Designing a spaceship with spin gravity - part 1

I’m working on designing a hard-ish sci-fi spaceship for a project with @mxmy. It’s a generation ship, but the journey time is reduced by some kind of magic thing to make wormholes, and the crew live inside spinning ring sections which provide gravity while the ship is not under acceleration.

By hard-ish, I mean I’d like it to seem ‘realistic’ to someone who has some idea how spaceships work, but not have everything calculated in full engineering detail, and allow some black-box-y future tech.

The builders of the spaceship are a future society of humans who live on Mars. Previous ships have travelled to other parts of the solar system, and this is the first one to go interstellar. It is, I imagine, a huge investment for the people who built it.

The destination is, if I recall right, the planet we currently call Kepler-186f, which is 558ly away - a huge distance, which makes it lucky they have the wormhole thingy.

The magic wormhole thingy can create a wormhole connecting a point ahead of the ship to a certain distance away, determined by how well it can be imaged by the ship’s instruments. This will allow, I presume, FTL. I don’t know all the details of this thing. (Heck, Maki just kinda vaguely said ‘space-folding’, but the way she described it sounded wormholey so I’m using that language). Not sure the targeted journey time, but with the wormhole thing we can probably finagle things pretty arbitrarily.

The ship otherwise accelerates like an ordinary spaceship, with some kinda rocket engine. Is it fusion? Antimatter? I don’t know, and it may not have been decided!

This post is mostly worrying about the spin gravity.

Turn those spin sections

Here’s a very rough not-to-scale initial design for such a spaceship. Engine at the back (it would need to be far enough away to not irradiate everything, since any interstellar engine is probably radioactive), some kind of instruments at the front (also maybe the wormhole machine), and a middle section with two counter-rotating spin sections on spokes. (I was told there’d be three spin sections, but for now I’m drawing two since it nicely balances angular momentum).

(apologies for the video quality and lack of looping. It does not appear to be possible to embed a Tumblr video in a text post, so we’re stuck with a suboptimal Youtube video. I rendered this video too small for Youtube to go above 360p, so it’s kinda blurry!)

There are various potential problems with such a design:

Spinning the spin sections up

The initial spin up can be achieved either using electromagnetic acceleration (similar to a maglev train) or using small rockets. Small rockets are limited by the amount of propellent, electromagnetic acceleration requires generation of strong electric currents.

To maximise torque, forces should be applied at the outer edge of the spin section rather than at the centre. This is easy for rockets: they can just be mounted on the outside of the spin sections.

For electromagnetic acceleration, the sections migh be spun up during construction (some kind of cradle), or else you might have a stationary boom holding EM acceleration devices like so:

In this picture, the yellow parts are supporting EM acceleration thingies at four points in the ring, and each ring has some kind of track running around it (also yellow) that responds to the accelerators. By applying electromagnetic forces through these accelerators, the wheels can be spun up. By accelerating both wheels at once, the central axle remains unspinning.

Why have this arrangement? It may be useful to adjust the spin of the wheels during flight, in order to e.g. compensate for friction losses, adjust the gravity in the ring, or store energy in the rotating wheels (like giant flywheels).

Frictionless bearings?

Unless the point where the rings join the axle is nigh frictionless, the rings will spin down over the flight.

One way to make a frictionless bearing might be to also make it magnetic - essentially each ring is something like a maglev train, or in another view, magnetic forces keep the axle from getting too close to the inside of the wheel hub. How hard would this be to engineer? I don’t know. The advantage is that there is no contact at all between the spin sections and the axle. A disadvantage is that the system only works where electricity is flowing, so it is a drain on the ship’s power systems.

I worry in such a configuration that when the ship is under acceleration, the electromagnetic bearings would not be able to hold the wheel in place. Perhaps the acceleration booms could also help keep the wheels spinning in place…

More of a problem is transforming people and materials between the rings and the hub.

It might be possible to generate electricity and carry water etc. inside the rings themselves without having anything pass anything out from the bearings. However, the greater the mass in the ring sections, the greater the moment of inertia of the rings - which means they would take more energy to spin up or down. So having heavy power plants or water tanks out in the rings may be a problem.

That said, having a water buffer around the edges of the ring sections would be a useful protection against space radiation. I suppose it may also be possible to transfer power using electromagnetic radiation to the spin sections, allowing electricity to be generated on the central axle and transferred to the rings without requiring contact.

This suggests a design where the ring sections are each nearly entirely sealed environments, with the only thing coming from outside being electrical energy. Food is grown, water is stored, used, cleaned and recycled, materials are processed and used entirely within the ring. The living sections are completely surrounded by tanks of water which absorb radiation from outside.

This is all great… until you want to get from one ring section to the other. That requires either a pressurised route from one section to the other, or a pressurised container - a capsule or a spacesuit - to crawl down one of the spokes, transfer to the other ring’s spokes, and crawl back out.

That doesn’t sound too unreasonable actually. Let’s add some little cars to our picture.

So in this arrangement, the car (strong green) crawls down the track on the inside of a spoke, waits until it’s lined up with a track on the axle, jumps over to the axle track, crawls over to the other hub, jumps over to the spoke track, and crawls back up to the outside of the other axle. It will be moving relative to the target track at each ‘jump’, so there will have to be some means of ‘catching’ it.

This will involve a few transfers of angular momentum, but because the car is much lighter than the rings, relatively minor ones.

Now I’m kind of wondering though… apart from the ‘car’ is there any reason to have the ‘spokes’ at all? Could the ‘wheels’ be entirely supported at the rim?

But as long as there’s reason to go down up to the centre, I guess we should stick to this arrangement.

As for what it’s like if you’re a passenger… Suppose you live in the aft ring. The outer surface of the ring is ‘down’, and the inner is ‘up’. If you want to go visit the centre or the fore ring, you climb up into a little car thingy. The car rises ‘up’, and as it gets closer to the centre, the ‘gravity’ you feel goes down to almost nothing. At the centre there’s a ‘clunk’ as the track passes you over to a suitable cradle on the hub. What little ‘gravity’ remains disappears: you are (assuming the ship is not under acceleration) in freefall.

What else is missing?

So without getting into the details of what happens inside the spin sections, we have the basic outlines of a spaceship.

But there are still important parts this ship is missing. In particular, there is nowhere to store reaction mass, and no way for the heat generated by the engine and living sections to be safely radiated into space.

Let’s add some stuff to the engine area. Big chunky reaction mass/fuel tanks, and some heat radiators.

I have no idea if the remass tanks or radiators are the right size, but then, this diagram has never been attempting to be to scale, and getting them right would require more effort at calculation. Assuming the material in the tanks is of a comparable density to the rest of the ship, and judging by eye that they’re of similar volumes this means it has a mass ratio of only about 2(ish). That’s not going to be a huge amount of delta-v.

Also, hold on a minute, wouldn’t the radiation from engine irradiate these fuel tanks? And also probably the crew rings The ship may need to be a lot longer and the tanks a lot narrower to employ a shadow shield, similar to the NASA Arthur C Clark design. I’ve also moved the angular magnetic accelerators to be on the aft side of the rings, since this would hopefully help transfer force to the rings when the ship is under acceleration.

Still this is probably a decent starting point. Here’s another spinaround with all the various additions.

Next update we’ll see about getting everything safely into the radiation shadow of the engines, and possibly eliminating the wheel hubs?

Comments welcome! @erin-space-goat and @bloodthreadsaltglassandtears you might have some ideas how to develop this further?

How to fly a rocket (in space!)

This is the first post in a series to try and cover all the main concepts of flying rockets around in space in a reasonably accessible way.

I am, sadly, not an astronaut, but I’ve done most of a physics degree, and Kerbal Space Program has convinced me that rocketry can actually be really simple and intuitive. Hopefully these posts will be helpful if you want to write about science fiction spaceships or the like.

There is a small handful of equations. I’m using MathJax, so click through to my blog if you want to see them properly.

Why are rockets a thing?

On Earth, if we want to move somewhere, we can do so by pushing off something else: the ground (for walkers, cars, etc.) or the air (for aeroplanes, birds, etc.). In space, there is nothing to push against.

There are, basically, two options. Either you can push against what little there is (such as sunlight, or a huge laser array back home), or you can shoot part of your spaceship away from the rest of it, causing the remaining parts of the spaceship to move in the opposite direction. All space engines, not matter how complicated, amount to doing one of these two things, usually the latter.

Here’s a very simple illustration “rocket”: the arm drive. An astronaut picks up bowling balls in throws them out the back of her rocket. This makes the rocket go forwards.

