Having explained the basic concept of gravitational barycentres, I can now get to the meat of what I wanted to talk about: Lagrange (or Lagrangian) points.

If we can find the gravitational barycentre between two objects, we should in theory be able to predict how they move based on their current position, velocity and their mutual gravitational position. Most of them time it’s not that simple, however. For instance, if we really want to plot the orbit of the Moon around the Earth, and do it accurately, we need to take other significant gravitational influences like the Sun into account. Predicting the orbital motion of the Earth-Moon-Sun system as they undergo mutual gravitational interactions is an example of what’s called the three-body problem, and it’s one that physicists and mathematicians have been trying to solve since the 17^th century. While solutions for special cases of the three-body problem have been found a general solution for all cases is still very much beyond us, and increasing the number of gravitational influences still further (up to n bodies, which is known as the n-body problem) makes things even worse. We can simulate the interactions of n-body systems using very powerful supercomputers but this is only an approximation, and it’s possible that the n-body form of the problem has no solution since the approximation involves an infinite series of expansions that has to be truncated at some point in order to get a finite answer; in other words, solving the n-body problem is kind of like trying to find the exact value of pi.

Lagrange points rise as part of the three-body problem, but happily it’s one of the special cases we have an exact solution for known as the restricted three-body problem. In the restricted version of the problem the third body has negligible mass in comparison to the other two – it’s a probe or an asteroid rather than a moon or planet – and so exerts practically no gravitational force on the other two bodies in the system. The problem therefore becomes one of figuring out how the two bodies that do have mass affect the movement of the third massless body, and since the solution to the two-body problem is known this is something we can do.

One quirk of the solution to the restricted three-body problem is that it indicates there are certain places where you can put the massless object that will cause it to – given the appropriate velocity – orbit the common barycentre of the system at the same rate as the two objects with mass. This is a fancy way of saying that it will be stationary with respect to the two massive objects and will be located at the same position relative to them at any point in their orbital system. These stationary points are called the Lagrange points, and there are five of them for every two-body system. For examples I’ll use the Earth and the Sun as my two-body system since they’re easy to understand, but Lagrange points also exist for the Earth and the Moon, the Sun and Jupiter and so on.

L1: Situated directly in between the Earth and the Sun, very close to the Earth. This is the point where the gravitational forces of the Earth and the Sun “balance”. Stuff orbiting closer to the Sun will do so at a faster velocity and over a shorter period, while stuff orbiting further away will orbit at a slower velocity and over a longer period. Ordinarily anything placed at the L1 point would orbit the Sun more quickly than the Earth and would thus be asynchronous with it; however, L1 is in just the right spot for the Earth’s gravity to provide a slowing force on any hypothetical body placed there that causes it to orbit the Sun with the same orbital period as the Earth.

L2: Like L1, but backwards. An object at L2 would ordinarily orbit the Sun more slowly than the Earth due to being further away; this time, though, the gravitational pull of the Earth works in the opposite direction, giving the L2 body a boost in speed and (again) causing it to orbit the Sun with the same orbital period as the Earth.

L3: Directly on the opposite side of the Sun to the Earth, L3 does not lie on the Earth’s orbital path as the diagram suggests; instead it’s just inside it as a result of the Earth-Sun barycentre being shifted slightly towards the Earth. L3 is the point where the combined gravitational pull of the Sun and the Earth act along the same line and provide a single unidirectional force that is equivalent to the gravitational force the Sun exerts on the Earth, causing an object placed there to orbit the barycentre just like the Earth does.

L4 + L5: The trickiest Lagrange points to explain, L4 and L5 lie at the apexes of a pair of equilateral triangles which have the Earth and the Sun as two of their corners. Objects at L4 and L5 are the same distance away from the Earth as they are from the Sun, forming an equilateral triangle! I was going to explain exactly how this happens but it involves trigonometry which I’m sure everyone is doing their damndest to forget, so I’ll just steal the following picture from Wikipedia.

Right, so, here the examples are the Earth and the Moon; the gravitational force of each combines into a single resultant force which is deflected to a point slightly away from the centre of mass of the Earth which just happens to be the barycentre of the Earth-Moon system, causing objects placed at L4 and L5 to – again – orbit the barycentre along with the Earth and the Moon.

