# Month: December 2016

## Gravity 2

Recall from “Gravity” that the central angle between Endymion and any girl at Furman University is $$\Delta\sigma= 0.20970853\ radians$$. If we assume the heights of their centers of mass to be negligible, then the distance between them is $$D_{woman} = 2*r_{Earth}*\sin(\frac{\Delta\sigma}{2}) = 1.335100*10^6\ meters$$. Knowing the gravitational constant $$G=6.7410^{-11}\ \frac{N*m^2}{kg^2}$$ and the masses of these two, we can calculate the force of gravity between Endymion and that Furman girl:

$$G*(74.0\ kg)*(51.7\ kg)*D_{woman}^{-2}$$

$$= 1.4*10^{-19}\ Newtons$$

Recall also from “Gravity” that Luna’s gravitational pull on Endymion was about 48 Newtons, about 3.4 * 1020 times as much. Forget Endymion’s location for a moment. If he stood under Luna’s feet, he would be as close as possible to her center of mass, 1.28*108 meters. At that distance, the force of gravity would be about the same, 48 Newtons. The farthest he could be from her is on the other side of the Earth, adding about 1.28*107 meters of Earth diameter to that distance. The force of gravity at its minimum is then 39 Newtons. These forces of gravity are very strong compared to the average girl.

Let’s think bigger. Looking at some data from the United States Census Bureau, it appears that of the 7,323,187,457 people alive in 2016, 1,945,617,771 were 20 to 60 year old women. Using the same average woman’s mass from before, we can calculate their gravitational pull. If we’re willing to pretend that the women are evenly distributed across the Earth, Newton’s Shell Theorem lets us calculate their gravitational pull as if their mass were all concentrated at the center of the Earth.

$$G*(74.0\ kg)*(51.7\ kg)*(1945617771\ women)*r_{Earth}^{-2}$$

$$= 1.22*10^{-11}\ Newtons$$

Of course, all the women alive in 2016 cannot compare to Luna’s attractive force. She still out-pulls them by a factor of anywhere from 3.23 trillion to 3.90 trillion depending where you are. If we want to out-gravitate Luna, we have to go all out.

In his paper “Modeling World Population,” Dalton Deshotel explores the use of three models of population growth, each more realistic than the last. I would like to have used the most realistic of them, Cohen’s augmented logistic, but it did not lend itself well to the problem I wanted to solve, namely finding how many adult women will ever live. Fortunately, the similar and more common (albeit less realistic) model of population, the Verhulst logistic model, does.

The model assumes that as population grows, birth rate decreases linearly $$B(t) = b – c*P(t)$$ while death rate increases linearly $$D(t) = d + a*P(t)$$. Population itself changes at a rate equal to the birth rate minus the death rate $$P'(t) = P*[B(t) – D(t)]$$. If we let $$r = b-d$$ and $$K = \frac{r}{a+c}$$, then the rate of population change simplifies to $$P'(t) = P(t) * r * (1 – \frac{P(t)}{K})$$. By doing some manipulations, we can arrive at a function for the number of people born within a given interval. We will call the function J(Q, W), and it will give the number of people born between the years Q and W.

Here be dragons

Since P'(t) is change in population, P(t) is population. Since B(t) is a birth rate and P(t) is the population, $$J(Q, W) = \int_{Q}^{W} P(t)*B(t) dt$$ gives the number of people born between years Q and W. All we have to do is solve the integral.

First we need P(t) from P'(t).

• $$P'(t) = P(t)*r*(1 – \frac{P(t)}{K})$$
• $$\frac{dP}{dt} = P(t)*r*(1 – \frac{P(t)}{K})$$
• $$rdt = \frac{dP}{P(t)*(1 – \frac{P(t)}{K})}$$
• $$rdt = \frac{K*dP}{P(t)*(K – P(t))}$$
• $$rdt = \frac{dP}{P(t)} + \frac{dP}{K-P(t)}$$

Integrate both sides.

• $$rt + C = \ln(P(t)) – \ln(K-P(t))$$
• $$rt + C = \ln(\frac{P(t)}{K-P(t)})$$

Let $$A = e^C$$. Exponentiate both sides.

• $$Ae^{rt} = \frac{P(t)}{K-P(t)}$$
• $$(K-P(t)) Ae^{rt} = P(t)$$
• $$KAe^{rt}-P(t)Ae^{rt} = P(t)$$
• $$KAe^{rt}=P(t)Ae^{rt} + P(t)$$
• $$KAe^{rt}=P(t)(1+Ae^{rt})$$

$$P(t) = \frac {KAe^{rt}} {1+Ae^{rt}}$$

Now we can start solving $$\int_{Q}^{W} P(t)*B(t) dt$$.