Jargon: the stuff that you throw out the back of your rocket is called reaction mass. This is distinct from fuel which provides the energy, though in the case of chemical rockets, the fuel and reaction mass are one and the same.

You use a rocket to change your velocity. Velocity is a thing that combines speed and direction: an object going 5 metres per second north, and an object going five metres per second west have the same speed but different velocities.

The good news about being in space is that nothing is going to slow you down. On Earth, friction and air resistance always conspire to make anything moving relative to the surface of the Earth, or at least to the air in the case of an aeroplane, slow down and stop. In space, there is nothing. So whatever your velocity is, it will stay the same unless affected by gravity.

So a rocket consists of two parts: the thing that you actually care about (sometimes called the payload) plus the structure of the rocket, and ‘reaction mass’ which will be launched out of the rocket to change its velocity.

Mass ratios and the almighty rocket equation

OK, here’s where it gets sticky. When you turn on the rocket, it doesn’t just have to move the rocket, but also all the reaction mass you’re still holding on board to throw out later. So the amount of reaction mass you need to change your velocity by a certain amount increases much faster than the velocity change you want to make.

Some more jargon: turning on your rocket to change your velocity is called a burn. The ratio of the mass of the rocket at the beginning of the burn \(M_i\) to the mass ratio at the end of the burn \(M_f\) (after losing some reaction mass) is called the mass ratio. The change in the velocity of your rocket is called the delta-v (“delta vee”), or \(\Delta v\).

(Reason for the name: the greek letter Delta \(\Delta\) represents a change in something: \(\Delta v\) is “change in \(v\)”. \(v\) is the usual symbol for velocity.)

There’s an equation called the Tsiolkovsky Rocket Equation which says that the change in the velocity of a rocket after a burn depends on the exhaust velocity \(v_e\) and the mass ratio \(M_i/M_f\) according to $$\Delta v = v_e \log \frac{M_i}{M_f}$$What this means is that the mass ratio is exponential in the delta-v: if you want to change your velocity by twice the amount, you need the mass ratio to increase by a factor of 7.4; if you want to change your velocity by three times the amount you need to increase your mass ratio by a factor of 20.1; it quickly becomes ridiculous.

If you’d like to know more about the maths, I work through the derivation of the Tsiolkovsky rocket equation and its relativistic counterpart here.

Practically speaking, only so much of your rocket can consist of reaction mass. The Saturn V used in the Apollo moon missions had a mass ratio of about 60, meaning 60 times the mass of the payload (the various modules used at the moon) was expelled either as reaction mass or as spent stages. In other words, the vast majority of material in your rocket is not actually useful, but just going to be thrown out the back.

This also places harsh limits on the amount of mass you can carry. Every kilogram of stuff you carry in your rocket requires you to carry tens or hundreds of kilograms of reaction mass to get the same delta-v. Rockets must be as light as possible and there’s no real way around that.

The two important numbers: thrust & specific impulse

To improve the amount of velocity change you can get from a given mass ratio, the only thing you can do is to increase the exhaust velocity, also known as your specific impulse (a name coming from the fact that it’s the force divided by the mass flow rate, 'specific’ being used by scientists to mean 'per unit mass’).

A chemical rocket, though easier to build than other kinds of space engine, has a relatively low exhaust velocity, around \(4,000\unit{m s^{-1}}\). Other propulsion systems, such as ion drives, can achieve much higher specific impulses; Wikipedia offers around \(30,000\unit{m s^{-1}}\) for a typical ion thruster, or more for more speculative designs such as the VASIMR.

The drawback of ion thrusters, and some other high specific impulse propulsion systems, is that they have very low thrust. Thrust is the other measure of a rocket: it’s just how much force a rocket applies to the ship when it is running, and hence how quickly it changes the velocity (the acceleration - note unlike everyday use of 'acceleration’, to physicists 'acceleration’ includes both changes of speed and direction). A low thrust rocket can achieve the same changes in velocity as a high thrust rocket, but it takes much longer to do it.

So in general the change of velocity in a burn depends on the thrust, mass of the rocket, and the duration of the burn.

In the distant future, much more speculative designs exist such as fusion rockets that have both high thrust and high specific impulse. But presently we are forced to choose.

High thrust is very important when leaving the ground. But we will come to that later.

Spaceships have their engines turned off most of the time

Because mass ratios increase so quickly, rockets can only change their velocity by so much before they expend all their reaction mass. This is called the delta-v budget of a mission.

Following the metaphor, a spaceship will 'spend’ its delta-v budget to perform maneuvers. Maneuvers - such as changing from one orbit to another - have a certain delta-v 'cost’.

The rest of the time, the rocket is inactive. Generally the rockets will only burn for a tiny portion of the mission.

This is unintuitive, since we’re used to Earth vehicles which are constantly losing energy to friction and air resistance, and therefore must have their engines running all the time. Sci-fi films frequently decide they’d prefer to show spaceships with the engines glowing, even if the ship is not performing a maneuver.

(Low-thrust engines such as ion drives are a semi-exception, in that they must burn for much longer than high-thrust engines to complete maneuvers, as mentioned in the previous section. But they still only activate their engines when they are changing their velocity.)

Orbits

Usually your rocket is not just hanging out in empty space; there is something big and heavy such as a planet or star in the vicinity that will change its trajectory through gravity. So lets talk about orbits.

When there are lots of planets, moons, etc. around it gets quite complicated, but we can start from building up from the simplest case: when your rocket is orbiting just one thing.

There are basically two kinds of orbits we’re concerned about: elliptical orbits and hyperbolic orbits. There is a third kind, parabolic orbits, which we won’t worry about because they only occur in ridiculously specific circumstances.

Elliptical orbits (space eggs)

Elliptical orbits are what you’d generally think of as an orbit, as in, your rocket’s going round a thing. They look like this:

Mathematically, this is an ellipse with one of the two foci (singular 'focus’) on the centre of the massive object (planet, star, etc.). For simplicity’s sake, we’ll call the thing it’s orbiting a planet.

Elliptical orbits are characterised by a parameter called the eccentricity, which is usually written \(e\). The eccentricity tells you how how long and narrow the ellipse is. A circular orbit has an eccentricity of zero. An orbit with a higher eccentricity moves further away from the thing it’s orbiting, and back, over time. Elliptical orbits all have eccentricities between zero and 1.

There are two important points on an orbit: the periapsis and the apoapsis. (These have special names in some cases: for example if something is orbiting Earth we call them the perigee and apogee, if something’s orbiting the sun the perihelion and aphelion, etc.). The periapsis is the point on the orbit closest to the planetg. The apoapsis is the point furthest away.

The second important parameter is the semi-major axis, usually written (a). This is just how big the orbit is: it’s half the length between the periapsis and apoapsis. For a circular orbit, the semi-major axis is the radius of the circle.

The periapsis is a distance \((1-e)a\) from the centre of the planet. The apoapsis is a distance \((1+e)a\) away.

The velocity of a rocket on an elliptical orbit is constantly changing, but its direction is always a tangent of the ellipse. And now an important point: the speed of a rocket is faster when it’s near the planet, and slower when it’s further from the planet. As the rocket comes in towards the periapsis, it picks up speed; as it moves out towards the apoapsis, it loses speed. This means the rocket spends a lot more time in the further away parts of the orbit than the nearby parts of the orbit. (The exact speed it has is precisely determined by for example the vis-viva equation, but I’m trying to avoid giving too much detail here.)

Here’s an animation to illustrate this principle:

[modified from this animation by Brandir on Wikimedia Commons; license CC BY-SA 3.0.]

Which orbit does a rocket actually follow? It depends on its position and velocity: every combination of position and velocity corresponds to a different orbit. For example, here’s a picture of a succession of orbits that each pass through the same point in the same direction, but with different speeds. For some of these orbits, this point is the apoapsis; for others it is the periapsis. Dividing these two groups of orbits is a circular orbit. There are two orbits shown for each eccentricity, with different semi-major axes.

So far, we’ve been showing two-dimensional representations of orbits, all in the same plane. However, an orbit can be in any plane containing the central planet.

Astronomers have a number of different ways to describe orbits. In general, you need five numbers to pick out a specific orbit, and a sixth number to identify a position on that orbit. One way is to pick a position (three coordinates) and velocity (three components). Another important way is the Keplerian orbital elements.

Hyperbolic orbits trajectories

Hyperbolic ‘orbits’, or if you prefer hyperolic trajectories, are what happens when your spaceship is going too fast to stay in orbit. This is the course you follow if you’re escaping from a planet’s gravity, or if you’re doing a flyby.