So that’s what Lagrange points are and why they are. Why are they important? Well, the examples I used of our massless third body being an asteroid or a probe weren’t just for fun since that’s exactly the sort of thing that tends to be present at a Lagrange point.

L1: In the Earth-Sun system a probe placed at L1 will never be out of sight of the Sun (i.e. no eclipses by the Earth), making it very useful for solar observatories. In the Earth-Moon system you can stick a space station up there and use it as a halfway house for lunar exploration. I mean, if you really wanted to.

L2: Opposite of L1 (again); a probe at L2 will have almost all of the solar radiation from the Sun blocked by the Earth, so this is where we want to put space telescopes for the best result since solar radiation is the space equivalent of light pollution from cities. The successor to the Hubble, the James Webb space telescope, is going to be put at the Earth-Sun L2 when they eventually get around to finishing the damn thing.

L3: Kind of boring! It does kind of amuse me though because this is where some people thought you might be able to hide a secret planet; since it’s on the other side of the Sun it’d always be out of sight of the Earth. These people did not take into account the massive gravitational perturbations such a secret planet would cause, nor the fact that L3 is inherently unstable and no object can persist there over solar system timescales.

L4 + L5: You know how Jupiter gives asteroids, comets etc. that have the temerity to venture inwards towards the Sun a massive gravitational slap that sends them careening out of the Solar System towards parts unknown? The L4 and L5 points in the Jupiter-Sun system is where the lucky survivors of this process tend to congregrate; each of them is populated by a cluster of asteroids (well, I say cluster, there’s like a million of them) which are collectively referred to as the Trojans, and which make the asteroid distribution in the Solar System look like this.

The reason the Trojans are spread out over a fairly large area is that the regions around the L4 + L5 points where you can get a stable orbit are actually pretty large. This image maps gravitational potential as contour lines; while the L1, L2 and L3 points are all tight gravitational troughs, the L4 and L5 points are broad gravitational peaks that an asteroid can, with a bit of luck, perch on top of over long timescales (millions of years) without too much difficulty.

The post Home On The Lagrange. appeared first on The Scientific Gamer.

Barycentre, barycentre. It’s the kind of word that sounds like it should be easy to pun, but really isn’t.

In preparation for the post on Lagrange points coming later on I should start by defining the concept of a gravitational barycentre (or centre of mass). For a single roughly-spherical object like a planet we can treat this as a single point mass located at the very centre of the sphere – that is, if we took the planet away and replaced it with a single infinitesimally small object with the same mass located at the planet’s centre of mass, this tiny object would (broadly) be gravitationally indistinguishable from the planet. If we add in another spherical object – say a moon – and attempt to calculate the centre of mass of the planet-moon system, things get a little trickier. The planet exerts a gravitational force on the moon and the moon exerts a gravitational force on the planet, meaning that the gravitational centre of mass of the combined system will be located somewhere in between.

This is actually kind of a bad picture for illustrating the concept because m₁ is the heavier mass, and yet it’s rendered as smaller than m₂. Never mind. Pretend m₁ is made of something really dense, like neutronium. The centre of mass of a two body system is defined by factoring each body’s mass by its distance from some arbitrary reference point – in this case x₁ and x₂ – and dividing by the system’s total mass. Since the reference point is entirely arbitrary we can cheat and say it’s located at the centre of mass of m₁; this means that x₁ is zero, x₂ is now the distance d between the two centres of mass and the equation reduces to

 In other words the centre of mass of the two body system is located a distance X_cm from the centre of mass of m₁along a line connecting the centres of mass of m₁ and m₂. This is the gravitational barycentre of the system, and it’s very important when considering orbital mechanics because in a two-body system like a moon and a planet, or a planet and the sun, both bodies will orbit the gravitational barycentre. If you have two masses of comparable size like a binary star system the barycentre will lie roughly in the middle of the two bodies, leading to very obvious mutual orbital behaviour like this:

(This is actually wikipedia’s example for Pluto + Charon, but I don’t like the one for the binary star system because it implies the stars have exactly the same mass, which would be unlikely.)

