I 'Oumuamua detection date and ramifications for a big asteroid collision with the Earth

  • Thread starter roineust
  • Start date
1,654
228
Light flash would be spectacular but won't be intense enough to do harm.
We are talking about several Gt TNT equivalent. That would be sufficient to raise the temperature of the entire atmosphere within a 10 km radius around ground zero by some hundred K. The temperatures in the upper atmosphere, where most of the energy is initially released, would be much higher. A major part of the resulting heat radiation would pass the atmosphere and reach the ground, resulting in a second temperature peak at ground level. I don't believe that this would be harmless until I see a corresponding calculation.

Shock waves would cause widespread window damage and may cause moderate structural damage in some buildings.
Even the Chelyabinsk meteor caused widespread window damage and moderate structural damage in some buildings in Chelyabinsk which is 40 km away from ground zero. A direct hit with several thousand times the energy would raze the city to the ground and let it go up in flames.
 
>> Light flash would be spectacular but won't be intense enough to do harm.

We are talking about several Gt TNT equivalent. That would be sufficient to raise the temperature of the entire atmosphere within a 10 km radius around ground zero by some hundred K.
...if all of it is absorbed. Which does not happen.

A direct hit with several thousand times the energy
This particular asteroid would not have "several thousand times the energy" of Chelyabinsk event. More like about 50 times more energy. (Chelyabinsk bolide estimated to be ~20m, Oumuamua is ~230m x 35m x 35m).
 
33,466
9,202
..if all of it is absorbed. Which does not happen.
The energy is still there - if it doesn't get absorbed by the atmosphere that just means it directly reaches the ground and heats that.

230*35*35/203=35. If we give the asteroid 3 times the speed it has 315 times the energy. If we give it 8 times the speed it has 2200 times the energy. Not necessarily thousands of times, but the difference is still huge.
 
The energy is still there - if it doesn't get absorbed by the atmosphere that just means it directly reaches the ground and heats that.
Half of the energy which has been converted to light goes up straight to space and has no effect. The other half shines onto the ground, and part of that (on average 30%) gets reflected.
 
1,654
228
This particular asteroid would not have "several thousand times the energy" of Chelyabinsk event. More like about 50 times more energy. (Chelyabinsk bolide estimated to be ~20m, Oumuamua is ~230m x 35m x 35m).
With 230m x 35m x 35m Oumuamua has a volume of about 2.8·105 m³ and with a density of 1500 kg/m³ a mass of 4.2·108 kg. With a speed of 50 km/s this results in a kinetic energy of 5.3·1017 J which corresponds to 1.3 Gt TNT eqivalent. That's like 25 Tsar bombs (50 Mt) or 2500 times the energy of the Chelyabinsk meteor (500 kt). What makes you think the resulting destructions at ground zero would be similar to a single Chelyabinsk meteor in a distance of 40 km?
 
What makes you think the resulting destructions at ground zero would be similar to a single Chelyabinsk meteor in a distance of 40 km?
Nothing makes me think so. I said no such thing.

I said that nuking this asteroid in space is (a) probably possible even with today's tech, and definitely possible if we'd finance a R&D program for nuclear-tipped asteroid interceptors; and (b) does significantly reduce the effects. The effects will still be severe, but many times less so than if we just let it impact the ground intact.
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
20,628
4,359
This paper, which studies the deflection and fragmentation of a near-Earth asteroid, may be of interest: http://web.gps.caltech.edu/~sue/TJA_LindhurstLabWebsite/ListPublications/Papers_pdf/Seismo_1621.pdf

Note that "fragmentation" here refers to the strategy of breaking up the asteroid so that the majority of its fragments are either too small to pose a threat to Earth or miss Earth completely. Interestingly, the paper claims that a surface burst is not more effective than a "stand-off" burst, which is detonated at a distance from the asteroid.

I said that nuking this asteroid in space is (a) probably possible even with today's tech, and definitely possible if we'd finance a R&D program for nuclear-tipped asteroid interceptors;
A short search on google netted me this paper: https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150011479.pdf
It discusses the suborbital interception and fragmentation of an asteroid with very short warning times using modern ICBM's. Interception of an asteroid at further distances is probably not possible without a specialized launch vehicle. I don't think ICBM's have enough fuel to escape Earth's orbit and reach their target. We certainly have the technology to create a launch vehicle and weapon system capable of intercepting and deflecting/fragmenting potential threats, we just haven't had the push to do so.
 

