Solar garbage disposal.

Why can't we drop our dangerous waste into the sun?

It's tempting to consider getting rid of dangerous (or undesirable) waste by shooting it into space and letting it fall into the sun; after all, it seems pretty unlikely that we could drop enough of anything into the sun to have an adverse effect on it.

So, just how hard is this?

Let's just start off by thinking about how hard it start off with something in the same orbit as the earth, and knock it out of that orbit into whose closest approach to the sun actually takes it inside the sun. We can worry later about getting it into this initial orbit.

Now, an orbit consists of an ellipse with one focus at the sun. Let's call the distance from the (centre of the) sun at perihelion (the closest point) \(r_1\), and the distance at aphelion (the farthest point) \(r_2\), and the orbital speeds at these positions \(v_1\) and \(v_2\) respectively. Then we have the (badly drawn) situation below:

Perihelion and aphelion convenient points to think about, because the orbital velocities at these points are perpendicular to the lines joining them to the focus. This makes it particularly easy to calculate the angular momentum there.

It's really quite hard to describe orbits properly: finding the position of an orbiting body as a function of time cannot be done in terms of the usual standard functions. However, we can avoid a lot of unpleasantness by being clever with two conserved quantities, the energy and angular momentum of the orbiting body. Since these are both conserved, they always have the value they have at perihelion.

At perihelion, then, the body has angular momentum and energy given by \[ \begin{split} L &= mv_1 r_1\\ E &= \frac{1}{2}mv_1^2 - \frac{GMm}{r_1} \end{split} \] where \(G \approx 6.67 \times 10^{-11}\text{m}^{3}\text{kg}^{-1}\text{s}^{-2}\) is Newton's gravitational constant, and \(M\) is the mass of the gravitating body (which for us will be the sun).

To reduce the algebra a little, we'll take advantage of the fact that everything is a multiple of \(m\), and think of the angular momentum and energy per unit of mass (i.e. per kilogram). That means we think about \[ \begin{split} l &= v_1 r_1\\ e &= \frac{1}{2}v_1^2 - \frac{GM}{r_1} \end{split} \]

A useful observation is that for a closed orbit, we must have \(e<0\); we'll save this fact up for later.

And now we'll do the sneaky thing of finding what the perihelion distance is for an orbit of a given angular momentum and energy.

The idea is simple enough: use the first equation to express \(v_1\) in terms of \(l\), then substitute that into the second equation to get a relationship between \(l\), \(e\) and \(r_1\).

This gives \[ e= \frac{l^2}{2r_1^2} - \frac{GM}{r_1} \] so that \[ r_1^2 +\frac{GMr_1}{e}- \frac{l^2}{2e} = 0 \]

So the solutions are given by \[ \begin{split} r_1 &= \frac{1}{2}\left( -\frac{GM}{e} \pm \sqrt{\frac{G^2M^2}{e^2} +2\frac{l^2}{e} }\right)\\ &= -\frac{GM}{2e} \left(1 \pm \sqrt{1+\frac{2l^2e}{G^2M^2}} \right) \end{split} \]

There are, of course, two roots to this quadratic: they are \(r_1\) and \(r_2\), the aphelion and perihelion values. (We could have gone through just the same process expressing \(l\) and \(e\) in terms of \(r_2\), and obtained the same quadratic.)

We should pay attention to signs here: we are taking a factor of \(|GM/e|\) out of the square root, but since \(e<0\), as mentioned above, this is \(-GM/e\).

Sanity check: for a circular orbit, the radius \(r\) is constant, and from Newton's law we obtain \(v^2/r = GM/r^2\). From this we find \[ e=-\frac{GM}{2r} \] and \[ l^2=v^2r^2 = GMr \] so that \[ \frac{2l^2e}{G^2M^2}=-1 \] and \(r_1=r_2\), i.e. the orbit is indeed circular.

Since \(r<0\), the square root term is smaller than \(1\), and we want the perihelion value, so we have \[ r_1 = -\frac{GM}{2e} \left(1 - \sqrt{1+\frac{2l^2e}{G^2M^2}} \right) \]

Now, let's see what the situation is for starting off with a body in the same orbit as the earth, and trying to adjust it so that it hits the sun. The earth's orbit is very nearly circular, so we'll just assume it is exactly circular, with constant radius and orbital speed.

By the magic of Google (or looking in a book), we find that the mass of the sun is\(M=1.99\times 10^{30}\text{kg}\), the radius of the sun is \(6.96 \times 10^{8}\text{m}\), the orbital radius of the earth is \(1.5 \times 10^{11} \text{m}\) and the orbital speed is \(2.98 \times 10^{4}\text{ms}^{-1}\).

In the diagram below, if the earth's orbit is the black one, we want to divert it to the red orbit by reducing the orbital speed from \(v\) to a smaller value \(u\), so that perihelion would occur inside the sun.

