Gravity

For this calculation, we need Luna’s location on the Earth, so let’s say she’s at Furman University in the Lakeside Housing. We also need a second hypothetical person. We’ll call him Endymion. He’s the average American male, nothing special. He studies at the University of Texas at Dallas and lives in the University Commons South Hall. That’s enough assumption for our calculation.

Let the coordinates for Lakeside Housing be given by \(\phi_L = 34.922753^\circ\ latitude\) and \(\lambda_L = -82.441248^\circ\ longitude\). Similarly, let the coordinates for South Hall be given by \(\phi_E = 32.989924^\circ \ latitude\) and \(\lambda_E = -96.751620^\circ \ longitude\). The central angle between these two points on Earth can be calculated. We’ll need this in a second.

 

\(\Delta\sigma = \arccos(\sin\phi_L*\sin\phi_E+\cos\phi_L*\cos\phi_E*\cos|\lambda_L-\lambda_E|)\)

\(= 0.20970853\ radians\)

 

Knowing that the height of the average United States female is 1.622 meters and that the average center of mass of a female is 54.3% of her height, the height of the average female’s center of mass can be calculated. Multiplying that by the Lunar Scale Factor gives the height of Luna’s center of mass.

 

\(R=\Lambda*0.543*1.622\ meters\)

\(=1.28*10^8\ meters\)

 

With the scale involved here, it’s safe to say that the height of Endymion’s center of mass is negligible. Now we just need the radius of the Earth \(r_{Earth}=6.378137*10^6\ meters\) and the Law of Cosines to calculate the distance between our subjects’ centers of mass. The number looks identical to the center of mass after applying significant figures, but there is a 1.47*105 meter difference.

 

gravity-diagram

\( D=\sqrt{ (r_{Earth})^2 + (r_{Earth}+R)^2 -2*(r_{Earth})*(r_{Earth} + R)*\cos(\Delta\sigma) } \)

\(=1.28*10^8\ meters\)

 

Now a brief tangent. The average height of the United States male is 1.763 meters (about 5 feet, 9.5 inches). The average weight for a man of this height ranges from 68 to 74 kg (about 150 to 163 lbs). For fun, let’s take Endymion’s mass to be the upper bound, \(M_E = 74\ kg\). Also, recall from “Sugar Content” that Luna’s mass is \(M_L=1.57*10^{26}\ kg\).

Finally, the fun bit. Knowing the mass of two objects and the distance between their centers of mass allows us to calculate the force of gravity between them with the Gravitational Constant \(G=6.7410^{-11}\ \frac{N*m^2}{kg^2}\).

 

\(F_g=\frac{G*M_L*M_E}{D^2}\)

\(=48\ Newtons\)

 

This deserves some perspective. With Endymion’s mass and the acceleration due to gravity \(g=9.8\ \frac{m}{s^2}\), we know that the Earth’s gravitational pull on him is \( M_E*g=725\ Newtons\), about 15 times stronger than the force of Luna’s gravity.

In fact, recalling from “Volume” that the average female surface area is 1.54 m2 and doing some quick calculations, we find that the force of air pressure on the average female is \((1.54\ m^2)*(1.013*10^5\ Pa)=156000\ Newtons\), about 3280 times the force of Luna’s attraction.

Despite her incredible mass 140 times that of the Earth, the fact that Luna’s center of mass is 10 times the diameter of the Earth away is quite a disadvantage for gravitational pull. So is her 48 Newton pull a disappointment? That’s a question for another week.

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