# Thread: Radioactive waste shot into the sun

1. ## Radioactive waste shot into the sun

Proposals for dealing with radioactive waste include shooting it into the Sun. Consider a waster container that is simply dropped from rest in the vicinity of the Earth's orbit. With what speed will it hit the Sun?

Here are some constants that we might need:
G = 6.67 x 10^-11Nm^2/kg^2
Radius (Earth) = 6.38 x 10^6 m
Mass (Earth) = 5.98 x 10^24 kg
Mass = 1.99 x 10^30 kg
1 A.U = 1.49 x 10^11 m

Answer: 6.2 x 10^5 m/s

2. Originally Posted by kenan
Proposals for dealing with radioactive waste include shooting it into the Sun. Consider a waster container that is simply dropped from rest in the vicinity of the Earth's orbit. With what speed will it hit the Sun?

Here are some constants that we might need:
G = 6.67 x 10^-11Nm^2/kg^2
Radius (Earth) = 6.38 x 10^6 m
Mass (Earth) = 5.98 x 10^24 kg
Mass = 1.99 x 10^30 kg
1 A.U = 1.49 x 10^11 m

Answer: 6.2 x 10^5 m/s
Sounds like a work-energy problem to me.

There is no friction, so there is no non-conservative work going on. Since we have a gravitational potential energy here we need to set a 0 point: I'm going to choose a 0 point out at an infinite distance from the center of the Sun. (Or you could choose one at the center of the Sun, it works out the same. Both choices are commonly used and are equivalent in all respects.)

So:
$W_{nc} = \Delta E$

$\Delta KE + \Delta PE = 0$

$\frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 + -\frac{GmM}{d} - -\frac{GmM}{d_0} = 0$
where m is the mass of the waste, M is the mass of the Sun, d is the radius of the Sun (remember we're measuring all distances from the center of the Sun here), and $d_0$ is the initial distance of the waste from the center of the Sun.

Since the container is "dropped" from rest
$\frac{1}{2}mv^2 - \frac{GmM}{d} + \frac{GmM}{d_0} = 0$

Solve for v.

-Dan