Now, we can set \(r_1 = 6.96 \times 10^{8}\text{m}\), and, noting that \(e\) and \(l\) are functions of the orbital speed, solve numerically for the orbital speeds which give a perihelion distance of less than \(6.96 \times 10^{8}\text{m}\). I asked a computer, and it told me that \(u\) must lie in \((-2858,2858)\), so we have to slow the orbit down by a change of speed \(\Delta v\) of at least \(29800-2858\approx 27000 \text{ms}^{-1}\).

For a kilogram of mass, that means we need to use \((\Delta v)^2/2 \approx 360\) MegaJoules to change its speed by enough. This is quite a lot of energy.

There's also the problem of getting the mass out of the earth's gravitational field: this takes about another \(62\) megajoules.

So in total, we need to use about \(420\) megajoules of energy to drop each kilogram of stuff into the sun.

For comparison, it requires about \(15\) megajoules to transport a kilogram of stuff to the ISS, in low earth orbit\(1\). An example from closer to home is that a megajoule is the kinetic energy of a vehicle weighing one tonne, travelling at about \(161\text{km}\text{h}^{-1}\).

So there we have it: we won't be just dropping our waste into the sun any time soon because apart from anything else, it's just so much work.

Afterthoughts

1. You might wonder if it's possible to do better by pushing the object at an angle to its original orbit, rather than just by slowing it down. Unfortunately, the option discussed above is the one requiring the least energy.
2. I've only considered the situation of giving the object an orbit which falls into the sun. Could one do better by arranging one which flies close past another planet and takes advantage of a slingshot orbit? Yes, and that was used to help send probes to Mercury. This can save energy, but it is a delicate manoeuvre; I'll leave you to try to work out if it might make this form of waste disposal more feasible.

Acknowlegements

Thanks to Colins @icecolbeveridge and @ColinTheMathmo for comments and feedback.

A matter of weight

What do you weigh? More to the point, what does that mean?

What is weight?

What it isn't: mass

Like many people, I find out what I weigh by standing on the bathroom scale. Usually, the answer is approximately 80 kilograms. Or so I try to convince myself.

Well, there's a problem with that. Kilograms is a measure of mass, not weight. But what's mass?

To answer this, we rely on Newton's laws. Mass is resistance to force. If you act on a mass, \(m\), with a force, \(\pmb{F}\), then it has an acceleration, \(\pmb{a}\), and the three are related by \[ \pmb{F} = m \pmb{a}. \]

Actually, things are a bit more complicated than that, but as long as the mass isn't changing, it's OK.

Then if I have a reliable way of exerting a known force, I can compare the masses of two different object by comparing the accelerations imparted to both. And if I have a reference mass (and there is one for the kilogram) then I can work out the mass of anything else I act on with this force in terms of the reference mass.

OK, so that's mass (well, the quick version) sorted out. But what is weight, and what is my bathroom scale really telling me?

What weight is absolutely

The first definition that you are likely to meet in a physics text is that it is the force exerted on a body by gravity: sometimes the actual force (as a vector), but more usually the magnitude of the force.

It is a remarkable fact that the force exerted by gravity on an object is exactly proportional to its mass, so that all objects, whatever their mass, fall with the same acceleration (in the absence of confounding factors like air resistance).

In SI units (i.e. those used by the civilized world), the unit of force is the Newton, and the unit of mass is the kilogram. At the surface of the earth, the force of gravity is approximately \(9.8 \text{N}/\text{kg}\); a mass of 1 kilogram weighs about \(9.8\) Newtons.

We can work with this. If I have access to materials which can be compressed or extended by acting on them with a force, then I can build a machine which measures the (magnitude of the) force I exert on it. And if I know that what people really want to know is their mass, I can calibrate it so that if the force acting on it is the force that gravity exerts on their body, then the scale shows the mass that gravity must be acting on, rather than the actual force.

This is typically how a bathroom scale works.

And is fine and excellent, as long as I know what the force of gravity is where I use the bathroom scale. There is a minor issue caused by the fact that the force of gravity isn't quite the same all over the surface of the earth; but the variation is fairly small (from \(9.78 \text{N}/\text{kg}\) to \(9.83 \text{N}/\text{kg}\), and for everyday purposes we can pretty much ignore this variation.

But now we bump up against a bit of a problem. For the scale to work properly, we have to stand still. You'll have noticed that if you jump up and down on it, the displayed weight oscillates along with you. It's less likely that you've taken it into a lift, but if you have you'll have noticed that when the lift has an upward acceleration you show a higher weight, and if it has downward acceleration you show a lower weight.

The snag is, of course, that to find your weight, the scale has to hold you stationary in the earth's gravitational field, and measure the force it takes to do that.

This raises a slightly awkward question: how do you know when the scale is holding you stationary?

