Supposing a Moon-like object approaches the Solar System on a collision path with Earth (or the Moon!). When and how do we detect it?
We can assume this object is spherical and roughly 3,500 km in diameter.
With my basic understanding of light and telescopy, you can detect such an object when enough light from our Sun is reflected off the object, into the mirrors of a telescope, and into your eyes (or a sensor). The unit of measurement of this light is called apparent magnitude (\text{m}), a reverse logarithmic scale where the brightest object is our Sun (-26.7\text{m}) and the dimmest known objects are small asteroids in the Kuiper Belt detected by the Hubble Telescope (31.5\text{m}).
It is intuitive to me that closer objects are brighter - that there is a direct relation between distance, light intensity, and apparent magnitude. Presumably this is because it is absorbed and scattered by air particles, or in space, by the interstellar medium. According to an answer on Quora, the Sun has an intensity of 3 octillion candelas.
The intensity would go down on the trip to the object (based on distance), when the surface of the object absorbs and scatters some of the light (based on size and surface material I think), and again on the trip to the Hubble Telescope (based on distance). So more precisely, what is the distance at which a Moon-like object’s apparent magnitude exceeds the limit (31.5\text{m}?) of what our best telescopes can detect?
But then there are other methods to detect celestial objects. I believe the way we detect extrasolar planets is not based on direct observation of light reflected from the planet’s surface, but rather by observing a slight drop in brightness as the planet passes between the star and the telescope. Perhaps by coincidence a similar observation could be made of a Moon-sized interstellar object - but how do you determine it is coming this way, or its distance?
We’d need to see it coming. How soon we would see it would depend on the albedo of the object and if we got lucky looking in that general direction.
The object you propose is bigger than Pluto (by about 1/3) and we can see Pluto with normal telescopes.
But…we need to be looking in that direction which is just luck on out part.
Seeing the object block some other light also means we are looking that way and, likely, we’d see it before we figured out it blocked some light.
It also can have a gravitational effect as it passes through but, again, I think we would see it long before we figured out it perturbed the orbits of our planets.
No, mostly it’s just because light spreads out. If a light source is a bazillion miles away from your telescope, most of its light is going to be going to places that aren’t your telescope.
And we can detect very faint things, if we’re looking at them. But we aren’t usually looking at most things. Unless we got lucky, the first instrument to detect this thing just might be a human eyeball.
We have a number of programs that are searching already. Of course knowing is not synonymous with doing.
According to Dr. Michael F. A’Hearn, a typical mission would take too long from approval to launch if there was an emergency:
REP. STEWART: … are we technologically capable of launching something that could intercept [an asteroid]? … DR. A’HEARN: No. If we had spacecraft plans on the books already, that would take a year … I mean a typical small mission … takes four years from approval to start to launch …
But that is for objects that produce their own light. A (non-stellar) object approaching the Earth is going to be relying on reflected sunlight, which itself falls off as it travels outwards. Ignoring the distance from the sun and Earth (which is good enough talking these distances) if you double the distance between the object and the sun, then the reflected light reaching the Earth will be cut by 16x.
you have a light source that just spreads in every direction in straight lines. Hold up a square sheet 1m on a side, 5m from the light. That’s how many photons hit the sheet each second.
To trap those same number of photons 10m away - twice as far - the sheet would have to be a square 2m on a side. 50m - 10m square, etc.
If I just used the same 1m square I used at 5m, at 10m it would receive 1/4 the light, at 50m 1/100 the ligth would fall on it.
Basically, if a ray of light diverges by X°, then to capture the same light x times as far away you need x^2 times the area. Or for the same area (your telescope or retina) you get 1/X^2 the light.
this applies to most sources. It’s also why gravity falls off as the inverse square from a mass.
One additional factor is that if the object is heading directly at us, it’s less noticeable because its transverse velocity is tiny (or zero). Objects that move across the sky are easier to spot.
Yes. You sometimes read authors who think that the Sun is “just another star” in the sky of the asteroid belt, which wildly underestimates the brightness of the Sun; after all, sunlight goes past the asteroid belt, reflects partially off Jupiter, and comes back to Earth - leaving Jupiter the fourth brightest object in Earth’s sky.