If one body is very much larger than the other, however, then X_cm will be smaller than the radius of the larger body, which is another way of saying that the gravitational barycentre will be located inside the larger mass. The existence of the barycentre is much less obvious in this case since the centre of mass of the combined system is so close to the centre of mass of one of the bodies. For example, in the case of the Earth-Sun system the barycentre is located

or about 450km away from the Sun’s centre of mass. Given that the Sun has a radius of 696,000km this does not produce much appreciable movement on the part of the Sun! I mention this so that the following gif doesn’t give you the wrong idea.

This is a very exaggerated portrayal of the sort of motion we’re talking about here; in actuality for most planets the motion on the part of the Sun will be tiny, and even a really big planet like Jupiter can only shift the barycentre of their mutual orbits to roughly the surface of the Sun, meaning that it’s less of a mutual orbit than it is a slight “wobble” on the part of the Sun. Nevertheless this motion can be detected, and as previously mentioned it’s one of the methods we use to detect exoplanets orbiting other stars. Hopefully by this point it should be obvious why we’ve mostly only found really big exoplanets, since they’re the ones which produce the biggest wobble in the motion of their parent star.

Now, remember, this applies for all two-body systems. The Earth and the Moon orbit a mutual gravitational barycentre. The Sun and Pluto orbit a mutual gravitational barycentre. Even tiny satellites like the Martian moons Phobos and Deimos will cause Mars to shift slightly in position, orbiting a mutual gravitational barycentre. This is an important factor to take into consideration when plotting celestial trajectories, and it also gives rise to some interesting side-effects which I’ll tackle on Thursday.

The post Barry’s Guide To Barycentres. appeared first on The Scientific Gamer.

Josh, 7, from London, writes:

Dear Hentzau,

            I went to the beach yesterday. The weather was nice. I had ice cream. When I was splashing around in the sea with my rubber ring wedged firmly around my waist to stop me sinking, I thought about the planets. Planets have rings, but they do not have to be prevented from sinking. Why do planet rings always go around the middle?

(And they do always go around the middle: see Uranus, which is tilted almost ninety degrees on its side but which has rings which still orbit around the equator.)

Well, little Josh, it’s because of three factors which shall be tackled in numerical order.

1)      Planetary formation from a protoplanetary disc.

2)      The Roche limit.

3)      Planets are fat.

When a planet forms, it does so out of a big cloud of stuff. Much of this stuff is fairly hefty planetesimals (stuff one kilometre plus on a side) but there is lots of smaller stuff as well. The planetesimals along with the smaller stuff collapse towards the centre of the cloud towards the proto-planet core. As we’ve seen in the post on the Nice model, when you have a big cloud of stuff that’s shrinking it will start to rotate. Rotation produces centrifugal force, and centrifugal force will flatten the cloud out into a disc. Eventually you end up with the proto-planet at the middle surrounded by a big disc of stuff. What happens to the leftover stuff is determined by how close it is to the proto-planet.

• Stuff close to planet: falls onto planet.
• Stuff far away from planet: forms moons and stuff.
• Stuff in the middle: forms rings.

Now, clearly there are some niggling factors which mean it isn’t quite as simple as that, chief of which is that only the gas giants have rings (yeah some people think Rhea might also have rings but honestly screw those people). Earth, Mars and the other terrestrials don’t. This is because the size of the “stuff in the middle” category is dictated by something called the Roche limit.

The Roche limit is, broadly speaking, the orbital radius inside which the gravitational tidal force of the planet is greater than the gravitational force keeping whatever object has been unwise enough to venture inside it in one piece. In other words, if you go inside the Roche limit and gravity is the major force holding you together¹ – so I don’t recommend this at all if you happen to be a planet or a moon — you get pulled apart. I like the Roche limit because it’s one of the few physical terms that I actually understand the derivation of, but I won’t go into that here. It’s expressed as an approximation by:

where R is the radius of the primary (i.e. the planet), ρ_M is the density of the primary, ρ_m is the density of the thing orbiting it, and d is the Roche limit.

