# What would the orbit of the Earth be like if we constantly sent our waste to the Sun?

Let's say we get a cheap technology to collect and then send (hypothetically all) our waste to the Sun by a rocket, and we choose this technology as the main (and only) method to utilize our waste.

What would the orbit of our planet be like after years of using this technology, as in the Earth-Sun kinetic-potential system, the Earth is continuously sending its mass to the Sun?

Or will that mass sent from Earth to the Sun return to the Earth in a form of irradiated energy and we (i.e. our descendants) will feel no effect?

• I think you're grossly underestimating the size of the sun and earth. Everything that exists on the surface is no more than the skin on an apple. The sun itself is about 333,000 times more massive than the entire Earth, no amount of garbage is going to do anything. – Dider Apr 4 '15 at 16:17

What Shihab Dider said. The Earth is monstrously massive compared to everything humans have ever built, and the Sun is monstrously massive compared to the Earth.

How monstrously massive are we talking about here? The mass of the Sun is about 1.99 $\times$ 1030 kilograms. The mass of the Earth is about 5.97 $\times$ 1024 kilograms. To put that in perspective . . . Well, perhaps it's best if we don't. Six orders of magnitude is quite a lot. Adding an entire extra Earth mass to Earth would barely change its orbit.

How much waste do we generate? Well, it depends on what you define as "waste." Garbage, feces, uneaten food - all could fall in this category. According to Duke University, a human (on average) generates 4.3 pounds (or 1.95 kilograms) of waste per day. I think that's only for Americans because stats from the EPA match it nicely, but we'll extrapolate it to the world, just to give a worst-case scenario (this is also interesting).

1.95 kilograms per person $\times$ about 7 billion people = 13.65 $\times$ 109 kilograms of waste per year. In other words, if we were to bring all this waste to the Sun, the Earth would lose only 1/1014th of its mass each year (not counting the spend rocket fuel). That translates to a change in orbital radius of . . . oh, something very small. You can do the exact math, if you want.

You can do the exact math, if you want.

~ HDE 226868

This is actually a type of orbital mechanics problem I haven't solved yet, so let's tackle it!

First, we need the amount of waste that we are going to shoot into the Sun. We can use HDE's figure:

$$2~\text{kg}/\text{person}/\text{day}\times 7\cdot 10^{9}~\text{people}= 5.6~\text{ton}/\text{y}=160\,000~\text{kg}/\text{s}$$

or Vincent's figure:

$$4000\cdot 10^{9}~\text{ton}/\text{y}=120\cdot 10^{6}~\text{kg}/\text{s}$$

These numbers are pretty different. I'm inclined to believe the smaller one, since (as jamesqf pointed out) we don't actually want to 'get rid' of most of our waste, as that wastes the energy expended in extracting the elements from the waste (no pun intended). However I'll use the larger number as a worst-case estimate.

Second, we need to know the speed at which we shoot out the waste, and in which direction. The obvious choice is to shoot it backwards, canceling out Earth's orbital velocity of around $30~\text{km}/\text{s}$ so that it will simply fall into the Sun. This is pretty inefficient (look to the upcoming Solar Probe+ mission for a more plausible trajectory), but again it represents a worst-case scenario.

The waste stream will act like rocket exhaust, pushing the Earth into a higher orbit (and, confusingly, slowing it down by speeding it up: but more on that later).

If an amount $\delta m$ of waste is expelled at a speed of $v_e$, the amount by which it increases the Earth's speed is:

$$\delta v = v_e \frac{\delta m}{M_\text{Earth}}$$

(This follows immediately from conservation of momentum.) Since the kinetic energy of Earth is:

$$E = \frac 1 2 M_\text{Earth}v^2$$

The increase in energy is:

$$\delta E = M_\text{Earth}v\ \delta v = v\ v_e\ \delta m \\ \dot E = v\ v_e\ \dot m = v^2\dot m$$

(Where in the second line I've turned the differentials into derivatives, and used the fact that the orbital speed and 'exhaust' velocity are the same.) This makes sense, as the amount of work that we did on the Earth is equal to the amount of force applied to it, times the distance that the Earth moved, or the amount of momentum transferred times the speed.

