## Posts Tagged ‘gravity’

### Three-Way

October 21, 2010

I’ve been told that when you and your sweetheart get accepted to different grad schools, you’ve encountered the “Two-Body Problem”. It’s an inapt analogy, because the two-body problem can be solved elegantly. What’s really difficult is the three-body problem.

If we have three massive bodies interacting through Newton’s gravitational law, with arbitrary initial positions and velocities, it turns out that exact analytic solutions are extremely difficult to find. That’s not to say there is no solution – the massive bodies have no trouble finding it. But if we want to predict their motion we’ll need to do some numerical integration.

However, we might wonder if there are some particularly simple or intriguing solutions, perhaps in very symmetrical situations. There are. Quite a few of them are cited in the Scholarpedia article. Here are some:

I’m not sure how these solutions were discovered. But if we consider the problem a moment it’s clear the center of mass cannot accelerate, and that energy and angular momentum must be preserved. An especially simple way to do this is to have the angular momentum and energy of each individual mass be constant by making them all rotate around the center of mass at a fixed frequency.

To see if this works, measure the positions of the three masses by vectors $\vec{r_1}, \vec{r_2}, \vec{r_3}$ from the axis of rotation, which is through the center of mass and perpendicular to the plane of the bodies. Let the masses of the bodies be $m_1, m_2, m_3$. Then the equation for the center of mass states

$\vec{r_1}m_1 + \vec{r_2}m_2 + \vec{r_3}m_3 = 0$.

If we look at a reference frame comoving with the center of mass and corotating with the masses, the masses will be stationary, so there must be no force on them. Let’s take $m_1$ as an example. The centrifugal force on it is

$F_c = \omega^2 m_1 \vec{r_1}$.

The gravitational force on it is

$F_g = G m_1 \left(m_2\frac{\vec{r_2} - \vec{r_1}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + m_3\frac{\vec{r_3} - \vec{r_1}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$

If those forces are going to add to zero to give $m_1$ zero acceleration, they better at least point the same direction. It’s not clear that they do. The centrifugal force points only in the direction of $\vec{r_1}$, while the gravitational force has all three position vectors in there. So we need the amounts of $\vec{r_2}$ and $\vec{r_3}$ to be such that they add up to point towards $\vec{r_1}$.

From the equation for the center of mass we get,

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

whence we know the proportions in which the $\vec{r_2}$ and $\vec{r_3}$ must appear in the gravitational force. We conclude

$m_2\vec{r_2} + m_3\vec{r_3} \propto \left(\frac{m_2\vec{r_2}}{\left|\vec{r_1} - \vec{r_2}\right|^3} + \frac{m_3 \vec{r_3}}{\left|\vec{r_3} - \vec{r_1}\right|^3}\right)$.

This will work perfectly iff the denominators on the right hand side of that equation are equal, meaning mass $m_1$ must be equidistant from masses $m_2$ and $m_3$. Repeating the argument while writing out forces on $m_2$ this time, the masses must be in an equilateral triangle.

We still don’t know if such a solution exists, only that it’s possible to get the centrifugal force to point opposite the gravitational force for arbitrary masses, as long as we put them in an equilateral triangle. Let’s continue, setting the centrifugal and gravitational forces equal for $m_1$ and see what the resulting rotation rate is. Also let’s let the sides of the triangle be length $l$

$F_c = \omega^2 m_1 \vec{r_1} = F_g = \frac{G m_1}{l^3} \left( m_2(\vec{r_2} - \vec{r_1}) + m_3 (\vec{r_3} - \vec{r_1}) \right)$

Applying the center of mass equation

$-m_1\vec{r_1} = m_2\vec{r_2} + m_3\vec{r_3}$

one more time and setting the total mass equal to $M$, this simplifies to

$\omega^2 = \frac{G M}{l^3}$.

A clean result that is the same for all three masses, meaning these orbits indeed solve the equations of motion.