So the Roche limit isn’t strictly a hard physical constant; it’s going to vary from body to body depending on the ratio of the densities of the primary and the orbiting body. However, because R is also in there, if you have a really, really big planet – like a gas giant, say – the Roche limit is going to be a factor no matter how dense your orbiting body is.

Using FABULOUS EXCEL TECHNOLOGY, I have created a graph which shows how the Roche limit varies with the density of the orbiting body for Saturn and Earth.

 
This makes it very easy to see that the Roche limit is big for less dense things and small for more dense things, and also that as you increase the density more and more it becomes practically static.

Now, to get the Roche limit for your proto-planet you have to put a number on the average density of protoplanetary material. I’m going to use the mass of a comet, which is about 500 kg m^-3. This gives Roche limits of:

• Saturn: 160,000 km.
• Earth: 35,000 km.

35,000 km is sod all, in orbital terms. It’s just slightly less than the altitude at which we orbit geostationary satellites; any protoplanetary material orbiting inside 35,000 km is overwhelmingly likely to fall onto the proto-Earth anyway, so that explains why the Earth has no rings. The Roche limit of 160,000 km for Saturn on the other hand matches very nicely the observed outer edge of Saturn’s ring system.

This is where the bulk of planetary ring material comes from, then; they’re formed from the leftovers of planet formation that are far enough away to avoid falling onto the planet outright, but too close to the planet to form themselves into moons. Since this disc was rotating around the “equator” of the proto-planet anyway thanks to the cloud collapse, that explains why the rings are found there. However, this doesn’t answer an important question: what about ring material that was captured after planet formation? Say a comet ventures inside the Roche limit on a highly inclined trajectory. Why isn’t it pulled apart into a ring orbiting the planet at that inclination?

Well, this is where we come to the third thing on our list: planets are fat. Literally. Because they’re rotating objects centrifugal force distorts their shape away from the perfect sphere of hydrostatic equilibrium, making them slightly fatter around the equator than they are around the poles. For example, Earth has a radius 6,378 km around the equator but only 6,356 km around the poles; those 22 km may not seem like much but they make a big difference as far as orbiting bodies are concerned, because it means things that aren’t orbiting around the equator will nevertheless be pulled in that direction thanks to the extra mass that’s present there. Over thousands of years they’ll slowly migrate to the equator, at which point they’ll probably collide with some of the other junk that’s orbiting there which robs them of their orbital momentum and stops them where they are. Planets may be a little bit flabby around the waist, but if you like to look at ring systems a little extra weight is no bad thing.

In conclusion:

• The majority of planetary ring material forms around the equator of the planet in a disc.
• The inner edge of the rings is created by the planet, which drags material that gets too close towards it.
• The outer edge of the rings is created by orbiting moons, which sweep up material outside of the ring area.
• The rings themselves are defined by the Roche limit, which dictates the nominal area where ring material cannot form itself into moons because the planet is gravitationally dominant.
• All material added to the rings after planet formation migrates towards the equator thanks to the planet not being a perfect sphere and having more mass – and therefore gravity – distributed around the equator.

BULLET POINTS.

1. You might think of the Earth as a solid object but it isn’t really. It’s held together by gravity; no material could possibly provide the same amount of bonding force to keep all of its mass in one place. This goes for most large bodies out there; they’re so big that the only thing that can keep them in one piece is gravity, so if they come across something with a larger gravitational force then it’s bad news for them.

The post The Rochefort Limit. appeared first on The Scientific Gamer.

Saying that I have a Ph.D elicits a fairly predictable reaction from most people. They will, in an attempt to appear interested, ask “What subject?” and then when informed that I did Astrophysics – one of the simpler branches of physics if you don’t tangle with cosmology or relativity but which appears to have a fearsome reputation in the eyes of the layman – their eyes glaze over and they either stop talking to me altogether, or else they desperately try to change the subject before I can get a chance to pounce on them, knock them to the ground and inject pure Science into their brains via their ear canal¹. There’s a second type of person out there, however; the freakish sort who are genuinely interested in science, and this second type will, after some circumspect small talk, eventually get around to asking me what my thesis was about. And this is a question to which I have gradually evolved a tried-and-tested one-sentence reply:

“I am trying to find out how much energy you need to blow up Pluto.”