Recall that $$B(t) = b – c*P(t)$$

• $$\int_{Q}^{W} P(t)*B(t) dt$$
• $$\int_{Q}^{W} P(t)*(b – c*P(t)) dt$$
• $$\int_{Q}^{W} bP(t) – \int_{Q}^{W} cP^2(t) dt$$
• $$b\int_{Q}^{W} \frac {KAe^{rt}} {1+Ae^{rt}} – c\int_{Q}^{W} (\frac {KAe^{rt}} {1+Ae^{rt}})^2 dt$$
• $$\frac{bK}{r} \int_{Q}^{W} \frac{rAe^{rt}}{1+Ae^{rt}}dt – \frac{cK^2}{r^2} \int_{Q}^{W} (\frac {rAe^{rt}} {1+Ae^{rt}})^2$$

Let $$H = \int_{Q}^{W} (\frac {rAe^{rt}} {1+Ae^{rt}})^2$$

• $$\left. \frac{bK}{r} * \ln (1 + Ae^{rt}) \right| _Q^W -\frac{cK^2}{r^2}H$$

Now we solve for H.

Let $$u = 1 + Ae^{rt}$$

Then $$du = rAe^{rt}dt$$

So $$u-1 = Ae^{rt}$$

And $$dt = \frac{du}{rAe^{rt}} = \frac{du}{r(u-1)}$$

• $$H = \int (\frac{r(u-1)}{u})^2 * \frac{du}{r(u-1)}$$
• $$H = \int \frac{r(u-1)}{u^2} du$$
• $$H = r \int u^{-1}du – r\int u^{-2}du$$
• $$H = \left. r ( \ln u + \frac{1}{u} ) \right|$$
• $$H = \left. r ( \ln (1+Ae^{rt}) + \frac{1}{1+Ae^{rt}} ) \right|_Q^W$$

So altogether:

• $$J(Q, W) = \int_Q^WP(t)B(t)dt =$$
• $$\left. \frac{bK}{r} * \ln (1 + Ae^{rt}) -\frac{cK^2}{r^2}(r ( \ln (1+Ae^{rt}) + \frac{1}{1+Ae^{rt}} )) \right|_Q^W$$
• $$\left. \frac{bK}{r} * \ln (1 + Ae^{rt}) -\frac{cK^2}{r^2}(r ( \ln (1+Ae^{rt}) + \frac{1}{1+Ae^{rt}} )) \right|_Q^W$$
• $$\left. \frac{1}{r} * \ln(1+Ae^{rt}) * (bK-cK^2) – \frac{cK^2}{1+Ae^{rt}} \right|_Q^W$$

We would be done except we have not defined values for a, b, c, d, r, K, and A. We will choose r and K to be 0.028 and 12.089*109. We get these values from section 2.2.2 of Deshotel’s paper, where he uses MATLAB to find the curve of best fit. Next, we use five pieces of information from the Population Reference Bureau. In 2004, the population was $$P_1 = 6.396*10^6$$ and the birth rate was $$B_1 = \frac{21}{1000} = .021$$. In 2016, the population was $$P_2 = 7418151841$$, the birth rate was $$B_2 = \frac{20}{1000} = 0.02$$, and the death rate was $$D_2 = \frac{8}{1000} = .008$$.

Since $$B = b-cP$$, we know $$B_1 = b-cP_1$$ and $$B_2 = b-cP_2$$. Subtracting one from the other gets $$B_1 – B_2 = c(P_2-P_1)$$, so $$c = \frac{B_1-B_2}{P_2-P_1}$$. Now that we have c, we can rearrange $$B_1 = b – cP_1$$ into $$b = B_1 + cP_1$$. Since $$r = b – d$$, we can solve for d using $$d = b – r$$. Then we can find a by rearranging $$D_2 = d-aP_2$$ into $$a = \frac{d-D_2}{P_2}$$. Finally, we get A from rearranging $$P = \frac {KAe^{rt}}{1 + Ae^{rt}}$$ into $$A = \frac{P}{Ke^{rt} – Pe^{rt}}$$. We might choose either P to solve for A, so I chose $$P_2$$.