### Gravity from a Galaxy

June 14, 2010

Newton was a very serious man – you might even say an expert in gravity. First he invented gravity so that guys who juggle apples at parties wouldn’t have to keep climbing up to the ceiling to get them back down. Then he invented math to prove that the moon wouldn’t fall, too.

I’ve heard Newton worked hard to prove the following about his gravitational theorem: Suppose you have a spherical planet made out of apples, and you are standing on it. Each of the individual apples pulls on you. When you add up all those little tugs, you get the total force of gravity from the apple planet. The force is just the same as it would be if you were standing in the same spot, but all the apples were mashed up on top each other at the center of the planet.

You are standing on a planet made of apples. Each apple pulls on you, and all their pulls combined are the gravitational force.

When the apples are concentrated at the center, some get closer to you and some get further away. It just works out so the total force of gravity stays the same.

That’s convenient because it makes calculating the gravity of the apple planet easy. You don’t have to add up all the apples. You just need to know how far it is to the center of the planet and how many total apples there are. Or, you could measure the force and the size of the planet, and use that to calculate the number of apples.

But what about a galaxy instead of a planet? A galaxy isn’t a sphere. Since it spins, it’s more of a disc, like a wad a dough that gets twirled repeatedly in the air and flattens out into a pizza. (Caveat: I have never observed a pizza to form spiral arms).

The Earth orbits around the Milky Way, so if we want to do some physics with that, can we assume that all the mass of the Milky Way is located at the center? In other words, does the same theorem that holds for spherically-symmetric matter distributions (a planet) hold for cylindrically-symmetric ones (a galaxy)? Think about it a moment, do an integral if you like, then read on.

Will the gravity from the disc galaxy also be the same as if all its mass were concentrated at the center?

In this post I won’t prove either theorem. Instead, I’ll show that if the theorem works for the sphere, then it does not work for the disc. To do this, imagine transforming a sphere into a disc. Start with the apple planet, then squash it down to two dimensions, keeping the radius the same, but moving all the apples into a disc shape. This is called projecting the sphere onto a plane.

The sphere gets projected onto the plane, making a disc. Imagine the apples are the same size, and there's the same number of them, and the center of the disc is filled. My drawing is not great.

Doing this, the force of gravity you feel must get stronger, because every apple that moved got at least a little bit closer to you. So the gravity from the disc is stronger than the gravity from the sphere. If we then collapsed all the apples to the point at the center, gravity would get weaker again. If the shell theorem works for spheres (and it does), it doesn’t work for discs.

One way of detecting dark matter is to look at the rate that stars orbit their galaxies and use that to infer how much matter there must be. If there’s lots of gravitating matter, but we don’t see that many stars, then we know the galaxy has dark matter somewhere. But when making that calculation, we’ll have to take into account the disc shape of the galaxy and its more-complicated gravity.

### Bounce, part 4

January 2, 2010

Previous parts: 1 2 3

Last time we made progress on figuring out how high a tennis ball can bounce in the classic experiment where we drop the tennis ball on top a basketball. We didn’t find the answer, but we said that if the tennis ball picks up a speed $v$ in falling, then immediately after bouncing off the basketball, it could have a maximum upward speed of $3v$.

Today we want to figure out what that means in terms of how high the tennis ball will bounce. It turns out that the tennis ball does not bounce three times as high as it started when it rebounds with three times the speed. In fact it bounces much higher.

After bouncing off the basketball, the tennis ball rises, but slows down under the influence of gravity until it comes to a stop at the top of its trajectory. To understand how high it goes, we must answer the question, “what does the influence of gravity do to the motion of the ball?”

One of the first people to understand this question and its answer was Galileo (although several people came to the correct conclusion before him). We’ll look at a few passages of his famous book, Dialogue Concerning Two New Sciences. (specifically this part)

Galileo begins by stating that he thinks “uniformly accelerated motion”, the motion of a tennis ball thrown into the air, should be very simple.

When, therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everybody? If now we examine the matter carefully we find no addition or increment more simple than that which repeats itself always in the same manner.