This could well seem like a rather stupid question. All you have to do is look around.

But what if you are in an enclosed room? How could you tell the difference between being (say) on the moon, with a weight of approximately one sixth of your weight on earth, and being in a lift which has a downward acceleration of about five sixths the acceleration due to earth's gravity, or indeed being in a rocket a long way away from any body of mass, accelerating at about one sixth of the acceleration due to earth's gravity?

It turns out, that's really hard to do. You can't do it by making local measurements, which is a heavily coded way of saying that no matter how good your measuring apparatus is, if you work in a sufficiently small laboratory, over a sufficiently small time span, you can't detect the difference. (If you allow non-local measurements, which effectively means that you work in a region big enough that the force of gravity varies appreciably, then you can measure the resulting tidal effects: here is a nice look at tidal effects from the point of view of an orbiting astronaut, from @ColinThe Mathmo.

This is a bit of a blow. We have what looks like a perfectly sensible definition, but it turns out to be hard to use. So we do that standard trick. We replace it by a somewhat different definition, which is easier to use, but doesn't look quite so sensible.

At least, it doesn't look quite to sensible to start with. Once you get used to it, you can convince yourself that it is in fact the sensible definition, and it's the original definition that's misled.

What weight is relatively

So what do we do instead?

We accept that since what we can measure (reasonably) depends on the state of motion of the equipment and body, we build that into the definition.

We now define the weight of an object in a given frame of reference to be the (magnitude of the) force required to hold it stationary in that frame of reference.

So, if your scale says you weigh a particular amount, that's what you weigh: you don't have to know whether you're on the moon, or whether you have to compensate because the scale is accelerating.

This does have it advantages. Principally, it means that you weight what you feel as if you weigh. If you're in a rocket taking off, or visiting a heavy-gravity planet (we can dream) you weigh more if you're standing still on the earth. If you're in a lift plummeting you towards an untimely end, or on the moon, you weigh less than if you're standing still on the earth. And that's all there is to it.

So, to the question I've been avoiding up to this point: is an astronaut in orbit, say on a space station, weightless?

There are two answers to this.

• No. The force on gravity on him is slightly less than it is on the surface of the earth, so he weighs a little less than he would at home.
• Yes. If he tries to stand on a scale, it reads zero; it takes no force to keep him stationary in his space-station, so in the frame of reference of the space-station, he is weightless.

And both answers are correct. Whether the astronaut is weightless or not depends on the definition of weight that you prefer.

But in either case, the astronaut is definitely not outside the gravitational influence of the earth; gravity is still pulling on him with a very significant force.

I prefer the second definition, because it's the one that doesn't require you somehow to know the situation outside your laboratory; it only requires you to make local measurements. But that doesn't make it right, only preferable from some points of view.

At least, that's the situation if you live in a universe where gravitation acts according to Newtonian theory. But in fact we don't really, and there's a better way of trying to understand gravity, and then we have a better reason for preferring the second definition.

What weight is relativistically

There's a great mystery in Newtonian mechanics: why is it that the mass which interacts via gravity is exactly the same as (or at least, exactly proportional to) the mass which resists acceleration? Very sensitive experiments have been carried out to test this, and they've always come out unable to detect a difference. They agree to no worse that one part in one hundred billion.

So, why should this be?

The current answer is that there is a better way to think about gravitational influence than as a force. In general relativity, we have a way of relating the distribution of matter to a distortion in space and time, which changes the notion of a straight line: we generalize this to a kind of curve called a geodesic. So a particle moving in a gravitational field is one which travels along a geodesic: there is no force acting on it, it's just (just!) that the nature of space and time mean that it doesn't travel along the same trajectory as it would in the absence of gravity.

Then any particle (with no external forces acting on it) travels along a geodesic which is determined purely by a starting point and a starting velocity; mass does not have a part in the trajectory of a the motion of a particle in a gravitational field. Of course, part of the joy of this is that it does give very nearly the same answer as Newtonian gravity.

Very nearly, but not quite the same as Newtonian gravity. And with careful observation we find that the general relativity version of gravity matches measurements better than the Newtonian one.

But now there is no force of gravity.

If I'm standing on the surface of the earth, the only force acting on me is the force of the earth pushing me off the geodesic that the curvature of space and time wants me to travel along. If I'm on the space station, there is no force acting on me at all.

So in this framework, the only definition of my weight that makes any sense is that it is the magnitude of the force required in some frame of reference to keep me stationary in that frame of reference.

This looks superficially the same as the second definition in Newtonian gravity, but it's really quite different in principle. In the Newtonian case, it's avoiding the issue that there's no easy way to access the actual force of gravity, and settling for what is easy to measure. In the general relativistic case, it's saying that weight is an illusion caused by being in an unnatural frame of reference.

I try to take some consolation from that as my weight seems to gradually, but inexorably, increase.