A hyperbolic orbit looks like this:

Mathematically, this is one of the two branches of a hyberbola. Like with the ellipse, its focus is on the centre of the planet. It has a periapsis, but no apoapsis. Many of the same rules apply: the spaceship gets faster as it gets closer to the periapsis, and slows down as it moves away.

As the ship moves further and further away from the planet in its hyperbolic orbit, it gets closer and closer (asymptotically) to moving on a straight line. This makes intuitive sense: in the absence of a planet, the ship would move in a straight line, and the planet’s gravitational influence gets less and less significant the further away you travel.

Like ellipses, hyperbolas have a parameter called the eccentricity \(e\). While for an ellipse, the eccentricity is always greater than or equal to zero, and less than 1, for a hyperbola it can be any number greater than 1. The reason both kinds of curves have this parameter is because hyperbolas and ellipses are both members of a family of curves called conic sections. (The parabolic trajectory, which we mentioned above but won’t discuss in detail, has an eccentricity of exactly 1. It’s nigh impossible to get onto a parabolic orbit in real life, and it’s pretty much the same as a hyperbolic orbit so we won’t worry about it.)

Here’s a quick summary of possible cases for the eccentricity:

For a hyperbola, the eccentricity basically measures how much the ship’s path curves around the planet. A spaceship following a hyperbola with a high eccentricity barely has its trajectory changed by the encounter with the planet. One with a low eccentricity can have its trajectory changed dramatically.

Like with an ellipse, the hyperbola also has a parameter called \(a\), again called the semi-major axis, though the reason for that name is less obvious. The periapsis is a distance \((e-1)a\) from the focus.

Here are some hyperbolic trajectories through a point:

The eccentricity can go as high as you like. The higher it goes, the closer it gets to a straight line. This corresponds to going very quickly past something very light that barely affects your course.

Energy & angular momentum: the unchanging numbers

Here’s a more mathematical bit.

So one important thing about orbits round a single thing like this is that there a two special things you can calculate at any point on your orbit, and you will get the same answer every time. These are the orbital energy and the angular momentum. Physicists will call these numbers conserved quantities.

These unchanging numbers have to do with the symmetry of the situation, which is something I will talk about in another post on Noether’s theorem at some point, eventually (if I keep promising to write this post, maybe it willl happen!)

Here’s the orbital energy $$E=-\frac{GMm}{r}+\frac{1}{2}mv^2$$where \(M\) is the mass of the thing you’re orbiting (keep calling it a planet), \(m\) is the mass of your rocket, \(r\) is your distance from the planet at a given moment, and \(v\) is your velocity at the same moment.

The first term is called the potential energy, and it gets more and more negative as you get closer to the planet. The second term is the kinetic energy and it gets bigger as you go faster. So as you go closer to the planet, your potential energy gets more negative, but your speed goes up by an amount that exactly counters that change so overall the total energy stays the same.

Divide out the mass of your rocket and you have the specific orbital energy (there’s that word specific meaning ‘per unit mass’ again). So the specific orbital energy is $$e=-\frac{GM}{r}+\frac{1}{2}v^2$$This is a handy quantity because it’s the same for anything on a given orbit, no matter how heavy the orbiting object.

The specific orbital energy is important because you can relate it to the size of the orbit, according to $$e=-\frac{GM}{2a}$$ for an ellipse, and $$e=\frac{GM}{2a}$$ for a hyperbola, where \(a\) is the semi-major axis in both cases. (This means an elliptical orbit has positive specific orbital energy, and a hyperbolic orbit has negative specific orbital energy). So an orbit with a lower specific orbital energy is smaller than one with a higher spefici orbital energy.

The orbital angular momentum is $$L=\frac{1}{2}mr\dot{\phi}^2$$ where \(\dot{\phi}\) is the rate of change of the angular coordinate, basically a measure of how quickly you’re going around the planet without regard for how much you’re going towards or away from the planet. So as you get closer to the planet, you don’t just need to be going faster in general, you need to be going faster around the planet. But we won’t talk about this one so much in this post.

The point of this is: these numbers are unchanging over a given orbit. So you can think of rocketry maneuvers and other things that affect the speed of your rocket changing these numbers, and therefore changing the orbit you’re on.

If you reduce your orbital energy, you will generally be put on a tighter orbit that goes closer to the planet.

For example: if you are flying through a thin atmosphere, you will lose orbital energy. This will put you onto a smaller, tighter orbit with a lower orbital energy. You might quickly regain the speed you lost as you move down to a lower orbit, but you’ve lost orbital energy. Since the orbital energy is unchanging under most circumstances, it’s easier to talk about that than your speed, which is constantly going up and down.

Six directions

Right, here are the things your rocket might do.

First of all we’re going to have to name some directions! The “mathematical” names are the terms used in the Frenet-Serret formulas. The other names are the ones used in Kerbal Space Program.

Prograde (mathematically, tangential) is the direction your rocket is going in in its orbit.

Retrograde (mathematically anti-tangential I guess) is opposite the direction your rocket is going in.

Radial and anti-radial, or radial-in and radial-out (mathemtatically, normal and anti-normal), are perpendicular to the direction of your orbit, in the plane of the orbit. It’s a matter of convention whether “radial” is inwards (roughly towards the planet) or outwards (roughly away from the planet). Kerbal Space Program uses the convention that “radial” is inwards. For clarity, I’m going to stick to “radial-in” and “radial-out”.

Normal and anti-normal (mathematically binormal and anti-binormal) are perpendicular to the plane of the orbit.

And yes, the makers of KSP decided to use a different meaning of the word “normal” to the Frenet-Serret thing.

These are independent of which way your rocket is pointing - they depend only on where you are, where the centre of the planet is, and which way you’re going.

Your rocket can in fact point in any direction. It does not have to fly “forwards” because there is no atmosphere to worry about streamlining. This is quite different from aircraft on Earth - an aircraft can fly in a way that’s not pointing straight forward, but it’s not usually a good situation to be in. A rocket, on the other hand, needs to point whichever direction is appropriate for the next maneuver, or to expose its solar panels to the sun, or spin to provide a sensation of “gravity” to the astronauts inside - all sorts of possibilities.

Since it can point in any direction, it can burn in any direction. But any burn can be broken into ‘components’ along these six directions. So we can talk about each of them individually.

The next post

We’re going to go into detail about the different directions you might want to burn in, and what they do to your orbit. And we’ll use that to talk about important maneuvers like circularisation burns, Hohmann transfer orbits, and how you go about making a rendezvous with another spaceship!

(And if you’re going to the Nine Worlds convention in London this year, I’m running a panel on rockets and rocket design with @darmokontheocean and @clockworkpipedreams covering basically the same ground as this post.)

Do planets close to the sun take more time to transit or less?

So yeah the transit of Mercury just happened, and I totally missed it. Oops. Not like I have a safe telescope projection rig anyway.

Nevertheless, this did make me wonder whether Mercury or Venus would take more time to transit.

Here’s some geometry, assuming a circular orbit for simplicity. We want to find the length of the orbital arc. \(x\) and \(y\) are angles because I couldn’t paste \(\theta\) and \(\phi\) into Inkscape for some reason. \(a\), \(r\) and \(l\) are known; the observer is stationary and coplanar with the circle.

So, peering at this diagram we say $$l \tan y = r$$ and also $$a \cos x = m$$ and also that $$m( \tan y + \tan x )=r$$ and the arc length is \(2ax\) (where \(x\) needs to be in radians of course).

OK, so let’s combine some things: $$a \cos x \left(\frac{r}{l}+\tan x\right)=r$$and that will allow a little rearrangement to $$\frac{ \cos x }{l} + \frac{\sin x}{r} = \frac{1}{a}$$

Well that’s like, probably impossible to solve analytically, but look we’re on a computer.

First of all here’s the angle \(x\):

So yeah \(x\) gets bigger as the orbit gets tight around the sun. I guess ultimately it’s going to go all the way up to \(\frac{\pi}{2}\) but Mathematica didn’t to show me that and it’s not important enough to fiddle around with Exclusions->None and all that.

OK let’s also check out the arc length

So the arc length is big when you’re in a really tight orbit round the sun, and otherwise just pretty much linearly decreases out towards the observer. cool

OK how much time does it take to cross this cone of vision? let’s grab Kepler. Kepler’s laws, which describe the motion of planets, say the square of the time it takes to go round the orbit (which we’ll call \(T\)) is proportional to the cube of the semi-major axis \(a\), which for a circular orbit is just the radius of the orbit. Neat. So that says $$T^2 \propto a^3$$

Of that time \(T\), a time \(t=\frac{Tx}{2\pi}\) is spent inside the transit visibility cone. That means $$t \propto x a^\frac{3}{2}$$ so let’s plot that thing.