In other words, the way the speed of a falling body changes shouldn’t depend on how high it is, or how long it’s been falling, or how far it’s fallen. It should depend on nothing at all – be always the same.

This may be a lot to swallow, but let’s look at one good reason (not due to Galileo) that we might expect the way gravity acts on a falling object not to change with how high the object is above the Earth’s surface. The radius of the Earth is very large compared to the heights we throw things. We expect that if the effects of Earth’s gravity do change with your distance from the center of the Earth, they ought to do so on a distance scale roughly equal to the radius of the Earth.

That is, if you want a significant difference in the force of gravity, you ought to change your position by something significant compared to the radius of the Earth, since it defines the only natural length scale in this problem. The radius of the Earth is roughly six million meters, so throwing a tennis ball up in the air six meters is completely negligible. We could calculate the effects of gravity using Newton’s gravitational law, but that is unnecessary. Any other reasonable gravity law ought to work out basically the same. Near the surface of the Earth, your height should not affect how gravity acts on you.

This is only one part of what Galileo said. For example, he also believes that how fast an object moves should not affect how gravity acts on it. This belief may have been stimulated by the relativity principle – that all laws of physics should be the same, even when you’re moving. Relativity does not absolutely preclude a force that depends on velocity, though (magnetic forces do this), but velocity-dependent forces are not as simple as velocity-independent forces, and for the time being Galileo is guessing that the way gravity acts ought to be very simple.

We continue with the G-spot’s wise words:

A motion is said to be uniformly accelerated, when starting from rest, it acquires, during equal time-intervals, equal increments of speed.

This is Galileo’s working idea of how things fall. If you drop something, and at the end of one second it goes speed $v$, then at the end of two seconds it will be going $2*v$, and at the end of three seconds $3*v$, etc. A plot of speed over time, if we drop an object from rest, should look like this:

This plot shows the speed of a falling tennis ball. The tennis ball is dropped from rest, and so starts at speed zero. Gaining equal speed in each moment of time, the speed is directly proportional to time.

Now that we have a theory for what the speed of the tennis ball does, we should be able to figure out how high it goes. The tennis ball reaches its highest height when its speed is zero, so we simply need to keep track of its speed until that speed falls to zero. If we know how fast it was going and for how long, we should also know how far it traveled.

I’ll paraphrase Galileo’s arguments here rather than quoting them, since he does not directly answer our exact question. The relevant pages are 171 – 178.

First, let us suppose it takes the tennis ball a time $t$ to fall before bouncing, and it acquires speed $v$ in that time. We know it bounces back up with speed $3v$. It loses speed in the same way it gained speed – the same amount per second. So after a time $t$, the ball loses speed $v$, and is down to moving at speed $2v$. The ball comes to a stop at the height of its trajectory after a time $3t$.

To summarize, if the ball gains and loses the same amount of speed in any moment of time, then if it two balls bounce upward, one three times as fast as the other, the fast one will take three times as long to get reach its apex.

The distance the ball travels just $speed * time$, which is the green area shaded in the previous drawing.

Here is a plot of the speed of the ball as it rises:

The tennis ball's return trip. This time it begins going quickly, three times as fast as before, and slows down. It takes three times as long to reach its peak as it took to fall.

It rises three times as long as it fell, and the distance it rises is purple the area in the above chart. Laying the two plots together, we see that the purple area is nine times as large as the green one – three times taller and three times wider.

The green area represents the distance the tennis ball fell (see first figure). The purple area is the area the tennis ball rises after bouncing off the basketball. The tennis ball rises nine times as high as it was dropped from.

