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.