There’s a classic problem in physics textbooks which asks you about astrology. It’s sometimes said – the problem will tell you – that the gravitational pull of the doctor delivering you is stronger than that of Jupiter, therefore it’s unlikely that the planets are exerting a whole lot of influence on your life. The problem asks you to check this.

Of course gravity is not generally purported to be the conduit of the supposed influence of the zodiac, but it’s an interesting problem. And in fact it will turn out that Jupiter generally has the doctor beat, but not by that much. It’s a neat little order-of-magnitude demonstration of the how gravity scales.

Still, I’m not a huge fan of the problem for different reasons. The problem almost always just uses Newton’s equation for the gravitational force due to a point object. A doctor isn’t – his head is farther from the baby than his hands, and so are his feet, and so on and so forth. As such it’s pretty hard to figure out what the “distance” is from the doctor to the baby. Using the center of mass would be a very good approximation if the distance was much longer than the doctor’s height, but it’s not.

So I propose we use this classic problem to learn a little more about approximations. What if we assume that instead of being a point mass, we assume the doctor’s mass is evenly distributed along a line? Sure it’s still a “spherical cow” type of situation, but it’s a lot better than the previous point approximation. The situation will look something like this, with the blue dot representing the baby and the black line representing the doctor. The distances are labeled:

Each little bit of mass on the line will contribute a bit of gravitational force. I’ve picked out one representative bit and labeled its distances specifically, and the sum of all the forces from all the bits will be the force on the baby. From Newton’s law of gravitation for a point mass, the force from one bit will be equal to the gravitational constant, times the mass of the first object, times the mass of the second object, divided by the square of the distance between those objects. So calling the baby’s mass M, the mass of that bit of the doctor dm, and plugging in the square of the distance:

However, this is the force toward that mass dm. Because I’ve drawn the situation symmetrically with the baby even with the doc’s center of mass, the upward component of the force from dm will be exactly canceled by the downward component of the force from the bit of mass located at -h. So only the horizontal forces contribute since they’re all pulling in the same way. To get that force, we multiply by the cosine of the angle between the line r and the hypotenuse. Cosine is equal to the length of the line opposite the angle divided by the length of the hypotenuse, so plugging that in we see that the total horizontal force is:

It is in general pretty difficult to add up this quantity mass by mass. We’d prefer to do it length by length, so let’s express dm in terms of the mass per length – ie, the density λ = m/L so that dm = λ dh, where dh is the tiny height of that bit of the mass. Doing that and collecting terms:

Now add up all the little lengths dh from the bottom to the top by integrating:

It would be a tremendous pain to typeset the solution to this integral, as it’s somewhat involved. I’ll just present the solution, but for those interested in trying it yourself the best method is trig substitution. Let r = h tan(u) and go from there. After the fireworks are done, you’ll get:

So assuming an 8 pound baby and a 180 pound doctor with a height of 6 feet standing 1 foot from the baby, the total gravitational force is about 3.36 x 10^{-8} newtons. Way too small to detect, but it’s a real number. At its average distance from earth, Jupiter would produce a force of about 7.6 x 10^{-7} newtons, or a little more than 20 times more force. Twenty times pretty much nothing is still pretty much nothing, but still – Jupiter wins.

I still can’t say I’d recommend going to an astrologer though…