Now we have our first answer to how high a tennis ball can bounce when dropped on top a basketball. It can bounce nine times as high, when we make the following assumptions:

• When things bounce off the ground, they change their direction keeping exactly the same speed and hence bounce back to the same height. (first post).
• A basketball is so much bigger than a tennis ball that it essentially acts as the ground – the tennis ball bounces off just the same as it would bounce off the ground. (third post)
• To understand the way something in motion works, we can imagine we are moving alongside it at the same speed so that it isn’t moving from our point of view, and understand it that way. Then we can imagine going back to the frame in which the thing is moving and translating over our new knowledge over. (third post)
• Gravity pulls an object down such that it gives it the same additional amount of speed in each moment of time. (this post)

My original claim was that I could have understood all these ideas as a child. I think that’s right. I was a pretty bright kid, and if someone had sat down to explain this reasoning to me, and answered my questions, I think I’d have gotten it. But I also hope I’d have realized there’s a problem. When you actually do the experiment, the tennis ball doesn’t bounce nine times as high, or anywhere near that. Three times as high is pretty good for this experiment. So I’d like to think I’d have noticed that, and asked for an explanation of the discrepancy.

We began to discuss this in part two, where we looked at why things bounce to a lower height than they’re dropped from. The assumption about reversing direction and speed when bouncing is simply not correct. It is also not correct to assume that the basketball is so much larger than the tennis ball that it acts like the ground, but this is a smaller source of error. It isn’t true that gravity is completely uniform, either, or that the only influence on the falling ball is from gravity. We’ll look at these things in more detail in a later post.

Before doing that, though, the next post or two will continue looking at the passage from Galileo. This passage isn’t interesting to me simply because it is an early source of someone understanding this fairly simple problem. It’s interesting because it’s an illustration of Galileo laying down a more sophisticated understanding of how we can understand nature. I want to look at what Galileo did and didn’t know, but also at how much he understood about what he did and didn’t know, and how he came to his conclusions.

There’s also a very surprising and egregious logical error in the passage, so we’ll talk about that, too, before returning to the tennis ball a little down the line.

### The Renaissance Man Uniform Gravitational Acceleration SMACKDOWN

October 9, 2009

Matt at Built On Facts posts about coriolis forces, and points out that a falling body is deflected by them one eighth as much as one tossed from the ground to the same height, and that they’re deflected in opposite directions. Here’s my attempt to explain this intuitively.

This makes me think of the competing da Vinci – Galileo laws for bodies (not their own I hope) falling freely under gravity. They stated their rules in the same basic way. I remembered these laws from watching The Mechanical Universe in high school – before taking physics from the real life version of David Goodstein three years later.

da Vinci said (or so I hear, I never met the guy) that if you fall one unit of distance in the first unit of time, you’ll fall two in the next unit, three in the one after that, then four, etc. So if you fall 5 meters in one second, in the next you’ll fall another 10 for 15 meters total.

Galileo said almost the same thing, but with odd numbers. If you fall one unit of distance in the first second, then in the second you fall three, then five more, then seven, etc. So if you again fall 5 meters in one second, in the next you’ll fall another 15, for 20 total.

Galileo was right; da Vinci wrong. But let’s not screw over our primitive-flying-device-making friend with such a cursory examination. They’re both awesome dudes, as Leonardo’s testudine counterpart would say.

Galileo was right because acceleration is constant, so the distance fallen is proportional to the square of the time. Adding Galileo’s odd numbers gives a square number. 1+3+5 = 9, for example. This is easy to see from a picture.

Each new section adds the next odd number worth of dots, and takes you to the next bigger square number when counted as a whole.

da Vinci, instead of the square numbers for total distance fallen, gave the triangular numbers. 1 + 2 + 3 = 6, which is triangular. This has its own picture.

According to da Vinci, each new row is how much you fall in one additional second.

da Vinci’s fub may have been in misunderstanding the relationship between speed and distance. If da Vinci’s rule had been giving the speed at the end on each second, rather than the incremental distance fallen, he’d have been right. If you’re going 10m/s after one second, you go 20m/s after two, and 30 m/s after three, etc. The problem is that you can’t find the distance traveled in a second by taking the speed at the end of that second and multiplying by time. If you do that, you get only an approximation to the correct integral, like this:

Don't worry about the numerical details. I stole this from the internet somewhere. da Vinci's law overestimates distance fallen every second by assuming your speed at the end of the second was you speed for the entire second.