Drakkith

Staff Emeritus
Science Advisor
2018 Award
20,628
4,359
1,654
228
Nothing makes me think so. I said no such thing.
You said it in #50:

Light flash would be spectacular but won't be intense enough to do harm.
Shock waves would cause widespread window damage and may cause moderate structural damage in some buildings.
That's exactly what happened in Chelyabinsk.

and (b) does significantly reduce the effects.
Egain, I don't believe that untill I see a corresponding calculation.
 
213
4
i guess people are not relating to my 'why not radar?' question, because somehow by just bringing this up, is demonstrated a lack of even basic understanding of how astronomical equipment is used for that goal. But yet can anyone please try and explain in simple words, how far away is current technology from being able to detect by radar at least 2 months ahead, an asteroid coming at an angle that optical and infra-red means are unable to detect ? Or are optical and infra-red means still a better possibility and if so, what improvements are needed to be able to do that in a case such as with 'Oumuamua? i mean, people are talking here about nukes and other very late time to impact options, but why does no one discusses ways to make it an earlier detection case? Is much earlier detection such an impossibility?
 
Last edited:
33,466
9,202
You need three things to discover an asteroid:

1) it has to be bright enough to be visible
2) it has to be in the field of view

These two sound trivial, but the combination is not. Hubble can see objects with an apparent magnitude of 31.5, about 10 billion times fainter than what we can see with the naked eye. But to do that it has to look at a tiny region, just 1/(30 million) of the sky, for one month. To cover the whole sky that way we would need 30 million Hubble telescopes - that doesn't work.
There are telescopes with larger fields of view and they typically operate with shorter exposure times. They can cover larger regions of the sky, but at a much lower sensitivity.
Radar astronomy is flexible, you can either use a wide beam and a very short range (for Earth orbits?) or a narrow beam and a still quite short range (to track asteroids nearby), but you still have the trade-off other methods have. You don't get a reasonable range at a reasonable field of view to detect new things. You can only track known things, and even then only if they are nearby or huge (e. g. planets).

Hubble should be able to follow Oumuamua well into the outer solar system, but to discover it at this distance Hubble would have had to make images of a very small region of the sky just by accident. Many asteroid detections were the result of these accidents - but you never get a comprehensive list that way.

3) It has to be recognized as asteroid.
Most spots of light that appear in telescope images are stars. If you just have a single image (e.g. from Hubble), you don't recognize asteroids in it. You need multiple images taken at different times. Everything that moves is nearby, everything that does not move notably is far away.

The Large Synoptic Survey Telescope is currently under construction. Once completed in ~2021 it will monitor the whole sky (apart from a small region near Polaris) every few days. It will increase our chance to detect something before it could hit Earth a lot.

Gaia is an interesting case. Its main focus is on objects outside the solar system, in particular mapping about 1 billion stars and other objects. To do that, it scans the whole sky where each spot comes into view typically once per month. Stars will appear at nearly the same spot every time, but near Earth asteroids move a lot during that time - they will appear at random-looking spots during the observation campaign. It will need clever algorithms to associate these spots to orbits of different asteroids. The list of stars can also help other telescopes. If they see spots of light, they can check the Gaia database if these correspond to known stars.
 
213
4
1) it has to be bright enough to be visible
2) it has to be in the field of view
Can you expand on objects coming from the direction of the sun, and the ability of existing equipment such as Gaia and possible future equipment, to detect in such a case?
 
33,466
9,202
That is challenging for all telescopes. Most near Earth asteroids should typically be not too close to the Sun during some time in their orbit, for extrasolar asteroids that is not necessarily true (but they are very rare). Gaia is expected to find various asteroids closer to the Sun that were missed in previous surveys. Its scanning procedure includes measurements down to a 45 degree angle with respect to the Sun (source: page 6).

Sentinel was a concept of a telescope in a Venus-like orbit. That way all the potentially dangerous asteroids would have been far away from the Sun as seen by the telescope. Didn't get funded.
 