As with all one-sentence descriptions it doesn’t even come close to summing up the totality of my work, but it is accurate and it does its job of seizing their interest in an unbreakable choke-hold so that they ask the follow-up question:“Why?” Today, you’re all going to find out. You lucky, lucky people.

What does it mean to destroy something? Gormless techno-fetishist Michael Crichton inadvertently touched on why this is a bit of a tricky question in one of his godawful dinosaur novels, in which the shameless author self-insert of Ian Malcolm tells another character that his statement that nuclear weapons could destroy the world is really, really goddamn stupid. The reasons for this should be fairly obvious given the nuclear weapons post – while nuclear weapons pose a significant threat to humans, they could do very, very little to imperil the Earth. They’d scorch some parts of the surface, irradiate others, screw up the atmosphere for years – things that would have dire consequences for the future survival of the human race — but all this would amount to nothing more than a mild skin rash as far as the Earth is concerned. It’d just keep on truckin’ quite happily while we killed ourselves off.

Clearly you need a different magnitude of threat altogether in order to stand a decent chance of destroying the Earth. The sort of exotic cosmic catastrophe so enamoured of the Hollywood disaster movie notwithstanding (solar flares, quasars, supernovae, mini-black holes etc.), stuff smashing into other stuff is a thing that happens fairly often in our Solar System². What happens if we hit the Earth with a really big space rock? Well, that’s actually happened. The Earth was hit by a Mars-sized object very early on in its lifetime. Mars-sized.  The impact was so destructive we have a constant reminder that it happened in the form of the Moon. The impact was so destructive it blew off a majority of the Earth’s surface material. Yet even after this colossal impact event which would have fit most people’s criteria for “destruction” the Earth got better.

Key to this is that while the physical structure of the Earth was shattered by the impact, with bits and pieces flying in all directions, the heavy iron core was left mostly intact which meant the gravitational centre of mass of the Earth remained largely unchanged. Unless the impact fragments had been boosted up to escape velocity they were gradually and inexorably pulled back towards that centre of mass, with the result that the Earth slowly reformed much like Robert Patrick in Terminator 2. The Mars-sized impactor also left a remnant that accumulated bits and pieces of ejected surface material from the Earth; this went on to form the Moon, and it’s why the Moon is mostly made up of elements we’d expect to find in the Earth’s crust and mantle.

So even if you hit a planet-sized object really, really hard you’re not guaranteed to “destroy” it in any real sense of the term; you will disrupt its physical structure temporarily, but it may eventually reform into a single homogenous body again over millions of years. This destruction thing is a tricky business when applied to planets, and even when dealing with smaller stuff it’s difficult to draw a line. How much of an object do you have to destroy in order to say that the object itself is destroyed? A third? Two thirds? Total destruction? This is the first fundamental question that faces scientists who want to model impact events, and the answer they’ve come up with may seem arbitrary but it does have logic behind it.

First, impact scientists do not say impacts destroy something. We say they disrupt it, for the very good reason that disrupt is a term that can apply to any quantity of blown-off material, whereas saying something is destroyed runs into any number of problems up to and including the rather large one that matter cannot be created or destroyed, merely converted into energy. We define disruption by how much of the original body is left after we’ve whacked it with an impactor; in the case of my experiments I weighed my targets before they went into the gun, shot the crap out of them, and then weighed the biggest remaining chunk I could find afterwards. If that chunk was less than half the mass of the original body I’d rip open my shirt and beat my chest while let out an Arnie-in-Predator-esque roar of “DISRUPTION!”

The fifty-percent mark delineates disruption from cratering. If an impact removes less than fifty percent of an object’s mass it doesn’t count as disruption since the majority of the original object is still intact. Technically it’s a crater. It might be a really, really, really big crater – a crater that’s almost larger than the remaining mass of the body – but it’s a crater nonetheless. If the impact blows off more than fifty percent, though, it’s a catastrophic disruption outcome. This is as close to the word “destroyed” as you’ll ever get an impact scientist to go; personally I think the work in my thesis showed there’s a more granular range of impact outcomes than the simple binary choice of disruption and cratering described here, but my work was never published so eh.