It’s possible that da Vinci was actually right on about the kinematics, but that he made a mathematical error in reporting his result. I wanted to follow up on this, so I checked online to see precisely what Leonardo said. I did not succeed. Fritjof Capra’s book quotes da Vinci:

The natural motion of heavy things at each degree of its descent acquires a degree of velocity. And for this reason, such motion, as it acquires power, is represented by the figure of a pyramid.

But when I search online texts of Da Vinci’s notebooks, I can’t find this passage. I can’t find the relevant passages in my Dover copy of Richter’s translation, either. In fact, I can’t find this passage anywhere else on the entire internet, except one book that doesn’t cite the source. So I’m not sure what to make of this. da Vinci’s writings on falling bodies must be somewhere, if we know about them. But as of now I’m still uncertain. Based on the preface to my translation of the notebooks, it looks like they decided to omit some of Leonardo’s physics, since that is obviously unimportant and uninteresting to readers of his notebooks.

Let’s assume Leo had the right idea, but brain farted on the integration thing. Considering how clever Da Vinci was, his mistake is very surprising, because his law is not only empirically wrong, it is logically impossible.

To see what I mean, let’s carry out Da Vinci’s argument a little further. According to his rule, in four units of time you fall 1+2+3+4 = 10 units of distance. But the choice of how long a unit of time is was arbitrary. So let’s do it again, but consider the unit of time to be twice as long as it was previously. We’ll call these “shmunits” of time. In one shmunit of time, you have to fall three units of distance to be consistent with the first calculation. Then you fall six units of distance in the second shmunit of time, because the second has you falling twice as far as the first. After two shmunits of time, you fall a total of nine units of distance. But we already said that with the same law you fall ten units of distance! Surely if Leonardo had considered his law carefully he’d have seen this error, right?

Unless it’s not an error. What if Leonardo actually meant that you have to take the limit as your unit of time becomes infinitely short? In that case, Leonardo’s law

$distance \propto t(t+1)$

can simply be reduced to the correct law

$distance \propto t^2$.

Could this really have been what Leonardo had in mind? I think it’s possible, but not likely. The Greeks explored the basic ideas here. They approximated $\pi$ using the method of exhaustion, and Archimedes is said to have been doing what amounted to integral calculus. If Leonardo was aware of this research, he might have stated such a law accurately. But it seems far-fetched.

### Reflections on the Moon

September 26, 2009

Q: Why don’t elephants play tennis?
A: They prefer squash.

Two players compete in a fast-paced game at the gym. I exercise across from them, watching as they smash a blue rubber ball in turns. The game is in a small indoor court, and the ball moves very quickly. Its motion, followed from a distance, appears nearly rectilinear. All of the walls, the floor, and the ceiling are in play, and the ball’s wild meanderings, jots of wild color, mesmerize me as I stretch.

The game is squash (Wikipedia), which I had heard of but not seen played before. Watching the squash players, I was at first surprised by their ability to predict the seemingly-chaotic motion of the ball. However, a geometric property of the court aids them.

The squash ball moves very quickly (faster than a major-leaguer’s fastball), so over the short distances of the court, we can approximate its motion as roughly a straight line. The court is a rectangular prism, and this shape has the property that if a player smashes the ball at a corner, the ball will pop back out right at them, parallel to the way it came in. In two dimensions this is shown in the following diagram:

The squash ball follows the dark blue path, bouncing off the corners at points A and B. The light blue line shows the continuation of the paths, with a hypothetical meeting point D if the incoming and outgoing lines are not parallel.

The ball comes into the corner bouncing at point $A$, making an angle $\theta$ with the wall. By hypothesis, it bounces off with the same angle, then comes into the next wall with an angle $\phi$ at point $B$. It bounces off with this angle as well and returns to the court. We’re trying to show that the incoming and outgoing paths are parallel.