70
20
A reasonably read and educated laymen, would response to a claim that science and technology still have no means to detect a big asteroid collision with earth, by saying that according to what he read and saw in communication channels reliable enough, an object of that size on course to hit earth, should be detected by radio telescopes around the world, in due time. Then comes into our solar system 'Oumuamua, which is detected only on 19 October 2017. Please let me on your thoughts and knowledge regarding this subject.
'Oumuamua (A/2017 U1) was brighter than apparent magnitude +22 only between 2 September 2017 and 13 September 2017. It wasn't an obvious night-sky object by any stretch of the imagination.

'Oumuamua's orbital elements

semimajor axis, a

a = −1.279792859772851 AU

eccentricity, e
e = 1.199512371721525

inclination to ecliptic, i
i = 122.6867065669057°

longitude of the ascending node, Ω
Ω = 24.59910692499587°

argument of the perihelion, ω
ω = 241.7023929688934°

time of perihelion passage, T
T = 2458005.9912606976 JD = 9 September 2017 @ 11:47:24.9 UT

The reduction of Keplerian orbital elements to position and velocity, for a specified time-of-interest, t.
For Hyperbolic Orbits.


mean anomaly, M, at time-of-interest t (for hyperbolic orbits)
M = 0.01720209895 (t−T) √[1/(−a)³]

eccentric anomaly, u, at time-of-interest t (for hyperbolic orbits)

u = 0.0
U = 999.9
while |u−U| > 1e-12:
U = u
f₀ = e sinh(U) − U − M
f₁ = e cosh(U) − 1
f₂ = e sinh(U)
f₃ = e cosh(U)
d₁ = −f₀ / f₁
d₂ = −f₀ / (f₁ + ½ d₁f₂)
d₃ = −f₀ / (f₁ + ½ d₁f₂ + ⅙ d₂² f₃)
u = U + d₃

heliocentric distance, r, at time-of-interest t (for hyperbolic orbits)
r = a (1 − e cosh u)

true anomaly, θ, at time-of-interest t (for hyperbolic orbits)

if M≥0, then
θ = arccos[(e − cosh u) / (e cosh u − 1)]
if M<0, then
θ = −arccos[(e − cosh u)/(e cosh u − 1)]

position in heliocentric ecliptic coordinates, [x, y, z], at time-of-interest t

x'' = r cos θ
y'' = r sin θ

x' = x'' cos ω − y'' sin ω
y' = x'' sin ω + y'' cos ω

x = x' cos Ω − y' cos i sin Ω
y = x' sin Ω + y' cos i cos Ω
z = y' sin i

velocity in heliocentric ecliptic coordinates, [Vx, Vy, Vz], at time-of-interest t (for hyperbolic orbits)

k = 29784.6918325927 m/s

Vx'' = k (a/r) √[1/(−a)] sinh u
Vy'' = −k (a/r) √[(1−e²)/a] cosh u

(Enter a and r in astronomical units. Their dimensions have already been extracted and converted into the constant k.)

Vx' = Vx'' cos ω − Vy'' sin ω
Vy' = Vx'' sin ω + Vy'' cos ω

Vx = Vx' cos Ω − Vy' cos i sin Ω
Vy = Vx' sin Ω + Vy' cos i cos Ω
Vz = Vy' sin i


Just to be complete about things, here's...


The reduction of Keplerian orbital elements to position and velocity, for a specified time-of-interest, t.
For Elliptical Orbits.


mean anomaly, M, at time-of-interest t (for elliptical orbits)

P = 365.256898326 a¹·⁵
m₀ = (t−T) / P
M = 2π [m₀ − integer(m₀)]

eccentric anomaly, u, at time-of-interest t (for elliptical orbits)

u = M + (e − e³/8 + e⁵/192) sin(M) + (e² − e⁴/6) sin(2M) + (3e³/8 − 27e⁵/128) sin(3M) + (e⁴/3) sin(4M)
U = 999.9
while |u−U| > 1e-12:
U = u
f₀ = U − e sin U − M
f₁ = 1 − e cos U
f₂ = e sin U
f₃ = e cos U
d₁ = −f₀ / f₁
d₂ = −f₀ / (f₁ + ½ d₁f₂)
d₃ = −f₀ / (f₁ + ½ d₁f₂ + ⅙ d₂² f₃)
u = U + d₃

true anomaly, θ, at time-of-interest t (for elliptical orbits)

x'' = a (−e + cos u)
y'' = a √(1−e²) sin u
θ = arctan(y,x)

