# Finding non-lit objects at light minute ranges

Let's say my hyperjumpwarpgate McGuffin drive goes bing and I find myself about half a light year away from the sun. This is right in the heart of the outer Oort Cloud. Wikipedia says that there are a trillion objects there larger than 1km in size, and they're about 30 light seconds apart.

So I want to find the closest one of these. How?

• Out there it's very dark, so I'm unlikely to be able to pick one up by reflected sunlight.

• Radar's possible, but it's going to be a very slow process scanning the complete sky with a narrow-angle high-power beam, particularly as I'm going to have to wait several minutes for each return ping. OTOH Venus has been mapped by radar from Earth, so it should at least work.

• I've seen suggestions of using infrared to look for very distant planets; but I suspect these are looking for large bodies with a certain amount of their own heat, while the small bodies I'm looking for are going to be very cold.

• Would it be feasible to look for shadows against the interstellar background? I'd have to photograph the entire sky at insane resolution, then move, photograph it again, and look for parallax. That would only work for objects at the right vector to my motion, and of course only where there was some sort of bright background --- but there's quite a lot of Milky Way.

What's the most practical real-world method of doing this?

Edit:

So to calculate the apparent brightness of the sun from that distance: half a light year is $180*24*3600$ light seconds, which is $\frac{180*24*3600}{8*60} = 32400$ AU. That means that it should be $\frac{1}{32400^2} = \frac{1}{10^{10}}$ the brightness that it is here. That's quite dim.

Given that apparent magnitude is:

$m_x - m_{x,0} = -2.5 log_{10} \frac {F_x}{F_{x,0}}$

That gives us, for the sun:

$m_x - -26.74 = -2.5 log_{10} \frac {1 / 10^{10}}{1}$

...or:

$m_x = -2.5 log_{10} (10^{-10}) - 26.74$

...so $m_x$ is -1.74, which makes the Sun about the same brightness as Sirius.

(I was totally not expecting the maths above to produce a meaningful result.)

Edit edit:

(Yes, I'm bored at work.)

The angular diameter of an object is $\tan \frac{\text {size}}{\text {distance}}$, which means that a 1km body at 30 light seconds is going to be $\tan \frac{1000}{9 \times 10^9}$, which is about $5 \times 10^{-6}$ degrees. Or about 0.02 arcseconds.

Wikipedia has a handy table of diffraction limits vs telescopes. Turns out that the Hubble could resolve that 1km body, provided it was radiating in ultraviolet, which is probably unlikely. For high infrared I'd need a telescope with an aperture of several hundred metres, which is doable, but for low infrared I would need something 10km across, which is harder.

It may, however, be possible to detect the body without having to resolve it (i.e. the same way we can see stars); but I don't know how that works.

• While at that distance the Sun will only appear to be a very bright object, there will be other sources of light out there. Still, an interesting question Hmmm... Mar 27 '15 at 15:16
• I'm inclined toward the radar plan. The round-trip time isn't really an issue if your receiving antenna is separate from and less directionally sensitive than your emitter. Mar 27 '15 at 17:19
• And it would be even harder to find one that has dilithium crystals. Aug 8 '15 at 4:16

First, everything in the Oort Cloud is really, really far apart. You could spend years looking and never see anything because of the reasons you gave.