So it looks like close to the sun is generally better, but there’s a local maximum some distance away from the sun. OK, what happens if we plot this in terms of more realistic relative sizes, i.e., set \(a\) to 1AU and \(r\) to the radius of the sun?

The hump is much bigger! And apparently for us Earth observers (or rather, a stationary observer at the orbit of Earth observing a coplanar circular orbit), the sweet spot is around 0.3AU.

Anyway that’s our answer for where it’s best to be if you want to spend a long time transiting the sun. Also, Mercury has a longer transit time than Venus. In reality it’s more complicated because orbits are elliptical and not coplanar, and the Earth moves during the transit too, so you only get transits sometimes and the time is a bit more complicated to calculate.

Time Travel

In a no-longer-recent ask (gosh it’s been five months now), I answered a question with a fairly narrow scope: if you brought back a modern phone or laptop, what use would it be in The Past? Could you reproduce it?

But ultimately questions of ‘bringing stuff back’ depend rather a lot on the model of time travel you adopt. So this is a post about time travel models, paradoxes, possible connections between time travel fantasies and actual physics (albeit rather flimsy ones), etc.

Contents:

Keep reading

if the earth suddenly vanished what would happen to the moon

common consensus seems to be that it would launch along the tangent to its orbit at a linear trajectory towards some other gravitational force. so what i propose is that we make all the other planets vanish so their moons launch to us and we can have all of them

Goals…

I’m pretty sure that that’s dead wrong but I don’t know the equations.

No, I think that’s right; without the force of the Earth’s gravitation to pull the moon into an eliptical orbit, linear momentum would take over and the moon would just fly off at a tangent from its current position,

hi, certified doing-absurd-things-to-the-moon expert here

So, this is roughly right, the moon would indeed continue in whatever path its momentum at the time of the disappearance gives, and this would initially be tangential to its original orbit.

That said, however, the moon would still be orbiting the sun. It would not follow a linear trajectory, but an elliptical orbit with a focus at the sun.

How different would the moon’s new orbit be from the earth’s current orbit? The earth’s orbital speed relative to the sun is about 30 km/s, whereas the moon’s orbital speed relative to the Earth is ‘only’ about 1 km/s.

This means that regardless which direction it’s going around the Earth, the moon would barely modify its orbit around the sun. It would follow a very very similar orbit to the Earth - probably slightly more eccentric, but basically the same.

I was going to open up Universe Sandbox and illustrate this, but setting up a clear illustration of the Earth, Moon and Sun is surprisingly difficult, so that may need to wait.

Here is the orbit of the Earth (blue trail) and Moon (white trail):

You can see the Earth and Moon’s orbits are almost indistinguishable.

Now, let’s see what happens to the orbit of the Moon when we delete the Earth at the point shown:

This is after one orbit of the sun. The fading-out trail is the moon’s original orbit as part of the Earth-Moon system; you can see it’s kind of bumpy due to the Moon orbiting the Earth. The stronger white trail is the Moon’s new, post-Earth orbit. It’s definitely more eccentric (less circular, more elliptical) than the Moon’s original orbit, but still very close to the Moon’s orbit before Earth disappeared.

people who don’t believe in aliens confuse me so much like you really think in all of space, which is infinite, we are the only planet with life on it. ok….

(psst I don’t disagree with you about aliens but there is a finite amount of stuff in space. it’s very large but it’s finite. space is infinite in direction in the sense that if you travel for long enough, you’ll end up back where you started.)

i don’t know what any of this means i’m a one direction blog

the one direction blog: Gemini, Aries, Sagittarius, Leo, Aquarius, Libra

the blog that knows about space: Cancer, Virgo, Pisces, Capricorn, Taurus, Scorpio

i hate this i’m a pisces

Actually, if I understand space right (maybe one of my followers who studies physics can confirm?) the thing about ending up where you started if you go far enough is only true if the universe has elliptic curvature, and last I heard evidence suggested that it was flat overall (some parts are slightly elliptical, some parts are slightly hyperbolic, depending on the local balance of mass and dark energy or something and how that works out with general relativity, and overall it adds up to flat. of course literally all of that, except some of the geometry, is way over my head so)

We could have a positive curve universe, negative curve, or no curve. If I recall correctly, positive curve means it’s like a sphere ( if I’m wrong, positive means a saddle shape and negative means a sphere), and we are confident it’s not a sphere because if it were, we’d be able to see ourselves if we looked far enough in any direction?

That’s true but the visualizations don’t make a whole lot of sense if you don’t know the properties they’re trying to describe. Positive curvature is elliptic and negative curvature is hyperbolic, yes.

Let’s say you and a friend are holding hands in a space with some curvature. There’s some distance between you, and you’re walking in the same direction at the same speed, exactly perpendicular to your arms, and each of you is following your own line. In a flat (Euclidean) space, which is the most intuitive, you remain the same distance apart no matter how far you walk.

In an elliptic space, no matter which direction you’re walking (as long as you’re walking the same direction), you’ll end up closer together just by walking forward, until you bump into eachother. If you and your friend are geometric points with zero width, you will eventually collide at the “pole” corresponding to whatever line your arms were on initially. Also, if you keep your arms on the shortest distance between you, the angle between your path and your arms will shrink as your arms end up leading you. Once you reach the pole, continuing onward and turning back around may be mathematically equivalent, depending on the construction of the elliptic space you’re in (though if it is, that’s because going in any direction and its opposite is always equivalent. believe it or not that makes the math simpler). If it isn’t, once you do you will again put space between you, reach a maximum distance at the furthest possible point from where you started, and then approach another pole (and then go back to where you started). You can sort of see how this will work by drawing great circles on the surface of a sphere (circles which share a center with the sphere; meridians are great circles, parallels are not), which is why this is said to be spherical

Hyperbolic curvature, on the other hand, does the opposite. If you and a friend walk like this, your arms will trail behind you, and will also have to get longer the more you walk; just walking distance adds space between the two of you (and the further you go, the faster it adds space). Like in euclidean geometry, and unlike in elliptical, in hyperbolic geometry this just continues without end. There is more and more space the further you go.

People call this “saddle-shaped” because drawing lines on an (idealized) saddle and then measuring distance along the surface of the saddle has some similar properties to this, but embedding something that infinitely expands inside a euclidean space that doesn’t is hard.

so a few things to add:

in general relativity, the “curvature of space” is determined by the mass density there. however, on enormous scales, we can safely model the universe as homogeneous and isotropic, which has the result that the curvature is basically the same everywhere.

when we talk about whether the universe is infinite or finite, we are talking about the topology of the universe rather than its curvature, which is a local property.

if you look up the shape of the universe on Wikipedia, you will find it awash with potentially unfamiliar mathematical terminology such as ‘metric space’, ‘compact’, ‘bounded’, etc. I’ll try and write it more simply here.

so it’s pretty much as @theomenroom said but that’s a bit of extra detail.

now, as for the global topology of the universe of a whole? we actually don’t know!

the problem is that the universe is really fucking big. according to current cosmological models, the parts of the universe we can see now comprise a big sphere 93 billion light years across. (hold on isn’t that bigger than the 13.8 billion year age of the universe? yeah! because space is expanding, and the light we’re receiving was emitted when the universe was a lot smaller. also, relativity makes ‘now’ a kind of complicated thing. that distance is the comoving distance for the present time, if you’re worried.)

now, if the entire universe was smaller than the observable universe, we might see stuff repeating in the sky, in increasingly early instances. but if the universe is bigger than the observable universe, we can’t necessarily tell whether it’s finite or infinite! awkward.

what we do know is that the curvature of the universe is really damn close to flat. so we probably don’t have to worry too much about that spherical and hyperbolic stuff. but is it a 3-torus, or an infinite space, or what? we don’t know and maybe we never will!

cosmologists are trying to work it out though. they try to work out what effect certain shapes of universe would have on the things we can observe, such as the cosmic micorwave background radiation. so far: inconclusive.

another cool fact: in an infinitely big universe, there are, statistically, infinitely many copies of everything, though they’re arranged randomly. how far away is the nearest exact copy of the observable universe? we can estimate this! i’m trying to find a good source for the calculation, but an old scientific american article by cosmologist Max Tegmark says a copy of you within 10^(10^28)m, a 100 light year sphere (corresponding to a matching 100 years of history) 10^(10^92)m, and the entire observable universe should have a copy within 10^(10^118)m. for comparsion the observable universe is a mere 10^27m across.

These numbers are bullshit big - if i wrote then out in this post as 10000… we’d need 10^28 bytes to write even the smallest distance (the one to an exact copy of you), which is to say 10,000,000,000 times the estimated amount of data on the entire Internet. I tried putting 10^(10^28) into Wolfram Alpha to convert it to light years, and the calculation returned no results.

