261,392,000 feet. We also know that Earth’s mass, M, is equal to 4.09 × 10^{23} slugs. Plugging these values into Newton’s Law of Gravitation equation, we calculate the acceleration of gravity exerted upon the moon in this, its normal orbit, to be: g = ((3.49 × 10^{8} ft^{3}/slug/sec^{2}) × (4.09 × 10^{23} slugs)) ÷ (1,261,392,000 ft)^{2} g = 0.0089 ft/sec^{2} Now let’s suppose that the moon’s orbit was caused to increase so that it became situated 400,000 miles, or 2,112,000,000 feet from Earth. Earth’s acceleration of gravity exerted upon the moon at this distance would be calculated as: g = ((3.49 × 10^{8} ft^{3}/slug/sec^{2}) × (4.09 × 10^{23} slugs)) ÷ (2,112,000,000 ft)^{2} g = 0.0032 ft/sec^{2} From these two examples we can see that the further the moon is positioned from Earth, the weaker Earth’s gravitational pull upon it is. This gravitational pull is the force of gravity, F_{g}, introduced in our last blog, a term which originates in Newton’s Second Law of Motion, as given by the equation: F_{g} = m × g (2) where, in this case, m represents the mass of the moon. Next time we’ll combine equations (1) and (2) and derive a third formula which will allow us to calculate the mass of the sun.
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Tags: acceleration of gravity, engineering expert witness, falling objects, forensic engineer, gravitational force, gravity, mass, Newton's Second Law of Motion, Newton's Universal Theory of Gravitation, pull of gravity