# When would we notice an object entering our solar system at 0.20c (20% speed of light)?

At what point would we notice or be able to detect an object entering our solar system at 0.20 speed of light?

What's the maximum range we would be able to detect with certainty beyond a reasonable doubt (e.g. edge of Oort Cloud)?

And how much time would it take for that object to reach the inner solar system, or up to 5.2 AU (Jupiter's orbit), from the moment of detection?

For the sake of this question, the object's trajectory is not expected to intersect with any other stellar body in the system and its course is not expected to deviate (direct line). The speed is not expected to deviate, i.e. no external forces are expected to act upon the object to decelerate or accelerate it.

Edit

Let's say it's a:

1. multi-generational ship,
2. constructed from known metals and materials (nothing exotic),
3. around 3km in length, with a 2km wide protective "asteroid-like" mass at the bow,
4. weighing in around, say, 8.5 billion metric tons,
5. with an albido of 0.5 - 0.8,
6. it does not rotate,
7. it would radiate in all the typical ways you would expect a ship of this nature to.
• This is unanswerable unless you specify the size and albedo of the object (not to mention if it comes from behind the sun then never). Even then it would seem to become a straight "how good are our telescopes and radars" question which might better fit on our Astronomy stack. There doesn't seem to be a worldbuilding problem here. Jun 26 at 16:20
• When you would notice an animal entering your property at night depends on which animal is it: an elephant wearing a reflective jacket and with bells on its legs, or a black mosquito in moonless and cloudy night would be detected at very different ranges. Same holds here. Can you tell us more about this object?
– L.Dutch
Jun 26 at 16:20
• @L.Dutch - I am relieved someone besides me has trouble with that elephant. I was starting to think I was imagining it. I call it the Bellephant for obvious reasons. Jun 26 at 16:45
• @Willk You're not imagining. That's my pet elephant. He sleeps in my bed. Needless to say, I sleep in a hammock in the garden, or those bells would keep me awake. Jun 26 at 18:19
• @InTheAbsenceOfFear - that's a strange coincidence - last night, I shot an elephant in my pyjamas. What he was doing in my pyjamas, I may never know. Jun 26 at 18:24

## Sorry!

Despite the undeserved check mark, the first calculation I jotted down had two errors that threw off the result... that I know of so far. Now I'm getting that it is a warm 786 K but nowhere near hot as a star. Because it would not leave a reflected trail like a comet, this might be hard to see.

Detecting it "when it enters the Solar system" is right, because the heliopause has a higher density of matter. The solar wind flows out from the Sun until it sort of stops, buffeted by the interstellar medium. So the density we're looking at for the heliopause is something like 0.12 electrons per mL, with I would assume a proton to go along with each one, whether as a hydrogen nucleus or (rarely) a part of something bigger.

So we've got, say, 3 km^2 of surface moving at 0.2 x 3e+5 km/s (CORRECTION). Each second we "clear" 1.8e+5 km^3 = 1.8e+20 cm^3 of volume. Everything in that volume impacts at 0.2 c, which is to say its relativistic mass will be 1/(1-0.2^2)^0.5 = 1.02 times higher than its rest mass. The rest mass (per the volume for one second) is (0.12 electron+proton / cm^3) x (1.8e+20 cm^3) x (1 g/mol) x (1 mol / 6.02e+23) = 36 mcg. So 2% of that, times c^2, is 6.5e+10 J. Per the second I mentioned, this emits 65 gigawatts of energy, continuously. This can be compared to the 23 watts used by Voyager 2 which obtained the data from the first link above, but bear in mind it's white noise in all frequencies, unmodulated, and no one is aiming a dish at it. Let's look further.

If I assume the asteroid is a black body (which seems anything but a safe assumption, since good designers would try to reflect radiation and heat away), radiating only on the putative 3 km^2 I approximated a circle to be, then T = (6.5e+10 J/s /( 3e+6 m^2 x 5.67×10−8 W⋅m−2⋅K−4)^0.25 = 786 K.

• That doesn't sound right; I'm getting luminosity on the order of a few hundred watts per square meter. I think this was the error: '0.2 x 3e+8 km/s' - the speed of light is 3e5 km/s, 3e8 m/s. Jun 28 at 0:12
• @rwallace - OOPS. I knew this, really. Jun 28 at 0:36

Didn't get an answer to a question I placed in the original post so I'll make a basic assumption here.

(Assuming for instance some form of realistic fusion drive?) Some time after it starts decelerating. With the length of time required before its discovery dependent on a lot of variables I don't have numbers for but in any case likely to be months if not years.

All the modelling I've seen suggest such drives operate at extremely high temperatures (1500K plus) but they are also capable of maintaining reasonable levels of thrust for extended periods of time, making them far superior to chemical rockets for deep space missions.

So since this is such a massive object and its starting velocity compared to the Sun is extremely high such a drive is going to have to start its breaking 'burn' years?? in advance of entering the inner solar system.

Astronomers: Hmmmm.... a new big/hot light in near interstellar space with a (slowly) diminishing blue shift. Curious!