(arctan is the two-dimensional arctangent function in which y=sin θ and x=cos θ.)

heliocentric distance, r, at time-of-interest t (for elliptical orbits)
r = √[ (x'')² + (y'')² ]

position in heliocentric ecliptic coordinates, [x, y, z], at time-of-interest t

x' = x'' cos ω − y'' sin ω
y' = x'' sin ω + y'' cos ω

x = x' cos Ω − y' cos i sin Ω
y = x' sin Ω + y' cos i cos Ω
z = y' sin i

velocity in heliocentric ecliptic coordinates, [Vx, Vy, Vz], at time-of-interest t (for elliptical orbits)


k = 29784.6918325927 m/s

Vx = −k sin θ / √[a(1−e²)]
Vy = k (e + cos θ) / √[a(1−e²)]

(Enter a in astronomical units. Its dimensions have already been extracted and converted into the constant k.)

Vx' = Vx'' cos ω − Vy'' sin ω
Vy' = Vx'' sin ω + Vy'' cos ω

Vx = Vx' cos Ω − Vy' cos i sin Ω
Vy = Vx' sin Ω + Vy' cos i cos Ω
Vz = Vy' sin i
 
94
29
The Barringer impactor (near Winslow, Arizona) was a nickel-iron meteorite and is estimated to have been only 50 meters in diameter. It left a hole 500 feet deep and about 3/4ths of a mile in diameter. Obviously, it did not blow up on entry to the atmosphere. Impact energy was about 10 megatons. Had such a meteorite impacted near a modern metropolitan area, well, the destruction would have been horrendous. Depending on the composition of Oumuamua, had it impacted, it could have been even more devastating, although not an extinction level event.
 
70
20
I forgot what I was leading up to. Using that algorithm, you can calculate that 'Oumuamua was 0.16162 AU from Earth at closest approach at 18h UT on 14 October 2017, where its speed relative to Earth was 60.225 km/sec.

On 14 October 2017, Earth's orbital elements were:

a = 1.000000394110650 AU
e = 0.0166942736943211
i = 0.0024086539688206°
Ω = 174.0573387205077°
ω = 288.8352443931458°
T = 2458122.078193375841 JD
 
429
117
The Barringer impactor (near Winslow, Arizona) was a nickel-iron meteorite and is estimated to have been only 50 meters in diameter. It left a hole 500 feet deep and about 3/4ths of a mile in diameter. Obviously, it did not blow up on entry to the atmosphere. Impact energy was about 10 megatons. Had such a meteorite impacted near a modern metropolitan area, well, the destruction would have been horrendous. Depending on the composition of Oumuamua, had it impacted, it could have been even more devastating, although not an extinction level event.
The speed of the object is also important. A 30 meter in diameter nickel/iron meteor (8.391 g/cm3) approaching Earth at an angle of 45° will explode in the atmosphere if its speed is in excess of 40 km/s, releasing 9.49 x 1016 Joules of energy (22.7 MegaTons TNT). But that exact same meteorite will impact with the Earth's surface if its speed is 20 km/s or less. At 20 km/s a 30 meter in diameter nickel/iron meteorite would create a final crater about 890 meters (just over half a mile) in diameter and release 6.96 x 1015 Joules of energy (1.66 MegaTons TNT) upon striking the ground.