Okay, that makes sense. So in pursuit of the question,
\text{Let the illuminance of the Sun } S = 3 \times 10^{27} \text{cd} \text{Let the distance from the interloper to the Sun be } d \text{Let the illuminance of the sun as measured on the surface of the interloper be } E
\text{As light follows the inverse square law, } E = \frac{S}{d^2} = \frac{3 \times 10^{27}}{d^2}
Apparently the albedo of the moon is 0.12, meaning 12% of light that falls upon the surface of the moon is reflected back. So let’s say the albedo of this interstellar object is also 0.12.
\text{Let the albedo of the interloper } A = 0.12 \text{Let the illuminance of the interloper (in candelas) be } I = E \times A = \frac{3 \times 10^{27}}{d^2} \times 0.12 = \frac{3.6 \times 10^{26}}{d^2}
As to the next step…? I would need to get an apparent magnitude of the interloper m as observed from Earth, then solve for distance d such that m fits the inequality m > 31.5.
Not sure it fits here because it’s not Moon sized but here is a video that shows a year before Chicxulub (the dinosaur killer) hit, it was the same magnitude as Neptune.
But because earth moves in its orbit, the night sky actually “moves” from our point of view. At a certain point we would at least see some parallax from the object.
By comparison, Tunguska is estimated to be an airburst of a meteor about 150’ to 200’ diameter, and nobody saw it coming back then. The recent Russian meteor that was captured on dashcams and the shock wave blew out windows and caused other damage - that was estimated to be 50’ diameter (“a five-storey building”) and again, nobody saw that coming.
The problem is the really big ones are difficult to deflect even if we see it coming.
Pluto was discovered with a 13-inch scope, which is really small by professional standards today. Lots of amateurs have larger scopes. But Tombaugh didn’t just happen to look in the right direction and say “hey, there’s a planet!”. He was doing a systematic survey of the ecliptic, taking two pictures of the same area two weeks apart, and using a blink comparator.
No one’s doing it that way anymore. As @DesertDog points out, surveys are automated these days and there’s a number of them going on at any one time. One of them is virtually certain to spot this interloper long before it gets into the inner Solar System.
But spotting it is the easy part. Predicting that it will actually collide with the Earth will take longer. It should be spotted years before any collision, which means there’ll be some uncertainty about whether it’ll actually hit or just do a near miss. But once we’re certain it’ll hit, there’s not much we can do about it. An object the size and mass of the Moon is impossible to deflect with our current tech. It’s bend over and kiss your ass good-bye time.
If we had a few years notice and knew it was impossible to prevent the earth from being hit I imagine we would use that time to set up a base on the moon or Mars to give the human race a chance at survival.
Agreed. It’s just that there are a lot of people looking for comets or for new planets, both of which would move across the sky. Looking for an unusually large (and growing) parallax circle is more of a specialty, I think.
Considering that a 5-mile diameter chunk basically destroyed almost every living thing bigger than a breadbox followed by years of nuclear winter level climate change, I would think auto-recto-osculation is the only course of action for a moon-sized object. That collision should reduce the earth to a ball of magma (or two or more balls) for a substantial period of time.
Meanwhile, it would be unlikely we could build a self-sustaining colony on Mars or anywhere in reasonable time. The level of tech required would be too much.
If an object the size of the Moon hits the Earth, being on the Moon won’t mean a damn thing (Space 1999 notwithstanding). A moonbase on the far side might last longer but there’s going to be a lot of loose rock floating around and a dramatic change in the gravitational pull of what’s left of the Earth plus the thing that hit it.
Mars is probably the closest one could be, and that’s assuming it wasn’t subsequently in line for detritus blown off Earth.
That depends on what direction it comes from. It’s possible for a very large object to sneak up on us if it comes from the opposite side of the solar system and comes whizzing around close to the sun.
A small space rock did this last October. No one detected this particular rock until it had already passed us.
To be fair, this wasn’t a moon-sized object or anything even close to it. News articles at the time described it as being the size of a refrigerator. But it does demonstrate the point that the sun creates a bit of a blind spot as we’re looking for things.
The inverse-square law applies here too. An object half the diameter is 1/4 as bright in the same lighting. An object 1/10 the diameter is 1/100 as bright. Etc.