If you got in a spaceship and tried to travel to the clone universe at a speed arbitrarily close to the speed of light, before you got even the unimaginably tiniest fraction of a percentage of the way, all the stars in the universe would burn out, the particles would decay (depending whether protons decay), and only black holes would remain; the black holes would decay into particles through Hawking radiation, and eventually the universe would contain nothing but photons that almost never encounter each other. You’re still not noticeably closer. By the time you get halfway to where the clone universe used to be, a new Big Bang might happen apparently (idk how they worked that one out). Meanwhile, the expansion of the universe is moving the clone universe even further away, not that there would be anything left of it by the time you get there.

An infinite space contains points separated by any distance, and it will contain infinitely many copies of our observable universe - as well as infinitely many near-exact copies that differ by only a few particles, and so forth.

Anyway, we don’t know the overall topology of the universe - it might not be infinite, or big enough to contain any such silliness as an exact copy of the universe.

A short(ish) post on Dyson spheres

“there is more tea in that sphere than the entire rest of the galaxy” - @rememberwhenyoutried

A Dyson sphere is a thought experiment about an engineering project: by encompassing a star in a sphere of solar collectors, all the energy it emits could be gathered. Dyson (who got the concept from earlier science fiction books, and wrote a paper about the possibility of detecting them) envisioned this as an inevitable creation of an ‘advanced’ civilisation with correspondingly enormous energy needs. For a star like our sun, that means trillions of times more than the current energy use of the Earth.

Dyson spheres show up in science fiction occasionally, usually as a solid shell of matter. (The post that inspired this one was a gifset of a solid Dyson sphere from Star Trek.)

That Wikipedia article is pretty substantial, and covers most of the things I might want to say about Dyson spheres. So all I want to do here is fill in some of the calculations the wiki article leaves out.

Statites (non-orbital satellites)

One of the proposed forms of a Dyson sphere is a swarm of ‘statites’ - enormous solar sail collectors held in place against gravity by radiation pressure. Wikipedia offers a value of the area density that would be necessary for a stable statite.

I have two questions about this: how was this value calculated, and is a statite stable against small peturbations to its orbit?

Calculating the value is pretty straightforward actually - I did it in physics in like second year?

So: solar radiation pressure. Photons carry momentum, depending on their frequency. To be precise, a photon carries momentum $$p=\hbar k=\frac{2\pi \hbar f}{c} =\frac{E}{c}$$ (where \(f\) is the frequency, \(k\) is the wavenumber, \(E\) is the energy of the photon).

Note that the momentum of the photon depends only on its energy. So we can just take the total energy flux on a surface, divide it by the speed of light, and get the force (rate of momentum change over time) per unit area. Sweet!

If a photon hits an object and is absorbed, to balance momentum, the object must gain an equivalent amount of momentum. If it’s reflected back in the opposite direction, the object has to gain twice the momentum of the photon. Which means the optimal situation for a solar sail is a perfect reflector that sends the incident photons exactly where they came from.

OK, so, now, how much force would the solar sail on our statite receive? Suppose the sun radiates a total power \(L\). At a distance \(r\) from the centre of the sun, this is spread over a sphere of area \(4\pi r^2\). So the power per unit area received at this distance from the sun is just $$P=\frac{L}{4\pi r^2}$$ and the force per unit area imparted on the surface, assuming perfect reflection, is $$F_\text{light}=\frac{2P}{c}=\frac{L}{2\pi c r^2}$$

Now, the solar sail is also attracted by gravity, which also follows an inverse square law! The force per unit area from gravity is merely $$F_\text{grav}=\frac{GM\sigma}{r^2}$$where \(\sigma\) is the surface mass density (i.e. mass per unit area).

If our statite is indeed static, the mass per unit area must equal the force of light per unit area, or else it would accelerate. So we can set \(F_\text{light}=F_\text{grav}\) which means $$\frac{L}{2\pi c r^2}=\frac{GM\sigma}{r^2}$$and after a little rearrangement we get that mass density we were looking for, $$\sigma = \frac{L}{2\pi c GM}$$

Plug in the solar luminosity and mass, and you get the mass per unit area of your statite as \(1.53\mathrm{gm^{-2}}\) which is really very extremely light!

A check: this is twice the Wikipedia quote, which cites this Dyson sphere FAQ which makes the same calculation we did, but reports the force of light as half what we had. Although they’re claiming this represents a perfectly reflecting sail, it looks more like they’re giving the result for a perfectly absorbing sail?

I checked my formulae against the numbers in the solar sail article and I believe the calculation above is correct, or at the very least, mistaken in the same way as whoever wrote the solar sail article.

Note also the solar sail would have to be lighter, per unit area, than that value, to account for rigging, electronics, etc. And this is already absurdly light: Wikipedia’s speculation is going as far as suggesting ‘single sheet of graphene’ as a solution.

Is it stable? Actually this is easily answered: the force balance above does not depend on the distance to the sun, so if it moved from its position slightly, it would not start drifting away. However, if the solar sail rippled or rotated slightly from directly facing the sun, it would fall towards the sun slowly; if the solar sail was slightly too big, it would start to blow away from the sun. The statite would have to monitor its position and adjust the sails constantly - but there’s no reason it couldn’t do that.

Solid dyson sphere

So the first thing to repeat is that as a result of the shell theorem, a solid spherical dyson sphere would not interact gravitationally with the sun, and could therefore drift until it hit the sun.

How strong would it have to be? This is discussed in the Dyson sphere FAQ above: they suggest considering the dyson sphere as two hemispheres joined at a seam. Each hemisphere, attracted by the sun, would press against the other: the dyson sphere must be strong enough to resist the pressure from both sides.

How strongly is a small piece of the sphere, of mass \(\rho t \dif A\), attracted towards the other hemisphere? Here I’m saying \(\rho\) is the density of the Dyson sphere material, \(t\) is the thickness of the sphere, \(\dif A=r^2 \sin \theta \dif \phi \dif \theta\) is an element of area with the spherical polar coordinates shown in this picture:

The force towards the sun is \begin{align}\dif g &= GM \rho t \dif A r^{-2} \\ &= GM \rho t \sin \theta \dif \phi \dif theta\end{align}which is, interestingly, independent of the size of the sphere! The net force in the direction of the other hemisphere is \begin{align}\dif F &= \cos \theta \dif g \\ &= GM \rho t \sin \theta \cos \theta \dif \phi \dif \theta \\ &= \frac{GM \rho t}{2} \sin 2\theta \dif \phi \dif \theta\end{align}

We now need to integrate this over the entire hemisphere to get the force at the join from one hemisphere: \begin{align}F&=\frac{GM \rho t}{2} \int\limits_{\theta=0}^{\pi/2} \int\limits_{\phi=0}^{2\pi}  \sin 2\theta \dif \phi \dif \theta \\ &= GM \pi \rho t \left[-\frac{1}{2}\cos 2 \theta\right]_0^{\pi/2} \\ &= \frac{GM \pi \rho t}{2}\end{align}so in fact all we did was insert a factor of pi, huh. Since both hemispheres are pressing with this force, the total compressive force at the join is \(2F=GM \pi \rho t\).

The area of the ‘seam’ between the two hemispheres is, assuming the sphere is thin compared to its size, just \(2 \pi r t\). So the pressure on the seam is $$P=\frac{GM\rho}{2r}$$ regardless of the thickness of the sphere (as long as it’s not so thick that the thickness is comparable to its radius!) We can compare this with the compressive strength of the material.

If the sphere is made of steel, for example, and the sphere is \(1 \mathrm{AU}\) in size (which seems to be the standard example, though it sounds unnecessariy large to me - but also note that the larger the sphere, the less strong the material must be), we find the pressure is \(3.5\mathrm{TPa}\) which is helpfully declared to be nine times the pressure at the centre of the Earth, and seven times the highest pressure sustained in a laboratory. It is, therefore, quite a lot more than the compressive strength of steel, which varies depending on how the steel is made and what materials are used, but might be in the range of hundreds of megapascals and therefore tens of thousands of times smaller what would be needed to hold up a Dyson sphere.

There are two options for making the strength requirement of your solid Dyson sphere more reasonable: make the sphere bigger, and make it less dense.

If you used one of the lightest materials known to humanity, carbon nanofoam, the compressive strength requirement goes down to \(0.28 \mathrm{TPa}\) which is still absurdly outside the realm of possibility.

On the other hand, if you doubled the size of the sphere, you’d halve the strength requirement, but quadruple the amount of material you need.