Oumuamua's speed as it flew past Earth was ~47 km/s. They also suspect that Oumuamua is made of metal rich rock because of its elongated shape and the fact that it did not display a cometary tail as it made its closest approach to the sun. Assuming Oumuamua has the density of iron (7.874 g/cm3), traveling at 47 km/s and impacting Earth at an angle of 45°, this would be the effect:
  • The meteorite begins to break up at an altitude of 31,300 meters (103,000 feet).
  • The meteorite strikes the surface at a velocity of 28.8 km/s. The impact energy is 3.52 x 1017 Joules of energy (8.41 MegaTons TNT). The broken projectile fragments strike the ground in an ellipse of 446 meters by 315 meters.
  • The initial crater is 2.03 km (1.26 miles) in diameter and 716 meters (2,350 feet) deep. The final crater will be 2.53 km (1.57 miles) and 539 meters (1,770 feet) deep. Roughly half of the melt remains in the crater.
  • The seismic effects at a distance of 100 km from the impact would be approximately a magnitude 5.9 that would arrive about 20 seconds after impact.
  • Ejecta would begin arriving 100 km from the impact site 2.4 minutes after impact and be 1.53 cm (0.604 inches) in diameter on average.
  • The air blast would arrive 100 km from the impact site 5.05 minutes after impact. The peak overpressure would be 2,870 Pa (0.408 psi) with a maximum wind speed of 6.69 m/s (15 mph) and be accompanied by a 69 dB blast (the sound of heavy traffic).
If such an impact were to occur in downtown Los Angeles, there would be 1,251,290 estimated fatalities and 1,845,600 estimated injured. Everything from San Fernando to Anaheim, and from Santa Monica to Pomona would be destroyed. Anyone within 16.5 miles (859 square miles) of the impact would suffer third degree burns, or worse.
 
Last edited:
33,466
9,202
Reducing the speed by a factor 2 (assuming you used 40 km/s) shouldn’t reduce the energy by a factor of more than 10 unless you are not considering the full energy.
 
429
117
Reducing the speed by a factor 2 (assuming you used 40 km/s) shouldn’t reduce the energy by a factor of more than 10 unless you are not considering the full energy.
I apologize, I should have been more clear. The 6.96 x 1015 Joules of energy (1.66 MegaTons TNT) released by the 30 meter meteorite traveling at 20 km/s was the energy released upon impact. Before entering the atmosphere that meteorite would have had 2.37 x 1016 Joules (5.67 MegaTons TNT) of energy. I have fixed my prior post.
 
4
3
Radar simple answer -- disregarding the power issue which is huge --
if I send a RF pulse out to an object months away ( not light months, just 2 months away at say 30Km/s or 60Km/s ) the travel to and return path is some number of
seconds. in those seconds the earth has rotated on its axis and the earth has moved in it orbit, the antenna is no longer in a position to see the return, and no separate antennas is not going to solve the problem. Off the top of my head, lets says two months is 150 million KM - the distance to the sun - that is about 8 light minutes each way, consider how much the earth rotated and what our orbital movement is in 15 minutes (hint the antenna is moving up to 450 m/sec and the earth moved in its orbit 30 Km/sec
 
33,466
9,202
Radar is scattered back in all directions. The motion of the antenna on Earth is completely irrelevant. You don't even need the same antenna as receiver, you can use one thousands of kilometers away if you want. It is just more convenient to do both at the same place as the antenna points in that direction anyway.

Absolute motion doesn't exist anyway. You can use the antenna as reference frame (an inertial frame to a very good approximation) and then it doesn't move.
 
4
3
Bistatic radar does work, but trying to do any real computation with side scatter radar reflections is challenging at best. Trying to do doppler on the return to see if the reflector was heading our way or even earth orbit crossing i think is beyond the limits of current technology. Something light minutes away outside the main bean would have a terrible S/N ratio.
 
213
4
Would it be better in anyway to have a radar in outer space at a convenient location, be it some kind of earth stationary orbit, Lagrange point or other?

As well, there is something i don't understand in the context of radars and detecting asteroids: In the Wikipedia entry for Albedo, under the title Astronomical albedo in the first paragraph, there is a reference to detecting asteroids Albedo by radar ('Radar albedo'). Why is a radar mentioned there and can such existing systems, contribute to asteroid collision early detection? Or perhaps radar albedo is a term byitslef, that actually has nothing to do with a radar machine? Or maybe it is only with asteroids of high metal composition, that radar can and are used for detection?
 
Last edited:

Related Threads for: 'Oumuamua detection date and ramifications for a big asteroid collision with the Earth

  • Last Post
2
Replies
25
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
23
Views
6K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
16
Views
14K
Top