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

  1. Dec 27, 2017 #61
    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: Dec 27, 2017
  2. Dec 27, 2017 #62

    mfb

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    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.
     
  3. Dec 28, 2017 #63
    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?
     
  4. Dec 28, 2017 #64

    mfb

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    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.
     
  5. Jan 1, 2018 #65
    '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
     
  6. Jan 2, 2018 #66
    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.
     
  7. Jan 2, 2018 #67
    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
     
  8. Jan 2, 2018 #68
    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: Jan 2, 2018
  9. Jan 2, 2018 #69

    mfb

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    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.
     
  10. Jan 2, 2018 #70
    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.
     
  11. Jan 3, 2018 #71
    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
     
  12. Jan 3, 2018 #72

    mfb

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    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.
     
  13. Jan 4, 2018 #73
    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.
     
  14. Jan 4, 2018 #74
    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: Jan 4, 2018
  15. Jan 4, 2018 #75

    mfb

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    We are talking about an angle of typically 0.1 millirad with respect to a direct reflection - it is not "side-scattering" in the conventional sense.
    No. For the same price you get a much smaller antenna and orders of magnitude lower power, which means the range gets much shorter. The probability that something would fly through the volume where the space-based radar happens to be is tiny.
    Don't misrepresent the article please. It is not talking about asteroid detection there, it is talking about measuring known asteroids. The amount of radar power they reflect tells us something about their composition. This requires them to be in range and at a known place, however.
     
  16. Jan 5, 2018 #76
    i was responding to:
    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,

    I am used to radar antenna with very small effective aperture - fighter aircraft -- so if i moved my receive antenna 350 Km east after i transmit a pulse straight up i would not expect much return signal to be received even though it is only 0.07 radians of earth rotation

    either way, you are a mentor, i am just a radar engineer - so I concede


     
  17. Jan 5, 2018 #77
    You need to consider the distance to the target. If you're looking at say 4 light minutes to the target, that's about 72 million km. Even if the antenna moves 350 km east, that's still a very small angle relative to the target distance, ~0.005 milliradian.
     
  18. Jan 5, 2018 #78

    mfb

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    If you produce a pulse straight up you are looking for things at a distance of 10 km, maybe 30 km - if you move your antenna by 350 km that is a huge difference, of course. The equivalent in the solar system would be to move the antenna to Saturn.

    The systems scale quite nicely. An interstellar asteroid in the inner solar system will have a typical distance of maybe 150 million km. Your radar pointing up looks for things with a typical distance of maybe 15 km, a factor 10 million closer. To keep the angular velocity the same, we have to scale down the typical velocity of 50 km/s by the same factor, and we get 5 mm/s. Surely a radar system can track an object moving at 5 mm/s. Or, exactly equivalently, a radar system moving at 5 mm/s can track a stationary object. Ships move more than a factor 1000 faster, and fighter jets move a factor 100,000 faster. The non-inertial reference frame of the antenna, scaled down as well, adds an acceleration of 3*10-9 m/s2, completely negligible as well.
    We didn't scale the light speed here, unfortunately scaling down everything doesn't work that nicely. To get that right, we have to keep the relative velocity. Expressed in the target frame, your antenna tracking the aircraft in 15 km distance would move by 5 meters. Expressed in the radar antenna frame, the target moves by 5 meters during the measurement. That is not an issue either.

    The signal intensity scales with the diameter squared divided by the distance to the fourth power. If we move our target a factor 10 million closer, we have to make its diameter a factor 100 trillion smaller. That reduces 'Oumuamua's size to 2.5 pm, smaller than an atom. Obviously we can't scale it that way, there are no solid objects smaller than an atom and even a full atom wouldn't have radar properties similar to an asteroid. But that gives you an idea of the size of the target you would have to find with radar here on Earth to make a radar detection of 'Oumuamua at this distance possible.
     
  19. Jan 6, 2018 #79
    "They are sensitive enough to hear a common aircraft radar transmitting to us from any of the 1000 nearest stars." Source: Breakthrough Listen.
    Would such a signal be less feeble, than a radar signal bouncing off a meteorite at a distance of 5-10 light minutes away?
     
  20. Jan 6, 2018 #80

    mfb

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    If we would use an aircraft radar as source in both cases, we get a factor 1/(star distance)2 for the first case and (radar albedo)(asteroid radius)2/(pi (asteroid distance)4) in the second case, making a rough assumption about the way radar is reflected for the numerical prefactor.
    There are 2000 stars within 50 light years, so let's be optimistic and use that as distance. I'll use 1 for the radar albedo, again very optimistic, typical values are in the range of 0.1 to 0.5.
    5 light minutes are 150 million km, and I'll use 200 m for the asteroid. In this case the first fraction is 4.5 *10-36 m-2, while the second one is 1.9*10-40, or a factor 20,000 weaker even with all these optimistic assumptions.
    It is unclear what exactly "aircraft radar" means. To find aircraft, or aircraft-mounted? Good aircraft-mounted radar seems to be in the range of tens of kW of power (example), ground-based radar seems to be similar (example). We don't have antennas that can send with more than a few MW, or a factor 100 more. We have radio dishes that can focus a beam better than these radar installations, however.

    You can probably install an even more powerful transmitter at Arecibo, or use multiple antennas to listen to the signal, or repeat the measurements over and over again, and it might become possible to detect a faint echo of the asteroid - if you aim the beam precisely at its position. It is still not useful to find it! You can only measure it once you know where to send the signal to.
     
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