# You’ve Launched Your Trash Into Space: Now What?

Unfailingly, in any social circle with more than a few nerds, there will be that conversation about launching trash or nuclear waste into the sun to get rid of it. From there, it usually goes into arguing about the safety of putting hazardous materials on top of a potentially explosive vehicle and lobbing it to an altitude where it could disperse over a huge area if anything were to go wrong.

What I want to know is why don’t we launch it into deep space? It would be easier. No, really. It would take less energy to send a payload into deep space than it would to drop it into the sun. A really fun set of gravitational mechanics problems called Hohmann transfers describe how to move from one circular orbit to another. Since we’re interested in being devoured by the heat of the sun or ending up in deep space, we only need to worry about half of a transfer.

A Hohmann transfer orbit is an eliptical orbit that crosses the current circular orbit and the desired circular orbit. For an object in a circular orbit, the velocity is the same at any point along it’s path, $v = \sqrt{\frac{\mu}{r}}$, since r is constant and $\mu$ is the gravitational parameter–the mass of the larger body times the gravitational constant G. We want to accelerate our payload such that it will be in an eliptical orbit where the semimajor axis–it’s furthest approach to the massive body–is the radius of the desired circular orbit. We can also drop into a lower orbit but the solution will be negative, indicating the satelite will be slowed down.

In an eliptical orbit the velocity is a function of r, the current distance from the massive body, $v(r) = \sqrt{\mu(\frac{2}{r}-\frac{1}{a})}$. So if our payload is in a circular orbit around the sun having just escaped Earth’s gravitational pull, we’d like to see what kind of energy we’ll need to enter an orbit where the payload will crash into the sun. Let’s use the solar radius, about 7e5 km. One AU, the distance from the Earth to the sun, is 150e6 km. The change in velocity to enter the eliptical transfer orbit is about -27 km/s.

To accelerate out of the solar system, let’s choose a distance of twice the semimajor axis of Pluto, about 1e10 km. The change in velocity at Earth’s orbit to enter this transfer orbit is only 12 km/s. So the change in kinetic energy to leave the solar system is smaller that that needed to crash into the sun. What’s going on is a little more subtle than a gravitational potential drawing objects nearer. Angular momentum adds an effective potential. While gravity goes like $\frac{1}{r}$ and is attractive, the effective potential from angular momentum goes like $\frac{1}{r^2}$ and is repulsive. So it takes more energy to overcome angular momentum moving towards the sun than it does moving away such that even though we’d be fighting gravity shooting for deep space, it’ll still be easier.

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