I will not bother to write the explicit values for these variables. It suffices to know that

$$J(Q, W) = \left. \frac{1}{r} * \ln(1+Ae^{rt}) * (bK-cK^2) – \frac{cK^2}{1+Ae^{rt}} \right|_Q^W$$

An article from the Population Reference Bureau estimates that 107,602,707,791 people had been born up to the year 2011. Since population trends in ancient times are unlikely to have been as regular and predictable as modern population, I will take their word for it and let Q=2011. To find W, we need a good estimate for when humans will stop being born. In true Wikipedia style, there does exist a page entitled “List of dates predicted for apocalyptic events.” In particular, I like the un-avertable asteroid theory which says the Earth will be hit by a humanity-extinguishing asteroid within 500,000 years. I’ll be optimistic and let W = 500,000. The total number of people ever born is thus:

$$T = 107602707791 + J(2011, 500000)$$

$$= 93\ billion\ people$$

I refer to the USCB data again. Of the 7,323,187,457 people alive in 2016, 2,449,933,457 were women over 20, so 33.4545% of the population were women who had been 20 years old at some point in their lives. Ancient life expectancy was lower than it is now and future life expectancy will be higher than it is now. With that in mind, we might pretend that 2016 had the average ratio of adult women to population. That is to say that if we took the ratio of living female adults to living humans every year and averaged all the ratios, we would get the same ratio that 2016 had. It’s a stretch, but if we’re willing to accept that it’s even close, then we can just multiply the 2016 ratio by the total number of people ever born and get the number of women who will ever have been adults:

$$W = T*.334545$$

$$= 31.1\ billion\ adult\ women$$

Surely all the adult women who ever lived will have a stronger gravitational pull than Luna?

$$G*(74.0\ kg)*(51.7\ kg)*(31.1\ billion\ adult\ women)*r_{Earth}^{-2}$$

$$= 1.953*10^{-7}\ Newtons$$

Their force of gravity it still a whole 202 million to 244 million times less than Luna’s, depending on where you stand. Maybe we were too picky to only consider women. 33.0352% of people alive in 2016 are adult men, so 30.7 billion adult men will ever live. Maybe if we include them too?

$$G*(74.0\ kg)*[(51.7\ kg)*(31.1*10^9\ women) + (74.0\ kg)*(30.7*10^9\ men)]*r_{Earth}^{-2}$$

$$= 4.713*10^{-7}\ Newtons$$

Luna is still 83.6 million to 101.2 million times more attractive than all adults who will ever live combined.

$$\frac{48\ Newtons} { 5 } * \frac{ 31.1*10^9\ women + 30.7*10^9\ men }{4.713*10^{-7}\ Newtons}$$

$$= 4.6627*10^{22}$$

If the average person were a 5, Luna would be a 46.6 sextillion. There’s just no way around it.

## Angular Diameter

The average woman has a distance of 10.5 cm from the top of the head to the eyes (Page 81, Table 9). The average female height is 1.622 meters, so the height of Luna’s eyes is:

$$\Lambda * (1.622\ m\ -\ .105\ m) = 2.20*10^8\ meters$$

The distance from the center of the Earth to the center of the Moon is called the Lunar Distance. At its closest, this distance is 356,500\ km. Subtracting the radius of the Earth $$r_{Earth} = 6.378137*10^6\ meters$$, we have the distance from the surface of the Earth to the center of the Moon:

$$356500\ km\ -\ r_{Earth} = 3.501*10^8 meters$$

Since Luna’s eyes are closer than the moon, they should appear bigger. Apparent size is measured by angular diameter $$\delta = 2\arctan(\frac{d}{2*D})$$, where d is the diameter of the object, and D is the distance to the object. Finding $$D_\Lambda$$, the distance to Luna’s eye, requires choosing a point of perspective.

From any point on Earth, the view to Luna’s eyes is blocked by the Earth, Luna’s toes, or 140 Earth volumes of giant girl. Let’s pretend that isn’t true for Endymion, the guy from Gravity. We can reuse some calculations from Gravity too: the central angle between him and Luna is $$\Delta\sigma = 0.20970853\ radians$$. We already know the height of Luna’s eyes off the surface of the Earth $$R = 2.20*10^8\ meters$$ from earlier. Knowing those two numbers and the radius of the Earth, we can use the law of cosines to calculate $$D_\Lambda$$.

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

$$=2.20*10^8\ meters$$

Just like in Gravity, the number looks the same with significant figures but is worth calculating. Now we can calculate angular diameter.

$$\delta_\Lambda = 2\arctan(\frac{d}{2*D_\Lambda})$$

$$= 1.58\ radians$$

In a similar way, we can use the maximum lunar distance (subtracting Earth radius) $$D_{max} = 400329\ km$$ to find the maximum angular diameter of the moon $$\delta_{max} = .00868\ radians$$ and the minimum lunar distance $$D_{max} = 350129\ km$$ to find minimum angular diameter $$\delta_{max} = .00993\ radians$$. So we know that Luna’s eyes can appear to be anywhere from 1.592 to 1.821 times the size of the moon depending on the time of month.

## 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.

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

$$=1.28*10^8\ meters$$

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.