So we’ve got a technical definition of “destroyed” now. In order to get there, we need to hit the object with enough kinetic energy to permanently disrupt more than 50% of its mass. Kinetic energy is worked out by the equation

where m is the mass of the impacting body and v is its velocity. Using kinetic energy allows us to make useful comparisons between an impact by a very small thing moving very quickly and an impact by a very big thing moving very slowly – i.e. it essentially makes the analysis independent of what the impactor actually is. We can further remove dependence on the size of the thing being hit by dividing the impacting kinetic energy value by the mass of the target; this gives the energy density of the impact, or Q, measured in joules per kilogram.

Q is, in theory, a very, very useful quantity. Say you have one target weighing 1 kilogram and another target weighing 1000 kilograms. By carrying out a series of impact experiments on 1 kilo targets in a lab environment you have determined that the energy density at which the 1 kilo target will lose more than 50% of its mass – the energy density at which it will suffer catastrophic disruption, or the critical energy density Q* — as 50 J kg^-1. In theory, it should then be possible to predict how much energy it’ll take to disrupt the 1000 kilogram target without having to manhandle any one-tonne weights around your laboratory; you just multiply by a thousand and bam, that’s how much energy you need to pump into it to remove 50% of its mass. In theory, we can predict how planet-scale masses will behave under impact using the same measurement we took from the 1 kilo target.

In practice, though, it doesn’t really work like this. You have a number of factors that alter Q* depending on the size of the target. At first, as you make a target larger, it actually becomes easier to disrupt, pound for pound, than a comparable smaller target. This is because cracks propagate far more easily through large objects than they do small ones, and it is crack propagation that breaks off the huge chunks of material required for disruption. So Q* will initially decrease as target size increases. Then, when you’ve increased the mass of the target to the point where its gravity starts to become a significant factor (as explained in the Earth impact example above), this trend reverses itself. From this point onwards the bigger – and therefore heavier – the target, the harder it is to disrupt, since you’ve not only got to blow the thing apart but you’ve also got to do it with enough force that you give the majority of the fragments escape velocity.

This is a graph from a paper published by Willi Benz and Eric Asphaug³ in 1999. Benz is a computer modeller while Asphaug does impact experiments in the laboratory; and they’ve combined their data in an attempt to put some numbers on how exactly Q* will change with target size. At first Q* decreases; the initial value of Q* and the rate at which it decreases will be heavily dependent on the material the target is made of – rock is harder to disrupt than ice, and so on – causing this to be referred to as the material regime. Then at target sizes of around 100 metres – 1 kilometre Q* starts to increase with target size; this is the gravitational regime. What’s really interesting about these graphs is that they’re for two wildly different materials – solid ice has a Q* of about 10 – 40 J kg^-1, while basalt is roughly twenty times that – and yet the shift to the gravitational regime occurs at exactly the same place and in exactly the same way for each. This means that once you hit a target size of 100-1000 m, the material strength of the target has entirely ceased to matter when calculating the result of an impact. It is instead the body’s self-gravity which must be overcome in order to disrupt it

My, that all got a little bit technical. If you didn’t understand the fine detail don’t worry; I didn’t explain it that well and it’s not necessary to grasp all of it to understand what I was doing with Pluto. All you need to take away from it is this:

• We can do impact experiments in the laboratory to get a value for the amount of energy we need to hit a target with to blow it up.
• We can then take that energy value and, allowing for variations in Q* with target size as per the above graphs, scale it up to larger bodies to predict how much energy we’d need to hit them with to blow them up.

So that’s how I was trying to find out how much energy we’d need to blow up Pluto. But why Pluto? And why is it important? Unfortunately I’ve already rambled on a little more than I really intended to, and thinking about it I probably need to explain some additional background before I get on to that. So it will have to wait a week or two. Sorry!

1. Presumably.
2. Over a timescale of hundreds of millions of years, that is.
3. Fun fact: I met Eric Asphaug at a conference in Spain and stole all his water.