Worse, if something causes your sphere to buckle towards the sun, gravity will tend to stretch out the distortion, potentially tearing a big hole. Big problem!

This is one of the major reasons a solid dyson sphere is considered one of the least possible kinds of dyson sphere, out of a range of things that are already pretty absurd.

How hot is a Dyson sphere?

Ultimately, whatever solar energy you gather from your Dyson sphere will eventually become unusable heat that you have to radiate away into space to avoid heating up. This is what Dyson was looking for in his paper: a Dyson sphere would radiate a different spectrum of radiation to a bare star, so you could potentially recognise one with a telescope. (People have tried, and not found any evidence of one.)

The radiative temperature of a Dyson sphere in equilibrium depends on its size. The usual rule for emission of something a bit like a black body is that the power emitted is $$L=4 \pi r^2\sigma \epsilon T^4$$where this time \(\sigma\) is the Stefan-Boltzmann constant, not the mass density, \(\epsilon\) is the emissivity, a number between \(0\) and \(1\) (with \(1\) being a black body) that tells you how well the material emits light, \(T\) is the temperature, and \(r\) is the radius of the spherical body.

The energy coming in to the Dyson sphere (from the star inside) is constant. So you have $$T\propto \frac{1}{\sqrt{r}}$$i.e. the temperature decreases as the inverse square root of the size of the sphere. If we put all the other constants in we get $$T=\sqrt[4]{\frac{L}{4\pi r^2 \sigma \epsilon}}$$and for a \(1 \mathrm{AU}\) Dyson sphere that is a perfect emitter, we find the temperature is about \(120^\circ \mathrm{C}\).

I want to make some metaphor about how like, if you poured water on the surface of the Dyson sphere, it would boil, but you’re in space so that would happen anyway. So instead um, if you had like, a cabin with 1 atmosphere pressure inside, and you had some water in there, it would boil.

If you want to reduce the temperature, make your Dyson sphere bigger, or make some holes in it to allow some of the heat of the sun to escape directly.

I think those are all the questions I had about Dyson spheres answered. For other questions, such as how much material it would take, have a look at the links above, they might cover you! (You are essentially talking about dismantling the entire solar system.) Or ask me. I like talking about stuff.

What are the planets up to?

Astrologically-minded people have been talking a bit about Mercury going into ‘retrograde’ recently.

I’m sure if you look you can pretty easily find an explanation of what that means in astronomical terms (I won’t try to speak to the astrological significance of it in whatever tradition you might subscribe to), but regardless I thought it would be nice to do a big long post about how the planets move around in the sky, how we know in advance what they’re going to do, etc.

This won’t be a post about the historical development of these ideas. I already did one of these. And I’m gonna include some maths, but also lots of words, so hopefully if you’d rather, you can just ignore all the equations!

Perspective view of the orbits of the main bodies in the solar system - image by Andrew Z. Colvin on Wikimedia commons

OK, so what’s going on up there then?

So in general when you want to work out the answer to some question like “what are the planets up to”, physicists want to be able to start with the where the planets are at one point in time (this is called the “initial state”) and get a description which tells them exactly where the planets are at every particular time before and after.

That’s the ideal.

The general problem of a bunch of objects (three or more) moving around and interacting through gravity is “chaotic”. That means that our prediction of where the objects go next can change a huge amount when the initial conditions change. Because of this, you can’t in general get a nice straightforward set of equations that tell you what‘s going to happen.

So you have to make some approximations in your description of the system, and use a computer to predict approximately what will happen. If you’re lucky, your approximations won’t lose anything important.

The lucky thing about the solar system is that the sun is much much much bigger than everything else. So while each planet is attracted not just to the sun but every other planet in the solar system, the effect of the sun is vastly bigger than that of every other planet.

A very good approximation, then, starts by assuming the sun is the only thing that matters, and just ignoring all the other planets. That’s really the only way you can go if you don’t have a computer simulation handy.

Kepler’s model, which is what happens if the planets all ignore each other

(hamiltonianflow has done a very good series on this subject, which you can read here. So if you want a bit more mathematical detail than I go into, try her series!)

Here’s an example of the kind of shape called an ellipse.

When you only have two bodies involved - for example, a planet and the sun - each body orbits along an ellipse. (Or some other kind of conic section - but if it’s not orbiting along an ellipse, the planet will fly off somewhere very far away, so let’s not worry about that.)

The point marked \(F_1\) is called a focus (plural foci) of the ellipse. \(F_2\) is the other focus; one definition of an ellipse is that if you take any point along the ellipse and add up the distances to the two foci, the total will be the same.

What’s important is that, for the ellipses that the objects two objects are moving along, each has one focus at a point called centre of mass.

The centre of mass is a point that’s nearer the more massive (heavier) of the two bodies (and bang in the middle if they’re the same mass). Now, the sun’s mass is \(1.99\times10^{30}\thinspace\mathrm{kg}\). The mass of the heaviest planet, Jupiter, is \(1.90\times10^{27}\thinspace\mathrm{kg}\). That means that the sun is a thousand times heavier than Jupiter; the difference between the sun and the other planets is even bigger. So the centre of mass is pretty much right in the centre of the sun, for every planet.

This means you can pretty accurately say each planet orbits along an ellipse, with its focus at the centre of the sun. (That’s called Kepler’s First Law, because Kepler came up with this model first.)

An ellipse tells you where the planet might be, but we want to know where the planet is at a particular time. Kepler had a complicated description that comes in terms of ‘areas swept out’ by a line between the planet and the sun. Nowadays, we treat it with various equations, though finding a simple equation to get the orbital state vectors as a function of time is surprisingly difficult!

But the important detail is basically that planets travel faster when they’re near the sun, and slower when they’re far away from the sun.

That particular relation - speed vs. distance from the sun - is expressed by quite a simple equation called the ‘vis-viva equation’,$$v^2=GM_\odot \left(\frac{2}{r} - \frac{1}{a}\right)$$In this equation, \(v\) is the speed of the planet, \(r\) is its distance from the sun, \(a\) is the semi-major axis which is a measure of how big the orbit is, \(M\) is the mass of the sun, and \(G\) is a constant that tells you how strong gravity is.

The third thing Kepler said is that the total time \(T\) that a planet takes to complete an orbit of the sun - its ‘year’ in other words - grows in a particular way depending on how far away the planet is orbiting:$$T^2=4\pi^2\frac{a^3}{GM}$$or the square of the period (year) is proportional to the cube of the semimajor axis.

So the length of the planet’s year depends how far away the planet orbits from the sun.

Those three rules are enough information to tell you exactly where the planets are going to be at any point in the future, assuming the sun is the only thing that matters.

So what about our planets specifically

Our sun formed from a giant disk of gas and dust that slowly shrank under its own gravity around a dense core. Because of a principle called ‘conservation of angular momentum’, as the disk shrank, the tiny motions of the gas within the disk are increased to form a large, flat, rotating disk.

A lot of that disk is picked up by the growing sun in the centre, but the leftovers form small lumps which grow into bigger lumps as they collide with each other. These eventually form the planets and the asteroids.

The result of this process is that the planets all rotate in the same direction around the sun, and they all orbit in almost exactly the same plane. This plane is called the ecliptic.

The orbits of the planets in our solar system are not very eccentric at all either. The eccentricity of an orbit is basically a measure of where the ellipse falls on a continuum from ‘circle’ to ‘straight line’, so a more eccentric orbit is one that’s more stretched out. So in other words, the orbits of the planets are, despite being ellipses, very near circular. This is part of why models of the solar system featuring circular orbits were very effective.

As for why the eccentricities are so low, there are a few possible answers - interactions with the other planets, interactions with dust in the protoplanetary disc, tidal interactions with the sun - but it’s still an area of active research.

A gif showing the above-described orbits from here, via Wikipedia

The sun is not the only thing that matters

You can get away with ignoring the influence of the other planets over a few centuries, but eventually the effects of their small gravitational pulls starts to add up. Which leads to surprisingly large consequences for things like the climate of the Earth!

Because the sun is still the most important thing in the picture by far, we usually look at these orbital peturbations as modifications to the picture we just described.

These can come from a few things:

Because a lot of the causes of these peturbations (i.e. other planets coming near) are periodic - they happen regularly with the same time between them - the peturbations are also periodic. (The planet Neptune was discovered as a results of the small periodic peturbations it was causing to the orbit of the planet Uranus).

For planets’ moons - particularly the Earth’s Moon - things get flipped round: the Earth is the main thing in the picture, but the sun causes the biggest peturbations and changes the Moon’s orbit quite significantly over time.

It is also the case, in theory, that these peturbations can add up and throw a massive spanner in the nice neat orbits we’ve been talking about so far, leading to planets colliding with each other.

