# 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.

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