The post I Am Become Q*, Destroyer Of Worlds. appeared first on The Scientific Gamer.

A load of old balls.

Kicking off the science portion of this blog, I’m going to start with an easy question I used to get asked a lot when I did Outreach for the university: why are the various planets, moons etc. round? It’s a fairly simple answer with some wide-reaching ramifications.

Baldly (or possibly badly) put, it’s because gravity. Gravity is constantly pulling every piece of matter that makes up a planet like the Earth towards its centre of mass – in this case the Earth’s core. The Earth is probably a bad example to use here because of its complicated internal structure, but it doesn’t matter that planets start out being made of what we might think are strong, non-malleable materials like rock; thanks to internal heating provided both by the compressive effect of the gravity itself and the slow decay of radioactive elements inside the planet in question, the material inside it is made just “soft” enough that it will deform under pressure and slowly flow inwards over millions of years.

But if gravity is constantly pulling every bit of a planet towards its centre of mass, what’s stopping the Earth from shrinking and shrinking until it collapses into a black hole? The answer is — again–  pressure. If you have a chunk of stuff, and you exert some external force to reduce the volume of that stuff, you consequently increase the pressure inside the stuff that pushes outwards and resists your external force. Think about compressing air in a bicycle pump: at first it is easy since the internal pressure inside the pump is low, but as the air is compressed into a smaller and smaller space working the pump becomes harder and harder until you eventually reach the point where the pump pedal won’t work even if you put all your weight on it. The pressure of the air inside the pump is now equal to the force exerted by your body weight – in other words, it has reached a state of equilibrium with your body.

The same principle applies to planets. As the planet is compressed under the force of its own self-gravity, the internal pressure pushing outwards and resisting this compression will start to rise. The smaller the planet gets, the greater the internal pressure. Eventually the outwards force provided by the internal pressure will equal the inwards force exerted by the planet’s gravity, and the planet will stop shrinking. The planet has reached what is called hydrostatic equilibrium, and since the two forces balance each other out at a fixed distance from the planet’s centre of mass the effect is kind of like taking a pair of compasses, setting them to a fixed distance, and drawing a big circle. Except in 3D, obviously.

Now, a few notes and corollaries. When the IAU introduced the new classification of dwarf planets a few years ago, one of the criteria a body needed to have in order to qualify was that it had to be in hydrostatic equilibrium – i.e. round. This was a handy way of delineating the boundary between planetoids (planet-like bodies such as Pluto) and regular bog-standard asteroids and comets, since the state of hydrostatic equilibrium requires a certain level of gravity in order to start the compression process, which in turn requires a minimum amount of mass. You won’t find any bodies below about 400 km diameter which have compressed themselves into this round shape; they’re not big enough. This is why even the larger asteroids retain their rocky, irregular shape, and it’s also why the one asteroid that was large enough to achieve hydrostatic equilibrium (Ceres) got reclassified as a dwarf planet.  However, even with a few of the smaller planetoids we find that they’re not perfectly round. Other forces besides their own self-gravity act on them and distort their shape. Tidally locked satellites like the Moon will always be facing the same way relative to their parent body and thus the parent body will always exert gravitational tidal forces on the same “side” of the satellite; this can warp the shape of the satellite if it is particularly small and the parent body is particularly large. Rotation also has an effect – even the Earth is 20 km fatter at the equator than it is at the poles thanks to its rotation, and in the further reaches of the Solar System you can find weird things like Haumea, which has a rotational period we scientists refer to as batshit insane and which has severely deformed it into an ellipsoid.

Hydrostatic equilibrium applies to stars too, except in their case they’re so big that the internal pressure due to compression is nowhere near enough to stop a star’s mass from collapsing in on itself. Instead it’s the fusion fire burning at the heart of every star which provides the outwards pressure necessary counterbalance the star’s immense gravity. But what happens when a star runs out of fuel and that fire winks out? That’s a process that can be described by several phrases. Interesting. Lethal to anyone standing within about a hundred light years. And very definitely a post for another day.

The post Why Are Planets Round? appeared first on The Scientific Gamer.