We can ask if, over millions and millions of years, the behaviour of planets in the solar system is stable (i.e. the only changes are the small peturbations, but everything remains much as described above). The answer is: we know pretty much exactly everything that’s going to happen over the next few million years, and we don’t have to worry about colliding with a planet.

For the next few billions of years, things are probably going to be ok, but if we give it five billion years, there’s a tiny chance that Mercury’s orbit will go wildly unstable, either plunging it into the sun, colliding it with Venus, or disturbing Jupiter in a way that fucks up the orbits of all the inner planets so we could collide with just about any of them. Fun. In any case, after that 5 billion year point, the sun will run out of hydrogen to effectively fuse in its core, and grow into a red giant star, swallowing up the innermost planets and destroying any remaining life on Earth.

(For comparison: life on Earth has existed for about four billion years, and we humans have been around for 50k-100k years, which is to say, 0.0025% of that. It is near certain that all humans will be billions of years dead by the time Mercury and the sun get around to doing any of those things. And if by some impossibility we have not, I would expect us to exist on other planets apart from Earth. So don’t worry too much :3)

One thing on that list, general relativity, only gets really important when you get near the sun. The very brief explanation is that Newtonian gravity is actually only an approximation itself to a much more mathematically complicated theory. On the scale of the solar system, that mostly doesn’t matter, but it does have a small but noticeable effect on the orbit of Mercury! That effect is one of the lines of evidence that formed some of the earliest experimental tests of the theory, so I thought it was worth a mention.

As for the climate thing - well, the peturbations to the Earth’s orbit and the way in which its spinning are generally periodic, repeating over thousands of years. Examining long-term historical climate data revealed these incredibly slow changes (exaggerated by feedback effects) lead to similar periodic changes called Milankovitch cycles in the Earth’s average temperature. Probably. This science is perhaps a bit less settled than some of the rest of this post.

What do we see of all this from the ground?

Since most of the people reading this post are probably on the surface of the Earth (any astronauts and/or extraterrestrial lifeforms reading, please say hi!), if we want to observe the planets, we look up at the night sky for little bright dots that look very much like stars. (If we are lucky enough to have a good telescope, we can resolve these dots into flat disks with some surface variation.)

Most of the things in the sky, as watched by human eyes from Earth, don’t do much, day to day and month to month. They sit in a fixed pattern in relation to the other stars. Now we have telescopes, we can notice that isn’t quite true - some of the more nearby stars move a bit over the course of the year due to ‘stellar parallax’ which lets us work out how far away they are, and some stars are moving through space quickly enough to notice in a telescope (particularly the stars orbiting the black hole Sagittarius A* at the centre of our galaxy!).

The planets are not like these ‘fixed stars’. They don’t stay in a set position compared to the fixed stars. The English word “planet” is derived from a Greek word meaning “wanderer” for this reason.

When something is odd and moving about, a natural thing to do for people across the world was to keep track of where it is each night. And if you do this, you can see the planets slowly make their way from place to place in the sky, reflecting their motion in space. Planets close to the sun, like Mercury, generally move much faster than planets far away from the sun, like Saturn. (Uranus and Neptune are harder to see without a telescope.)

We can imagine the fixed night sky as being an enormous sphere centred on the Earth, with all the “fixed” stars as little dots attached to it. This is helpful for describing how the planets move in the night sky. (The sphere is basically a map of the angular direction of each star from the centre of the Earth.)

So in this picture, the Earth is sitting in the middle of the frame, with the celestial sphere centered on it. I’ve drawn two planets, and two lines through those planets to the celestial sphere, to show how the positions of planets in the celestial sphere is determined by the relative positions of these planets.

How do the planets move? Here’s a picture where I’ve drawn this diagram for two different days, one over the top of the other…

The positions of the planets on the first day and its celestial sphere is faded out slightly. I’ve also drawn the day-1 positions of the planets on the day-2 celestial sphere, again faded out. I’ve made Venus move further than Earth move further than Jupiter, but I’m not doing this at all exactly.

Now all three planets have moved around the sun, both Jupiter and Venus have moved to slightly different places on the celestial sphere.

That’s the general principle of how planets appear to ‘wander’. You can see it’s pretty complicated.

Now, a very large part of the motion, especially for the inner planets, comes from the Earth’s own orbit. Because of the Earth’s orbit, these planets appear to travel all the way round the night sky over the course of a year, even though their movement in space is much smaller than that of the Earth. This is why it made a lot of sense to people, at first, to say the planets travelled round the Earth.

Because the planets’ orbits are largely confined to a plane (the Ecliptic mentioned above), they all pass through roughly the same areas of the night sky. As a result of the Earth’s motion, so does the sun. (Based on this property, the predominant Western astrological system divides the night sky into 12 equal-sized regions, into which the planets pass over the course of a year. Patterns of stars in each of these regions (as seen from Earth) give each one its symbolism).

And retrograde motion?

One of the interesting aspects of living in a heliocentric solar system like this is that, on occasion, the planets seem to temporarily change the direction of their motion relative to the fixed stars in the night sky. They can follow a pattern such as this over the fixed stars:

(this squiggle shows the motion of Mars in 2009-2010, via Wikimedia commons)

This presented a big headache to people trying to work out a geocentric model of the solar system (one in which everything rotated around the Earth) according to Aristotle’s principle . In a heliocentric solar system, such as Kepler’s one described above, there is a straightforward explanation, though the situation looks slightly different for planets orbiting inside the Earth and planets orbiting inside.

For outer planets such as Jupiter, the situation is this: Jupiter is sedately carrying on its orbit, and the Earth zooms past on the inside.

This image (by Rursus on Wikimedia commons) shows the positions of two planets, the Earth (the ‘T’ planet) and another planet (’P’) as well as the other planet’s position on the celestial sphere, at each of five points in time over the coures of half an Earth year. Unlike my diagrams, Rursus has centred the celestial sphere on the sun, and not the Earth. The difference stops being important as you make the celestial sphere bigger, and it does make it a lot easier to show everything on one diagram.

Anyway, as the inner planet moves up towards the outer planet, they are moving at quite different angles, so the outer planet appears to move a large distance on the celestial sphere. As the inner planet catches up, being faster, the outer planet appears to move ‘backwards’, until eventually the inner planet is moving more or less directly away from the outer planet and it appears to move forwards again.

To show that this still happens if we use a properly Earth-centered celestial sphere, I did a series of images :)

…which, however rough they are, hopefully show the general principle.

For a very slowly moving outer planet, this happens nearly every Earth year. For a more quickly moving planet such as Mars, it will happen slightly less than once a year.

For the planets closer to the sun than Earth (Mercury and Venus), you can essentially draw the reverse of this diagram, fixing the outer planet in the frame as your ‘Earth’ and drawing lines through the inner planet. That said, to see retrograde motion when I tried this, I had to tune the speeds a little bit - but with the rigt speed, it can be quite pronounced. (I’m not going to post the whole series, but this image was constructed the same way as the last set!) The red squiggle is a rough trace of the positions visited in order, in order to make the retrograde motion clearer.

Anything else I should know about those planets?

While I was talking, Neptune stole your wallet. Sorry.

More seriously, there’s a lot more mathematical detail to be had if you want to learn more about the planets. I represented stuff as 2D circles and the celestial sphere as a circle, but of course we live in 3D space (3+1D spacetime if you want to get all relativistic on me), and the sky forms a 2D plane, so you have to invoke a lot more parameters to describe the motion of the planets of the sky in all its precise grisly detail. To fully describe an elliptical orbit, you need six numbers. To precisely state the position of objects in the sky (in terms of “point your telescope here”), there are many coordinate systems. Like, really a lot.

Also I guess… I have no wish to be the ‘no fun allowed’ person, but if you’re genuinely worried about Mercury going into retrograde (I can never tell quite how serious people are about that, but… I know what it’s like to have anxiety :/), astronomically there’s nothing more going on at present than an interesting visual effect that becomes apparent if you very carefully keep track of the position of Mercury in the sky over a long period of time. There’s no special reason why things are more likely to go wrong while Mercury and the Earth are in these particular parts of their orbits.

The moon passed between ‪NASA‬’s Deep Space Climate Observatory and the Earth, allowing the satellite to capture this rare image of the moon’s far side in full sunlight. As the moon is tidally locked to the Earth and doesn’t rotate, we only ever see the one face from the Earth. Awesome shot!

This isn’t the only picture that DSCOVR (that is the official acronym, I know…) took of the moon! Here’s a gif of the pictures of the moon transit, from Wikimedia Commons:

This picture makes the Earth and Moon look really close. You can see a to-scale orthographic picture here that shows how far they are apart on average.