The orbital energy of Earth is

$$\mathcal{E}=-M_\text{Earth}\frac{\mu_\text{Sun}}{2a}$$

Where $\mu=MG$ and $a$ is the radius of the orbit. Taking the derivative and substituting in $\dot E$:

$$\dot{\mathcal{E}}=M_\text{Earth}\frac{\mu_\text{Sun}\dot a}{2a^2} \\ v^2\dot m = M\frac{\mu\dot a}{2a^2}$$

(From here on out I'll drop the subscripts since they won't change.) Assuming the orbit is close to circular (in fact the waste stream will tend to circularize the orbit), we have:

$$v = \sqrt{\frac \mu a}$$

Note that the speed will decrease as our orbital radius increases. This is because the orbit won't be exactly circular: the stream of waste will keep pushing the aphelion (highest point in the orbit) ahead of us, so that we're continually 'climbing' out of the gravity well. The speed increase isn't quite enough to counter this, so the net effect is a slowdown. Substituting into our differential equation:

$$\frac \mu a \dot m = M\frac{\mu\dot a}{2a^2} \\ \dot a = \frac{2\dot m a}{M}$$

Now we can substitute in our numbers:

$$\dot a = \frac{2\times 4000\cdot 10^9~\text{t}/\text{y}\times 1~\text{AU}}{6.0\cdot 10^{24}\text{kg}}=180~\text{m}/\text{y}$$

This is a pretty minor change, about one part per billion per year. The fact that the differential equation is looks exponential is promising. The solution, taking into account the decreasing mass of the Earth is:

$$a(t) = \frac{a(0)}{\left(1-(\dot m/M)t\right)^2}$$

We can use this to calculate that the Earth's orbital radius will increase by 1% after 8 million years, and double after 500 million years. So, not something that we'd have to worry about.

The second point you ask about is the effect on the Sun. The Sun already looses a billion kilograms per second to the solar wind. Even the large rate is only a tenth of that (and the 'reasonable' rate is one ten-thousandth), so it's not likely to have a significant effect.

According to this article, Earth could produce as much as [4000] billion tons of waste per year in 2100. It's not a very optimistic scenario. That seems a lot but it's 0.00000007% of Earth's mass. At that rate, it would take more than 14 million years to send the equivalent of the mass of Earth toward the Sun. And this is ignoring that Earth receives debris from space like meteor that increases it's mass on the long term.

The mass of the rockets would have to be considered as well. If the rockets are propelled with fuel and destroyed in the Sun, all of this is lost to the star as well as the wastes. I don't have hard numbers because it would depend on the technology involved but this doesn't seems like an efficient method. We could consider rockets with renewable solar energy, but I'm not sure it's a realistic alternative with the time-frame.

Lastly, the Sun represents 99.8 % of the solar system mass. What happens if we add the remaining 0.2%? Not much.

• Also, consider that most 'waste' really isn't, it's valuable feedstock for someone else. Your crap is fertilizer for plants. – jamesqf Apr 4 '15 at 18:24
• And how much of that 0.2% of the solar system mass does the Earth make up for? Without having looked up the exact numbers, I'd expect Earth to be closer to on the order of 0.002% of the solar system mass. Not much we can do with our puny planet is going to have much of an effect in the grand scheme of things, unless we can somehow slam ourselves into Venus or something along those lines (which would take out both). – user Apr 5 '15 at 17:41

The only way to really figure this out would be to make some assumptions and plug in the figures into the rocket equation.

The assumptions (if we really are looking to change the orbit of the planet) would be that the waste is being launched from the surface of the Earth using a mass driver, railgun or related technology, in which case the waste is the reaction mass of a rocket system.

We would also need to know at what velocity the waste is being launched (values would be all over the place from just past Earth escape velocity, where the waste spirals into the Sun over a very long period of time, to just under solar escape velocity, after which the waste will exit the Solar system entirely), and have at least an average mass of the waste stream (the rocket equations can vary greatly depending on the molecular weight of the exhaust stream; a hydrogen/oxygen rocket like the Space Shuttle has an ISP of 450, while a nuclear thermal rocket using straight hydrogen at an exhaust temperature similar to the Space Shuttle has an ISP of 800 to 1200, depending on other factors).

Of course the ration of mass between the waste stream and the Earth will still result in a very slow change in orbit, even if the waste was somehow converted to a stream of hydrogen blasting off the earth at just under solar escape velocity.....