To do this, I’ve drawn in light blue the continuation of the incoming and outgoing paths. If they’re parallel, they never meet, and the angle drawn as $\delta$ should be zero. Notice that $\theta$ and $\phi$ are the small angles of a right triangle $ABC$, so they add to a right angle. Angle $CBD$ is opposite $\phi$, and so equal to it. That means angle $ABD$ is $2\phi$ and similarly angle $BAD$ is $2\theta$. Those two angles, along with $\delta$ form the triangle $ABD$. However, they add to a straight angle by themselves, and so we must have $\delta = 0$, showing that the ball pops out parallel to the way it came in, allowing the players to predict its motion easily.

In three dimensions, this is just the same, except that you have to work through the argument over again. A student of mine pointed out a different argument to come to the same result. Set up and $x-y-z$ coordinate system at the corner along the intersections of the planes. Then one wall works by flipping the $x$-coordinate of the incoming ball’s velocity vector, and the other two flip the ball’s $y$ and $z$-coordinates, so that after bouncing off all three, the velocity vector is reversed.

In squash, this result is far from perfect because gravity affects the ball’s motion, and its rotation, along with friction from the walls, may affect its angle of reflection. Energy is lost in each reflection as well, and the ball will slide some against the wall, so all in all it’s a rough approximation.

For light the approximation is much better as long as the wavelength of light is much smaller than the size of the mirrors. The setup with three orthogonal plane mirrors like this is a called a retroreflector because it reflects light back the way it came. If you look into one from any angle, you will see your own pupil at the corner, because at the corner the incoming and outgoing rays are not only parallel, but on top of one another, so light must start and end at the same place after reflecting there. All the light you see ends at your pupil, so that’s what you see in the corner. If you have three hand mirrors, it’s an easy experiment to try.

One interesting application of this idea is shown here:

A retroreflector on the moon's surface. The Apollo missions left this array of hundreds of reflectors intended for extremely accurate measurements of the Eart-Moon distance.

This is a retroreflector array on the moon. When an Earth-based laser sends a pulse of light at the moon, the retroreflectors send the pulse back to Earth. If you could measure the trip time very precisely, you can multiply by the speed of light to find the distance to the moon. The APOLLO project (not the lunar orbiters, but the ground-based Apache Point Observatory Lunar Laser-ranging Operation) is trying to do this to an accuracy of one millimeter.

I’ve sometimes heard silly things like “the Campbell’s soup cans thrown out by Americans in a single month could stretch to the Moon and back three times.” I say this is silly because

• Why would you want to do that?
• I totally just made the statistic up because it is meaningless and nonmemorable. Things like per-capita consumption, percentages of usable land being turned into dumps, and statistics about ecological impact actually mean something.
• No they can’t. They would fall down if they tried.

Regardless, if you know the distance to the moon with one millimeter accuracy, then your estimate of how many Campbell’s soup cans away it is is limited by how accurately you know the size of a Campbell’s soup can until you measure the can to an accuracy of single atom (and Campbell’s soup cans vary from one to another by a lot more than that, and don’t stack perfectly regularly, and shift around, and get hotter and colder, etc).

There are a few good reasons you’d want to know the position to the moon so precisely. Perhaps the most striking is as an extremely tight test of the equivalence principle of general relativity. The Earth and Moon have different densities, and so might conceivably fall towards the sun at different accelerations, even when they’re the same distance away. Modern cosmology and theoretical physics frequently explore theories of modified gravity in attempts to explain the acceleration of the universe’s expansion or create quantum theories of gravity. If the equivalence principle doesn’t hold, watching the acceleration of the moon very closely could be the first place we’d get a hint of it.

### New Problem: The Kepler Exhibit

October 17, 2008

At the Exploratorium in San Francisco, you can play with this exhibit:

What should the height profile of the bowl be so that balls that roll without slipping (or, so that blocks sliding without friction) would reproduce 2-D Kepler orbits when viewed in projection from above?