The DSCOVR satellite orbits at the Sun-Earth L1 Lagrange point, which allows it to stay more or less directly between the Earth and Sun.

(A Lagrange point is a place in relation to two orbiting bodies where a third small body will orbit in a fixed position relative to them, i.e. an object exactly at the L1 point will stay fixed between the Earth and Sun. There are five Lagrange points, but three of them - the L1 point included - are always unstable, so if something isn’t quite at the Lagrange point, it will move away, slowly at first and faster the further away it gets. The DSCOVR satellite makes use of a trick called a Lissajous orbit, which allows it to effectively orbit the Lagrange point itself, staying fairly near it. But over a long period, this orbit is unstable, and will require corrections.)

The L1 point is about 1.5 million km from Earth. The Moon orbits the Earth with a semi-major axis of 384 thousand km. So on average, the DSCOVR satellite is about four times further away from the Earth than the Moon.

A time traveller comes to a small deserted town. She lives in this town for a month, then travels back in time one month. She does this again, six hundred times.

To an observer, a traveller arrives, and suddenly - for one month - the town is bustling with the activity of six hundred people. Then, it all shuts down. The traveller continues on, now fifty years older than when she entered.

you can also use this trick to overwhelm an opponent, but make sure they don’t get a hit in because you are vulnerable to an attack on all of your instantiations.

Not necessarily! Only the first couple of instances need to avoid hits. The “older” (traveller pov) versions who have already seen that fight multiple times will definitely have an advantage, knowing how things already transpired.

(Also the hits won’t be too bad if there ARE older versions around…)

Going, as physicists who speculate about time travel often do, by the Novikov self-consistency principle, there cannot be a timeline where someone is rendered unable to time travel in the presence of their future self.

So, if one of the ‘yous’ gets injured in the fight, it will never be enough to make them no longer able time travel. The injury will afflict every future copy, of course.

So if you’re about to fight someone and five copies of you appear and they all have the same injury that you don’t, you will know that you’re about to suffer that injury in the fight. This knowledge will not allow you to avoid the injury! All your future copies knew about the injury as well, and still suffered it.

If say 2/5 of your future copies have the injury, you’ll know that you’ll suffer the injury on your fourth loop through the fight.

You might also see one of your copies die (implicitly, the final copy). In this case, despite knowing it will prove fatal, you will necessarily still time travel through the fight several times before dying. Likewise if you see one of your copies become injured so they can’t time travel: that one is the final copy, and you can look forward to that.

You won’t ever see two copies of yourself die, because that would be inconsistent. In a natural-seeming way, the timeline will simply never include such an event, despite anyone’s best efforts.

What happens if i go back in time to try and change a thing, does the time machine just not work if that would happen were i to go back? 

either your attempt to change the thing is already part of the timeline, and unsuccessful, or else yes, the time travel doesn’t work

novikov self-consistency says only that the timeline is consistent - it doesn’t say anything about how that consistency is maintained. it might be that time travel could only work in tightly controlled conditions to ensure consistency. which could still allow you to implement time travel based algorithms!

objection

not a literal objection just me being difficult but 

novikov self-consistency is a conjecture – nay, a coping mechanism – to help physicists who don’t wanna go “but what if causality is arbitrary”

have we ever actually demonstrated it? maybe we only think causality works because it’s like gravity, and we’re still figuring out airplanes and spaceships

what does this scenario look like if the time traveller isn’t bound by closed loops

what now science side of tumblr

oh, absolutely. there’s no evidence of novikov self-consistency - it’s just an option for when you say “ok, suppose imagine we allow time travel and throw out causality, what rule could we have?”

replacing novikov self-consistency with causality allows for all sorts of weird shit, too! at any moment, an object inside a stable time loop could appear.

so while we tend to speculate along the lines of ‘what if you could meet yourself time travelling from the future’, you might also bump into an automatic time machine that only exists inside a time loop.

What rule would govern when time travellers spontaneously appear? Usually I think we say ‘while what if the conservation laws - energy, momentum and so on - are more fundamental than causality, and apply even to time travellers’.

This is where my questionable analogy with Feynman diagrams comes in. But I’ll elaborate on that below a readmore…

So you might imagine - a time traveller can spontaneously appear, but only as long as all the relevant conservation laws are maintained. Perhaps then to achieve significant time travel you need a ‘receiver’ time machine that produces a large amount of energy that turns into a time traveller, and a ‘sender’ time traveller that (to outside observers) disintegrates a time traveller.

That still doesn’t include anything at all getting out of the ‘time machine’, as long as it gets back in the time machine later.

Novikov self-consistency can’t be the only way to think about time travel, but we do need some handle for our speculations. So without Novikov, we need some other way to constrain what happens… I don’t know what that would be.

Keep reading

Real quick question for a different kind of scientist. How quickly and easily do you think someone could construct a device capable of flinging my body into the sun?

To be clear,

Do you think if I fling myself into the sun I’ll have to write this final tomorrow?

Well, if you’re not too concerned about surviving, you could, in principle, just use a really big bomb. But I doubt you’d be able to scrounge up the parts, and you wouldn’t make it all the way to the sun by tomorrow.

So we’ll assume that the sun is the only safe place in the universe where you will not be asked to complete the final. If so, you need to reach the sun within a day.

The sun is small compared to the distance between Earth and Sun, so the minimum average speed you’ll need is about 1AU/1day, which is 1700km/s or about 0.6% of the speed of light.

In general, just pointing your rocket at the sun and firing for that much delta-v won’t get you to the sun, because it will instead perturb you onto a very elliptical orbit that will swing past the sun fairly close before orbiting on a wide arc that will leave you far too much time to complete your final. So you need to fire in a way that cancels the Earth’s orbital speed in addition to increasing your velocity in a Sun-wards direction, increasing the total delta-v budget.

That said, the Earth’s orbital speed is “only” about 30km/s, which is very small compared to what we need to reach the sun on time. So we’ll ignore this correction.

As you approach the sun, you will gain additional speed because of gravitational acceleration, but let’s assume we want to achieve that minimum speed using a rocket, bomb, etc. And we want to do it quickly: if the thrust is to low, even if you eventually achieve that speed, you’ll do it too late.

Basically what I’m saying is we probably need to resort to something like Project Orion, or antimatter propulsion, or else an extremely powerful railgun. I’ll come back to this later.

So checking Wikipedia’s article on Project Orion - the nuclear bomb pushed rocket design - it seems that Freeman Dyson suggested a ‘momentum limited’ variant of the design that could accelerate at 1g up to a delta-v of 10,000 km/s (over the course of ten days), with the rocket itself weighing fifty million kg empty, and carrying a fifty million kg structure and payload. (The nuclear bombs make up another 300 million kg.)

This spaceship could not be built in a day; with present technology it could not even be built at all. But, if it had already been built, the first day’s acceleration would get you about 1000 km/s, not quite enough to meet your final deadline. At least, if you had the full 50 gigagram structure on board. But you only need to be on the spaceship for a day! By dropping everything but basic life support (or even that), we could double the acceleration, and quadruple the final speed, which should get us to the sun on time.

Dyson estimated the cost of this machine to be $0.367 trillion, or 10% of the entire US annual GDP.

We’re also ignoring the difficulties of getting you into orbit in the first place, which would make a difference to the delta-v budget - but not a big one, considering, as the Earth escape velocity is “only” 11km/s.

Building Project Orion is of course currently outside our technical capabilities, not to mention banned by international nuclear testing laws. The same goes for massive railguns and antimatter. Could we instead use any past or currently existing rockets to get you en route to the sun, even if you won’t get there in a day?

A payload-less Saturn V has a delta-v of about 18km/s. Your mass is probably negligible compared to that of the rocket, so this applies. But that is not enough to cancel the Earth’s 30 km/s orbital speed and reach the sun. That said, the Saturn V’s first two stages are designed to operate in the Earth’s atmosphere, so you could get better efficiency by using rockets more like its third stage.

I’m not sure any mission has ever gone particularly close to the sun, for the straightforward reason that if you get to close to the sun you will be vaporised by incident radiation. But we have sent spacecraft to Mercury, which has similarly extreme delta-v requirements. The Mariner 10 probe made flybys of Mercury, making use of a gravitational slingshot around Venus to dump excess velocity. The MESSENGER probe, after several flybys of Earth and Venus to adjust its trajectory gravitationally, was able to enter the orbit of Mercury after being launched with a Delta II rocket.

So your best bet for reaching the sun would be to use Venus, and indeed Mercury, for gravity assists. Working out a suitable trajectory is a bit beyond me at the moment!